(a) Expand \(f\relax (x) =\left \{ \begin {array} [c]{ccc}\frac {2xh}{L} & & 0\leq x\leq \frac {L}{2}\\ \frac {2h\left (L-x\right ) }{L} & & \frac {L}{2}\leq x\leq L \end {array} \right . \) in an exponential Fourier series. (b) What do you think is the value of \(\sum _{odd}\frac {1}{m^{2}}\) where sum is over all positive odd integers?
Solution
The period is \(L\). The function \(f\relax (x) \) looks like the following (where \(L\) is choosing to be \(2\) and \(h=1\), for illustration only)
The period is \(L\). The exponential Fourier series for periodic \(f\left ( x\right ) \) is given by the expansion\begin {equation} |f\rangle =\sum _{m}f_{m}|m\rangle \tag {1} \end {equation} Where \begin {align} f_{m} & =\langle m|f\rangle \nonumber \\ & =\int _{0}^{L}\frac {1}{\sqrt {L}}e^{-i\frac {2\pi m}{L}x}f\relax (x) dx \tag {2} \end {align}
And \(|m\rangle \) are the basis functions given by \[ |m\rangle =\frac {1}{\sqrt {L}}e^{i\frac {2\pi m}{L}x}\qquad m=0,\pm 1,\pm 2,\cdots \] Putting these together gives\begin {equation} f\relax (x) \sim \sum _{m=-\infty }^{\infty }f_{m}\frac {1}{\sqrt {L}}e^{i\frac {2\pi m}{L}x} \tag {3} \end {equation} Now \(f_{m}\) is found\begin {align} f_{m} & =\int _{0}^{L}\frac {1}{\sqrt {L}}e^{-i\frac {2\pi m}{L}x}f\left ( x\right ) dx\nonumber \\ & =\int _{0}^{\frac {L}{2}}\frac {2xh}{L}\frac {1}{\sqrt {L}}e^{-i\frac {2\pi m}{L}x}dx+\int _{\frac {L}{2}}^{L}\frac {2h\left (L-x\right ) }{L}\frac {1}{\sqrt {L}}e^{-i\frac {2\pi m}{L}x}dx \tag {4} \end {align}
For \(m=0\)\begin {align} f_{0} & =\int _{0}^{\frac {L}{2}}\frac {2xh}{L}\frac {1}{\sqrt {L}}dx+\int _{\frac {L}{2}}^{L}\frac {2h\left (L-x\right ) }{L}\frac {1}{\sqrt {L}}dx\nonumber \\ & =\frac {2h}{L\sqrt {L}}\int _{0}^{\frac {L}{2}}xdx+\frac {2h}{L\sqrt {L}}\int _{\frac {L}{2}}^{L}\left (L-x\right ) dx\nonumber \\ & =\frac {2h}{L\sqrt {L}}\left (\int _{0}^{\frac {L}{2}}xdx+\int _{\frac {L}{2}}^{L}\left (L-x\right ) dx\right ) \nonumber \\ & =\frac {2h}{L\sqrt {L}}\left (\left (\frac {x^{2}}{2}\right ) _{0}^{\frac {L}{2}}+\left (Lx-\frac {x^{2}}{2}\right ) _{\frac {L}{2}}^{L}\right ) \nonumber \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {1}{2}\frac {L^{2}}{4}+\left (L^{2}-\frac {L^{2}}{2}-\frac {L^{2}}{2}+\frac {1}{2}\frac {L^{2}}{4}\right ) \right ) \nonumber \\ & =\frac {\sqrt {L}h}{2} \tag {5} \end {align}
And for \(m\neq 0\), the first integral in (4) is\[ \int _{0}^{\frac {L}{2}}\frac {2xh}{L}\frac {1}{\sqrt {L}}e^{-i\frac {2\pi m}{L}x}dx=\frac {2h}{L\sqrt {L}}\int _{0}^{\frac {L}{2}}xe^{-i\frac {2\pi m}{L}x}dx \] Integration by parts. Let \(u=x,dv=e^{-i\frac {2\pi m}{L}x}\), then \(du=1,v=\frac {e^{-i\frac {2\pi m}{L}x}}{-i\frac {2\pi m}{L}}=\frac {iL}{2\pi m}e^{-i\frac {2\pi m}{L}x}\). The above now becomes\begin {align*} \int _{0}^{\frac {L}{2}}\frac {2xh}{L}\frac {1}{\sqrt {L}}e^{-i\frac {2\pi m}{L}x}dx & =\frac {2h}{L\sqrt {L}}\left (\frac {iL}{2\pi m}\left [ xe^{-i\frac {2\pi m}{L}x}\right ] _{0}^{\frac {L}{2}}-\int _{0}^{\frac {L}{2}}\frac {iL}{2\pi m}e^{-i\frac {2\pi m}{L}x}dx\right ) \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {iL}{2\pi m}\left [ \frac {L}{2}e^{-i\frac {2\pi m}{L}\frac {L}{2}}\right ] -\frac {iL}{2\pi m}\int _{0}^{\frac {L}{2}}e^{-i\frac {2\pi m}{L}x}dx\right ) \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {iL}{2\pi m}\left [ \frac {L}{2}e^{-i\pi m}\right ] -\frac {iL}{2\pi m}\left [ \frac {iL}{2\pi m}e^{-i\frac {2\pi m}{L}x}\right ] _{0}^{\frac {L}{2}}\right ) \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {iL}{2\pi m}\left [ \frac {L}{2}\left ( -1\right ) ^{m}\right ] -\frac {i^{2}L^{2}}{4\pi ^{2}m^{2}}\left [ e^{-i\frac {2\pi m}{L}x}\right ] _{0}^{\frac {L}{2}}\right ) \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {iL^{2}}{4\pi m}\left (-1\right ) ^{m}+\frac {L^{2}}{4\pi ^{2}m^{2}}\left [ e^{-i\frac {2\pi m}{L}\frac {L}{2}}-1\right ] \right ) \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {iL^{2}}{4\pi m}\left (-1\right ) ^{m}+\frac {L^{2}}{4\pi ^{2}m^{2}}\left [ e^{-i2\pi m}-1\right ] \right ) \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {iL^{2}}{4\pi m}\left (-1\right ) ^{m}+\frac {L^{2}}{4\pi ^{2}m^{2}}\left (\left (-1\right ) ^{m}-1\right ) \right ) \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {iL^{2}}{4\pi m}\left (-1\right ) ^{m}+\frac {L^{2}}{4\pi ^{2}m^{2}}\left (-1\right ) ^{m}-\frac {L^{2}}{4\pi ^{2}m^{2}}\right ) \\ & =\frac {i2hL^{2}}{L\sqrt {L}4\pi m}\left (-1\right ) ^{m}+\frac {2hL^{2}}{L\sqrt {L}4\pi ^{2}m^{2}}\left (-1\right ) ^{m}-\frac {2hL^{2}}{L\sqrt {L}4\pi ^{2}m^{2}}\\ & =\frac {i2hL}{\sqrt {L}4\pi m}\left (-1\right ) ^{m}+\frac {2hL}{\sqrt {L}4\pi ^{2}m^{2}}\left (-1\right ) ^{m}-\frac {2hL}{\sqrt {L}4\pi ^{2}m^{2}}\\ & =\frac {i2h\sqrt {L}}{4\pi m}\left (-1\right ) ^{m}+\frac {2h\sqrt {L}}{4\pi ^{2}m^{2}}\left (-1\right ) ^{m}-\frac {2h\sqrt {L}}{4\pi ^{2}m^{2}}\\ & =\frac {ih\sqrt {L}}{2\pi m}\left (-1\right ) ^{m}+\frac {h\sqrt {L}}{2\pi ^{2}m^{2}}\left (-1\right ) ^{m}-\frac {h\sqrt {L}}{2\pi ^{2}m^{2}}\\ & =\frac {ih\sqrt {L}\pi m\left (-1\right ) ^{m}}{2\pi ^{2}m^{2}}+\frac {h\sqrt {L}\left (-1\right ) ^{m}}{2\pi ^{2}m^{2}}-\frac {h\sqrt {L}}{2\pi ^{2}m^{2}}\\ & =\frac {ih\sqrt {L}\pi m\left (-1\right ) ^{m}+h\sqrt {L}\left (-1\right ) ^{m}-h\sqrt {L}}{2\pi ^{2}m^{2}} \end {align*}
Hence\begin {equation} \int _{0}^{\frac {L}{2}}\frac {2xh}{L}\frac {1}{\sqrt {L}}e^{-i\frac {2\pi m}{L}x}dx=h\sqrt {L}\frac {\left (i\pi m\left (-1\right ) ^{m}+\left (-1\right ) ^{m}-1\right ) }{2\pi ^{2}m^{2}} \tag {6} \end {equation} Now the second integral in (4) is evaluated\[ \int _{\frac {L}{2}}^{L}\frac {2h\left (L-x\right ) }{L}\frac {1}{\sqrt {L}}e^{-i\frac {2\pi m}{L}x}dx=\frac {2h}{L\sqrt {L}}\int _{\frac {L}{2}}^{L}\left ( L-x\right ) e^{-i\frac {2\pi m}{L}x}dx \] Integration by parts. Let \(u=L-x,dv=e^{-i\frac {2\pi m}{L}x}\), \(du=-1,v=v=\frac {e^{-i\frac {2\pi m}{L}x}}{-i\frac {2\pi m}{L}}=\frac {iL}{2\pi m}e^{-i\frac {2\pi m}{L}x}\). The integral becomes\begin {align*} \frac {2h}{L\sqrt {L}}\int _{\frac {L}{2}}^{L}\left (L-x\right ) e^{-i\frac {2\pi m}{L}x}dx & =\frac {2h}{L\sqrt {L}}\left (\left [ \left (L-x\right ) e^{-i\frac {2\pi m}{L}x}\right ] _{\frac {L}{2}}^{L}+\int _{\frac {L}{2}}^{L}\frac {iL}{2\pi m}e^{-i\frac {2\pi m}{L}x}\right ) \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {iL}{2\pi m}\left [ 0-\frac {L}{2}e^{-i\frac {2\pi m}{L}\frac {L}{2}}\right ] +\frac {iL}{2\pi m}\int _{\frac {L}{2}}^{L}e^{-i\frac {2\pi m}{L}x}\right ) \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {-iL^{2}}{4\pi m}\left [ e^{-i\pi m}\right ] +\frac {iL}{2\pi m}\left [ \frac {iL}{2\pi m}e^{-i\frac {2\pi m}{L}x}\right ] _{\frac {L}{2}}^{L}\right ) \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {-iL^{2}}{4\pi m}\left (-1\right ) ^{m}+\frac {i^{2}L^{2}}{4\pi ^{2}m^{2}}\left [ e^{-i\frac {2\pi m}{L}x}\right ] _{\frac {L}{2}}^{L}\right ) \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {-iL^{2}}{4\pi m}\left (-1\right ) ^{m}-\frac {L^{2}}{4\pi ^{2}m^{2}}\left [ e^{-i2\pi m}-e^{-i\pi m}\right ] \right ) \\ & =\frac {2h}{L\sqrt {L}}\left (\frac {-iL^{2}}{4\pi m}\left (-1\right ) ^{m}-\frac {L^{2}}{4\pi ^{2}m^{2}}\left [ 1-\left (-1\right ) ^{m}\right ] \right ) \\ & =\frac {-i2hL}{\sqrt {L}4\pi m}\left (-1\right ) ^{m}-\frac {2hL}{4\sqrt {L}\pi ^{2}m^{2}}\left [ 1-\left (-1\right ) ^{m}\right ] \\ & =\frac {-i2hL}{\sqrt {L}4\pi m}\left (-1\right ) ^{m}-\frac {2hL}{4\sqrt {L}\pi ^{2}m^{2}}-\left (-1\right ) ^{m}\frac {2hL}{4\sqrt {L}\pi ^{2}m^{2}}\\ & =\frac {-i2h\pi mL}{4\pi ^{2}m^{2}}\left (-1\right ) ^{m}-\frac {2hL}{4\pi ^{2}m^{2}}-\left (-1\right ) ^{m}\frac {2h\sqrt {L}}{4\pi ^{2}m^{2}}\\ & =\frac {-i2h\pi mL\left (-1\right ) ^{m}-2h\sqrt {L}-2h\sqrt {L}\left ( -1\right ) ^{m}}{4\pi ^{2}m^{2}} \end {align*}
Therefore\begin {equation} \frac {2h}{L\sqrt {L}}\int _{\frac {L}{2}}^{L}\left (L-x\right ) e^{-i\frac {2\pi m}{L}x}dx=h\sqrt {L}\frac {\left (-i\pi m\left (-1\right ) ^{m}-\left ( -1\right ) ^{m}-1\right ) }{2\pi ^{2}m^{2}} \tag {7} \end {equation} Therefore, using (5,6,7) gives \[ \,f_{m}=\left \{ \begin {array} [c]{ccc}\frac {\sqrt {L}h}{2} & & m=0\\ h\sqrt {L}\frac {\left (i\pi m\left (-1\right ) ^{m}+\left (-1\right ) ^{m}-1\right ) }{2\pi ^{2}m^{2}}+h\sqrt {L}\frac {\left (-i\pi m\left ( -1\right ) ^{m}-\left (-1\right ) ^{m}-1\right ) }{2\pi ^{2}m^{2}} & & m\neq 0 \end {array} \right . \] The above can be simplified more to\[ \,f_{m}=\left \{ \begin {array} [c]{ccc}\frac {\sqrt {L}h}{2} & & m=0\\ h\sqrt {L}\frac {\left (-1\right ) ^{m}-1-\left (-1\right ) ^{m}-1}{2\pi ^{2}m^{2}} & & m\neq 0 \end {array} \right . \] Now, for \(m=\pm 2,\pm 4,\cdots \) even, the above becomes\[ \,f_{m_{even}}=\left \{ \begin {array} [c]{ccc}\frac {\sqrt {L}h}{2} & & m=0\\ 0 & & m\neq 0,even \end {array} \right . \] And for or \(m=\pm 1,\pm 3,\cdots \) odd, it becomes\[ \,f_{m_{odd}}=\left \{ \begin {array} [c]{ccc}\frac {\sqrt {L}h}{2} & & m=0\\ -h\sqrt {L}\frac {2}{\pi ^{2}m^{2}} & & m\neq 0,odd \end {array} \right . \] Therefore only the odd terms survive. From (3)\begin {align*} f\relax (x) & \sim \sum _{odd}f_{m}\frac {1}{\sqrt {L}}e^{i\frac {2\pi m}{L}x}\\ & =\frac {\sqrt {L}h}{2}\frac {1}{\sqrt {L}}+\sum _{odd}\frac {-2h\sqrt {L}}{\pi ^{2}m^{2}}\frac {1}{\sqrt {L}}e^{i\frac {2\pi m}{L}x}\\ & =\frac {h}{2}-h\sum _{odd}\frac {2}{\pi ^{2}m^{2}}e^{i\frac {2\pi m}{L}x}\\ & =\frac {h}{2}-\frac {h}{2}\sum _{odd}\frac {4}{\pi ^{2}m^{2}}e^{i\frac {2\pi m}{L}x} \end {align*}
Or\begin {equation} f\relax (x) \sim \frac {h}{2}\left (1-\sum _{odd}\frac {4}{\pi ^{2}m^{2}}e^{i\frac {2\pi m}{L}x}\right ) \tag {8} \end {equation}
From Eq (8), by letting \(x=\frac {L}{2}\), it becomes\begin {align*} f\left (x=\frac {L}{2}\right ) & \sim \frac {h}{2}\left (1-\sum _{odd}\frac {4}{\pi ^{2}m^{2}}e^{i\pi m}\right ) \\ & =\frac {h}{2}\left (1-\left (\sum _{n=-\infty ,odd}^{-1}\frac {4}{\pi ^{2}n^{2}}e^{i\pi n}+\sum _{k=1,odd}^{\infty }\frac {4}{\pi ^{2}k^{2}}e^{i\pi k}\right ) \right ) \end {align*}
Replacing \(m=-n\) in the first sum above gives\begin {align*} f\left (x=\frac {L}{2}\right ) & \sim \frac {h}{2}\left (1-\left ( \sum _{m=\infty ,odd}^{1}\frac {4}{\pi ^{2}\left (-m\right ) ^{2}}e^{-i\pi m}+\sum _{k=1,odd}^{\infty }\frac {4}{\pi ^{2}k^{2}}e^{i\pi k}\right ) \right ) \\ & =\frac {h}{2}\left (1-\left (\sum _{m=1,odd}^{\infty }\frac {4}{\pi ^{2}m^{2}}e^{-i\pi m}+\sum _{k=1,odd}^{\infty }\frac {4}{\pi ^{2}k^{2}}e^{i\pi k}\right ) \right ) \end {align*}
Combining the terms and calling the common index \(n\) gives\begin {align*} f\left (x=\frac {L}{2}\right ) & \sim \frac {h}{2}\left (1-\left ( \sum _{n=1,odd}^{\infty }\frac {4}{\pi ^{2}m^{2}}\left (e^{i\pi n}+e^{-i\pi n}\right ) \right ) \right ) \\ & =\frac {h}{2}\left (1-\left (\sum _{n=1,odd}^{\infty }\frac {8}{\pi ^{2}m^{2}}\cos \left (n\pi \right ) \right ) \right ) \end {align*}
But \(\cos \left (\pi n\right ) =-1\) since \(n\) and odd. The above becomes\[ f\left (x=\frac {L}{2}\right ) \sim \frac {h}{2}\left (1+\sum _{n=1,odd}^{\infty }\frac {8}{\pi ^{2}m^{2}}\right ) \] but \(f\left (x=\frac {L}{2}\right ) =\left [ \frac {2xh}{L}\right ] _{x=\frac {L}{2}}=h\). Hence the above becomes\begin {align*} h & =\frac {h}{2}\left (1+\sum _{odd}\frac {8}{\pi ^{2}m^{2}}\right ) \\ 2 & =1+\sum _{odd}\frac {8}{\pi ^{2}m^{2}}\\ 1 & =\frac {8}{\pi ^{2}}\sum _{odd}\frac {1}{m^{2}} \end {align*}
Therefore\[ \sum _{odd}\frac {1}{m^{2}}=\frac {\pi ^{2}}{8}\]
(i) Obtain the series in terms of sines and cosine for \(f\relax (x) =e^{-\left \vert x\right \vert }\) in the interval \(-1\leq x\leq 1\). (ii) repeat for the case \(f\relax (x) =\cosh x\). Show that\[ f\relax (x) \sim \frac {\sinh \pi }{\pi }\left (1+2\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos nx-n\sin x\right ) \right ) \] represents \(e^{x}\) in the interval \(-\pi \leq x\leq \pi \) (and its periodicized version outside). Show how you can get the series for \(\sinh x\) and \(\cosh x\) from the above.
solution
The period is \(L=2\) in this case. The exponential Fourier series for periodic \(f\relax (x) \) is given by the expansion\begin {equation} |f\rangle =\sum _{m}f_{m}|m\rangle \tag {1} \end {equation} Where \begin {align} f_{m} & =\langle m|f\rangle \nonumber \\ & =\int _{-\frac {L}{2}}^{\frac {L}{2}}\frac {1}{\sqrt {L}}e^{-i\frac {2\pi m}{L}x}f\relax (x) dx \tag {2} \end {align}
And \(|m\rangle \) are the basis functions given by \[ |m\rangle =\frac {1}{\sqrt {L}}e^{i\frac {2\pi m}{L}x}\qquad m=0,\pm 1,\pm 2,\cdots \] Putting these together gives\begin {equation} f\relax (x) \sim \sum _{m=-\infty }^{\infty }f_{m}\frac {1}{\sqrt {L}}e^{i\frac {2\pi m}{L}x} \tag {3} \end {equation} Now \(f_{m}\) is found, using \(L=2\)\begin {align} f_{m} & =\int _{-1}^{1}\frac {1}{\sqrt {2}}e^{-i\pi mx}f\relax (x) dx\nonumber \\ & =\frac {1}{\sqrt {2}}\int _{-1}^{0}e^{-i\pi mx}e^{x}dx+\frac {1}{\sqrt {2}}\int _{0}^{1}e^{-i\pi mx}e^{-x}dx \tag {44} \end {align}
The first integral in (4) gives\begin {align} \frac {1}{\sqrt {2}}\int _{-1}^{0}e^{-i\pi mx}e^{x}dx & =\frac {1}{\sqrt {2}}\int _{-1}^{0}e^{\left (-i\pi m+1\right ) x}dx\nonumber \\ & =\frac {1}{\sqrt {2}}\left [ \frac {e^{\left (-i\pi m+1\right ) x}}{-i\pi m+1}\right ] _{-1}^{0}\nonumber \\ & =\frac {1}{\sqrt {2}\left (-i\pi m+1\right ) }\left [ e^{\left (-i\pi m+1\right ) x}\right ] _{-1}^{0}\nonumber \\ & =\frac {1}{\sqrt {2}\left (-i\pi m+1\right ) }\left [ 1-e^{-\left (-i\pi m+1\right ) }\right ] \nonumber \\ & =\frac {1}{\sqrt {2}}\frac {e^{i\pi m-1}-1}{i\pi m-1} \tag {5} \end {align}
And the second integral in (4) gives\begin {align} \frac {1}{\sqrt {2}}\int _{0}^{1}e^{-i\pi mx}e^{-x}dx & =\frac {1}{\sqrt {2}}\int _{0}^{1}e^{\left (-i\pi m-1\right ) x}dx\nonumber \\ & =\frac {1}{\sqrt {2}}\left [ \frac {e^{\left (-i\pi m-1\right ) x}}{-i\pi m-1}\right ] _{0}^{1}\nonumber \\ & =\frac {1}{\sqrt {2}\left (-i\pi m-1\right ) }\left [ e^{\left (-i\pi m-1\right ) x}\right ] _{0}^{1}\nonumber \\ & =\frac {-1}{\sqrt {2}\left (i\pi m+1\right ) }\left [ e^{\left (-i\pi m-1\right ) }-1\right ] \nonumber \\ & =\frac {1}{\sqrt {2}}\frac {1-e^{-i\pi m-1}}{i\pi m+1} \tag {6} \end {align}
Putting (5,6) together gives\begin {equation} f_{m}=\frac {1}{\sqrt {2}}\frac {e^{i\pi m-1}-1}{i\pi m-1}+\frac {1}{\sqrt {2}}\frac {1-e^{-i\pi m-1}}{i\pi m+1} \tag {7} \end {equation} For \(m=0\), eq (7) becomes\begin {align*} f_{m} & =\frac {1}{\sqrt {2}}\left (\frac {e^{-1}-1}{-1}\right ) +\frac {1}{\sqrt {2}}\left (1-e^{-1}\right ) \\ & =\frac {1}{\sqrt {2}}\left (1-e^{-1}\right ) +\frac {1}{\sqrt {2}}\left ( 1-e^{-1}\right ) \\ & =\frac {1}{\sqrt {2}}-\frac {1}{\sqrt {2}e}+\frac {1}{\sqrt {2}}-\frac {1}{\sqrt {2}e}\\ & =\frac {1}{\sqrt {2}}\left (2-\frac {2}{e}\right ) \\ & =\frac {1}{\sqrt {2}}\left (\frac {2e-2}{e}\right ) \\ & =\frac {2}{\sqrt {2}}\left (\frac {e-1}{e}\right ) \end {align*}
And for \(m\neq 0\) , eq (7) becomes\begin {align*} f_{m} & =\frac {1}{\sqrt {2}}\left (\frac {e^{i\pi m-1}-1}{i\pi m-1}+\frac {1-e^{-i\pi m-1}}{i\pi m+1}\right ) \\ & =\frac {1}{\sqrt {2}}\left (\frac {e^{i\pi m}e^{-1}-1}{i\pi m-1}+\frac {1-e^{-i\pi m}e^{-1}}{i\pi m+1}\right ) \end {align*}
Since \(m\) is integer, then the above becomes\begin {align*} f_{m} & =\frac {1}{\sqrt {2}}\left (\frac {\cos \left (\pi m\right ) e^{-1}-1}{i\pi m-1}+\frac {1-\cos \left (\pi m\right ) e^{-1}}{i\pi m+1}\right ) \\ & =\frac {1}{\sqrt {2}}\left (\frac {\left (-1\right ) ^{m}e^{-1}-1}{i\pi m-1}+\frac {1-\left (-1\right ) ^{m}e^{-1}}{i\pi m+1}\right ) \\ & =\frac {1}{\sqrt {2}}\frac {\left (\left (-1\right ) ^{m}e^{-1}-1\right ) \left (i\pi m+1\right ) +\left (1-\left (-1\right ) ^{m}e^{-1}\right ) \left (i\pi m-1\right ) }{\left (i\pi m-1\right ) \left (i\pi m+1\right ) }\\ & =\frac {1}{\sqrt {2}}\frac {2\left (-1\right ) ^{m}e^{-1}-2}{-\pi ^{2}m^{2}-1}\\ & =\frac {1}{\sqrt {2}}\frac {2-2\left (-1\right ) ^{m}e^{-1}}{1+\pi ^{2}m^{2}}\\ & =\frac {2}{\sqrt {2}}\frac {1-\left (-1\right ) ^{m}e^{-1}}{1+\pi ^{2}m^{2}} \end {align*}
Hence (3) becomes\begin {align*} f\relax (x) & \sim \frac {2}{\sqrt {2}}\left (\frac {e-1}{e}\right ) \frac {1}{\sqrt {2}}+\frac {1}{\sqrt {2}}\sum _{\substack {m=-\infty \\m\neq 0}}^{\infty }\frac {2}{\sqrt {2}}\frac {1-\left (-1\right ) ^{m}e^{-1}}{1+\pi ^{2}m^{2}}e^{i\pi mx}\\ & =\frac {e-1}{e}+\sum _{\substack {m=-\infty \\m\neq 0}}^{\infty }\frac {1-\left ( -1\right ) ^{m}e^{-1}}{1+\pi ^{2}m^{2}}e^{i\pi mx}\\ & =\frac {e-1}{e}+\sum _{n=-\infty }^{-1}\frac {1-\left (-1\right ) ^{n}e^{-1}}{1+\pi ^{2}n^{2}}e^{i\pi nx}+\sum _{k=1}^{\infty }\frac {1-\left (-1\right ) ^{k}e^{-1}}{1+\pi ^{2}k^{2}}e^{i\pi kx} \end {align*}
Let \(m=-n\) in the first sum above. This gives\begin {align*} f\relax (x) & =\frac {e-1}{e}+\sum _{m=\infty }^{1}\frac {1-\left ( -1\right ) ^{\left (-m\right ) }e^{-1}}{1+\pi ^{2}\left (-m\right ) ^{2}}e^{i\pi \left (-m\right ) x}+\sum _{k=1}^{\infty }\frac {1-\left (-1\right ) ^{k}e^{-1}}{1+\pi ^{2}k^{2}}e^{i\pi kx}\\ & =\frac {e-1}{e}+\sum _{m=1}^{\infty }\frac {1-\left (-1\right ) ^{m}e^{-1}}{1+\pi ^{2}m^{2}}e^{-i\pi mx}+\sum _{k=1}^{\infty }\frac {1-\left (-1\right ) ^{k}e^{-1}}{1+\pi ^{2}k^{2}}e^{i\pi kx} \end {align*}
Now the two sums can be combined using one index, say \(n\), since they sum over the same interval\[ f\relax (x) =\frac {e-1}{e}+\sum _{n=\infty }^{1}\frac {1-\left ( -1\right ) ^{n}e^{-1}}{1+\pi ^{2}n^{2}}\left (e^{-i\pi nx}+e^{i\pi nx}\right ) \] But \(\left (e^{-i\pi nx}+e^{i\pi nx}\right ) =2\cos \pi nx\). The above becomes\begin {align*} f\relax (x) & =\frac {e-1}{e}+2\sum _{n=\infty }^{1}\frac {1-\left ( -1\right ) ^{n}e^{-1}}{1+\pi ^{2}n^{2}}\cos \pi nx\\ & =\frac {e-1}{e}+2\sum _{n=\infty }^{1}\frac {e-\left (-1\right ) ^{n}}{e\left (1+\pi ^{2}n^{2}\right ) }\cos \pi nx \end {align*}
Now \(f\relax (x) =\cosh x\). Therefore\begin {equation} f\relax (x) \sim \sum _{m=-\infty }^{\infty }f_{m}\frac {1}{\sqrt {L}}e^{i\frac {2\pi m}{L}x}\tag {3} \end {equation} Where now \(f_{m}\) is found, using \(L=2\)\begin {equation} f_{m}=\int _{-1}^{1}\frac {1}{\sqrt {2}}e^{-i\pi mx}\cosh \relax (x) dx\nonumber \end {equation} But \(\cosh x=\frac {1}{2}\left (e^{x}+e^{-x}\right ) \). The above becomes\begin {align} f_{m} & =\int _{-1}^{1}\frac {1}{\sqrt {2}}e^{-i\pi mx}\frac {1}{2}\left ( e^{x}+e^{-x}\right ) dx\nonumber \\ & =\frac {1}{2\sqrt {2}}\left (\int _{-1}^{1}e^{-i\pi mx}e^{x}dx+\int _{-1}^{1}e^{-i\pi mx}e^{-x}dx\right ) \nonumber \\ & =\frac {1}{2\sqrt {2}}\left (\int _{-1}^{1}e^{\left (-i\pi m+1\right ) x}dx+\int _{-1}^{1}e^{\left (-i\pi m-1\right ) x}dx\right ) \tag {4} \end {align}
The first integral is\begin {align*} \int _{-1}^{1}e^{\left (-i\pi m+1\right ) x}dx & =\frac {1}{-i\pi m+1}\left [ e^{\left (-i\pi m+1\right ) x}\right ] _{-1}^{1}\\ & =\frac {1}{1-i\pi m}\left (e^{\left (-i\pi m+1\right ) }-e^{-\left (-i\pi m+1\right ) }\right ) \\ & =\frac {1}{1-i\pi m}\left (e^{-i\pi m}e-e^{i\pi m}e^{-1}\right ) \end {align*}
Since \(m\) is integer, then \(e^{-i\pi m}=\left (-1\right ) ^{m}\) and \(e^{i\pi m}=\left (-1\right ) ^{m}\). The above becomes\begin {equation} \int _{-1}^{1}e^{\left (-i\pi m+1\right ) x}=\frac {\left (-1\right ) ^{m}}{1-i\pi m}\left (e-e^{-1}\right ) \tag {5} \end {equation} The second integral in (4) becomes\begin {align*} \int _{-1}^{1}e^{\left (-i\pi m-1\right ) x}dx & =\frac {1}{-i\pi m-1}\left [ e^{\left (-i\pi m-1\right ) x}\right ] _{-1}^{1}\\ & =\frac {-1}{1+i\pi m}\left (e^{\left (-i\pi m-1\right ) }-e^{-\left ( -i\pi m-1\right ) }\right ) \\ & =\frac {-1}{1+i\pi m}\left (e^{-i\pi m}e^{-1}-e^{i\pi m}e\right ) \end {align*}
Since \(m\) is integer, the above becomes\begin {equation} \int _{-1}^{1}e^{\left (-i\pi m-1\right ) x}dx=\frac {-\left (-1\right ) ^{m}}{1+i\pi m}\left (e^{-1}-e\right ) \tag {6} \end {equation} Substituting (5,6) in (4) gives\begin {align*} f_{m} & =\frac {1}{2\sqrt {2}}\left (\frac {\left (-1\right ) ^{m}}{1-i\pi m}\left (e-e^{-1}\right ) +\frac {-\left (-1\right ) ^{m}}{1+i\pi m}\left ( e^{-1}-e\right ) \right ) \\ & =\frac {1}{2\sqrt {2}}\left (\frac {\left (-1\right ) ^{m}}{1-i\pi m}\left ( e-e^{-1}\right ) +\frac {\left (-1\right ) ^{m}}{1+i\pi m}\left ( e-e^{-1}\right ) \right ) \\ & =\frac {\left (-1\right ) ^{m}\left (e-e^{-1}\right ) }{2\sqrt {2}}\left ( \frac {1}{1-i\pi m}+\frac {1}{1+i\pi m}\right ) \\ & =\frac {\left (-1\right ) ^{m}\left (e-e^{-1}\right ) }{2\sqrt {2}}\left ( \frac {\left (1+i\pi m\right ) +\left (1-i\pi m\right ) }{\left (1-i\pi m\right ) \left (1+i\pi m\right ) }\right ) \\ & =\frac {\left (-1\right ) ^{m}\left (e-e^{-1}\right ) }{2\sqrt {2}}\left ( \frac {2}{\pi ^{2}m^{2}+1}\right ) \\ & =\frac {\left (-1\right ) ^{m}\left (e-e^{-1}\right ) }{\sqrt {2}\left ( \pi ^{2}m^{2}+1\right ) } \end {align*}
Hence for \(m=0,\)\[ f_{0}=\frac {\left (e-e^{-1}\right ) }{\sqrt {2}}\] Therefore \(f\relax (x) \sim \sum _{m=-\infty }^{\infty }f_{m}\frac {1}{\sqrt {L}}e^{i\frac {2\pi m}{L}x}\) becomes\begin {align*} \cosh \relax (x) & \sim \frac {\left (e-e^{-1}\right ) }{\sqrt {2}}\frac {1}{\sqrt {2}}+\sum _{\substack {m=-\infty \\m\neq 0}}^{\infty }\frac {\left ( -1\right ) ^{m}\left (e-e^{-1}\right ) }{\sqrt {2}\left (\pi ^{2}m^{2}+1\right ) }\frac {1}{\sqrt {2}}e^{i\pi mx}\\ & =\frac {\left (e-e^{-1}\right ) }{2}+\frac {1}{2}\sum _{\substack {m=-\infty \\m\neq 0}}^{\infty }\frac {\left (-1\right ) ^{m}\left (e-e^{-1}\right ) }{\left (\pi ^{2}m^{2}+1\right ) }e^{i\pi mx} \end {align*}
As was done in part(i), \(\sum _{\substack {m=-\infty \\m\neq 0}}^{\infty }e^{i\pi mx}\) can be rewritten as \(\sum _{n=1}^{\infty }2\cos \left (n\pi x\right ) \). The above reduces to\[ \cosh \relax (x) \sim \frac {\left (e-e^{-1}\right ) }{2}+\sum _{n=1}^{\infty }2\frac {\left (-1\right ) ^{n}\left (e-e^{-1}\right ) }{\left (\pi ^{2}n^{2}+1\right ) }\cos \left (n\pi x\right ) \] But \(\frac {\left (e-e^{-1}\right ) }{2}=\sinh 1\). Therefore the above becomes\[ \cosh \relax (x) \sim \sinh \relax (1) \left (1+2\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{\left (1+\pi ^{2}n^{2}\right ) }\cos \left (n\pi x\right ) \right ) \]
I think now the book is asking to find the Fourier series for \(e^{x}\) over \(-\pi \leq x\leq \pi \) in this last part. Therefore, as before, starting with\begin {equation} f\relax (x) \sim \sum _{m=-\infty }^{\infty }f_{m}\frac {1}{\sqrt {L}}e^{i\frac {2\pi m}{L}x} \tag {1} \end {equation} Where now, using \(L=2\pi \) as the period, then\begin {align*} f_{m} & =\int _{-\pi }^{\pi }\frac {1}{\sqrt {2\pi }}e^{-i\frac {2\pi m}{2\pi }x}e^{x}dx\\ & =\frac {1}{\sqrt {2\pi }}\int _{-\pi }^{\pi }e^{\left (-im+1\right ) x}dx\\ & =\frac {1}{\sqrt {2\pi }\left (1-im\right ) }\left [ e^{\left (-im+1\right ) x}\right ] _{-\pi }^{\pi }\\ & =\frac {1}{\sqrt {2\pi }\left (1-im\right ) }\left [ e^{\left (-im+1\right ) \pi }-e^{-\left (-im+1\right ) \pi }\right ] \\ & =\frac {1}{\sqrt {2\pi }\left (1-im\right ) }\left [ e^{-im\pi }e^{\pi }-e^{im\pi }e^{-\pi }\right ] \end {align*}
But \(e^{im\pi }=\left (-1\right ) ^{m}\) and \(e^{-im\pi }=\left (-1\right ) ^{m}\) since \(m\) is integer. The above becomes\begin {align*} f_{m} & =\frac {\left (-1\right ) ^{m}}{\sqrt {2\pi }\left (1-im\right ) }\left [ e^{\pi }-e^{-\pi }\right ] \\ & =\frac {\left (-1\right ) ^{m}\left (2\sinh \pi \right ) }{\sqrt {2\pi }\left (1-im\right ) }\\ & =\frac {2}{\sqrt {2\pi }}\frac {\left (-1\right ) ^{m}}{\left (1-im\right ) }\sinh \pi \\ & =\frac {\sqrt {2}}{\sqrt {\pi }}\frac {\left (-1\right ) ^{m}}{\left ( 1-im\right ) }\sinh \pi \end {align*}
For \(m=0\) the above gives\[ f_{0}=\frac {\sqrt {2}}{\sqrt {\pi }}\sinh \pi \] Therefore \(f\relax (x) \sim \sum _{m=-\infty }^{\infty }f_{m}\frac {1}{\sqrt {L}}e^{i\frac {2\pi m}{L}x}\) becomes, where \(L=2\pi \) now,\begin {align} e^{x} & \sim \sum _{m=-\infty }^{\infty }f_{m}\frac {1}{\sqrt {L}}e^{i\frac {2\pi m}{L}x}\nonumber \\ & =\frac {\sqrt {2}}{\sqrt {\pi }}\sinh \pi \left (\frac {1}{\sqrt {2\pi }}\right ) +\sum _{m=-\infty ,\neq 0}^{\infty }\left (\frac {\sqrt {2}}{\sqrt {\pi }}\frac {\left (-1\right ) ^{m}}{\left (1-im\right ) }\sinh \pi \right ) \frac {1}{\sqrt {2\pi }}e^{imx}\nonumber \\ & =\frac {\sinh \pi }{\pi }+\frac {\sinh \pi }{\pi }\sum _{m=-\infty ,\neq 0}^{\infty }\frac {\left (-1\right ) ^{m}}{\left (1-im\right ) }e^{imx}\nonumber \\ & =\frac {\sinh \pi }{\pi }+\frac {\sinh \pi }{\pi }\sum _{m=-\infty ,\neq 0}^{\infty }\frac {\left (-1\right ) ^{m}\left (1+im\right ) }{\left (1-im\right ) \left (1+im\right ) }e^{imx}\nonumber \\ & =\frac {\sinh \pi }{\pi }\left (1+\sum _{m=-\infty ,\neq 0}^{\infty }\frac {\left ( -1\right ) ^{m}+i\left (-1\right ) ^{m}m}{1+m^{2}}e^{imx}\right ) \nonumber \\ & =\frac {\sinh \pi }{\pi }\left (1+\sum _{m=-\infty ,\neq 0}^{\infty }\frac {\left ( -1\right ) ^{m}}{1+m^{2}}e^{imx}+\frac {i\left (-1\right ) ^{m}m}{1+m^{2}}e^{imx}\right ) \nonumber \\ & =\frac {\sinh \pi }{\pi }\left (1+\sum _{m=-\infty ,\neq 0}^{\infty }\frac {\left ( -1\right ) ^{m}}{1+m^{2}}e^{imx}+\sum _{m=-\infty ,\neq 0}^{\infty }\frac {i\left ( -1\right ) ^{m}m}{1+m^{2}}e^{imx}\right ) \tag {2} \end {align}
The first sum above becomes\begin {equation} \sum _{m=-\infty ,\neq 0}^{\infty }\frac {\left (-1\right ) ^{m}}{1+m^{2}}e^{imx}=2\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+m^{2}}\cos \left ( nx\right ) \tag {3} \end {equation} And the second sum in (2) becomes\begin {align*} \sum _{m=-\infty ,\neq 0}^{\infty }\frac {i\left (-1\right ) ^{m}m}{1+m^{2}}e^{imx} & =\sum _{m=-\infty ,\neq 0}^{\infty }\frac {\left (-1\right ) \left ( -1\right ) ^{m}m}{1+m^{2}}\frac {e^{imx}}{i}\\ & =\sum _{k=-\infty }^{-1}\frac {\left (-1\right ) \left (-1\right ) ^{k}k}{1+k^{2}}\frac {e^{ikx}}{i}+\sum _{r=1}^{\infty }\frac {\left (-1\right ) \left (-1\right ) ^{r}r}{1+r^{2}}\frac {e^{irx}}{i} \end {align*}
Letting \(m=-k\) in the first sum above gives\begin {align*} \sum _{m=-\infty ,\neq 0}^{\infty }\frac {i\left (-1\right ) ^{m}m}{1+m^{2}}e^{imx} & =\sum _{m=\infty }^{1}\frac {\left (-1\right ) \left (-1\right ) ^{-m}\left (-m\right ) }{1+\left (-m\right ) ^{2}}\frac {e^{-imx}}{i}+\sum _{r=1}^{\infty }\frac {\left (-1\right ) \left (-1\right ) ^{r}r}{1+r^{2}}\frac {e^{irx}}{i}\\ & =\sum _{m=1}^{\infty }\frac {\left (-1\right ) ^{m}m}{1+m^{2}}\frac {e^{-imx}}{i}-\sum _{r=1}^{\infty }\frac {\left (-1\right ) ^{r}r}{1+r^{2}}\frac {e^{irx}}{i} \end {align*}
Merging the two sums back together since now on same interval, and using \(n\) for the common index\begin {align} \sum _{m=-\infty ,\neq 0}^{\infty }\frac {i\left (-1\right ) ^{m}m}{1+m^{2}}e^{imx} & =\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}n}{1+n^{2}}\left (\frac {e^{-inx}}{i}-\frac {e^{inx}}{i}\right ) \nonumber \\ & =-\sum _{n=1}^{\infty }2\frac {\left (-1\right ) ^{n}n}{1+n^{2}}\left ( \sin \left (nx\right ) \right ) \tag {4} \end {align}
Substituting (3,4) back in (2) gives\begin {align} e^{x} & \sim \frac {\sinh \pi }{\pi }\left (1+2\sum _{n=1}^{\infty }\frac {\left ( -1\right ) ^{n}}{1+n^{2}}\cos \left (nx\right ) -\sum _{n=1}^{\infty }2\frac {\left (-1\right ) ^{n}n}{1+n^{2}}\left (\sin \left (nx\right ) \right ) \right ) \nonumber \\ & =\frac {\sinh \pi }{\pi }\left (1+2\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left (nx\right ) -n\sin \left (nx\right ) \right ) \right ) \tag {5} \end {align}
The question is now asking to show how to use (5) to obtain the series for \(\sinh x\) and \(\cosh x\). Since \[ \sinh x=\frac {e^{x}-e^{-x}}{2}\] Then substituting (5) in the RHS of the above gives\begin {align} \sinh x & \sim \frac {1}{2}\left (\frac {\sinh \pi }{\pi }\left (1+2\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left ( nx\right ) -n\sin \left (nx\right ) \right ) \right ) -\frac {\sinh \pi }{\pi }\left (1+2\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left ( \cos \left (n\left (-x\right ) \right ) -n\sin \left (n\left (-x\right ) \right ) \right ) \right ) \right ) \nonumber \\ & =\frac {1}{2}\left (\frac {\sinh \pi }{\pi }\left (1+2\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left (nx\right ) -n\sin \left (nx\right ) \right ) \right ) -\frac {\sinh \pi }{\pi }\left ( 1+2\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left ( \cos \left (nx\right ) +n\sin \left (nx\right ) \right ) \right ) \right ) \nonumber \\ & =\frac {1}{2}\left (\frac {\sinh \pi }{\pi }+2\frac {\sinh \pi }{\pi }\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left ( nx\right ) -n\sin \left (nx\right ) \right ) -\left (\frac {\sinh \pi }{\pi }+2\frac {\sinh \pi }{\pi }\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left (nx\right ) +n\sin \left (nx\right ) \right ) \right ) \right ) \nonumber \\ & =\frac {1}{2}\left (\frac {\sinh \pi }{\pi }+2\frac {\sinh \pi }{\pi }\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left ( nx\right ) -n\sin \left (nx\right ) \right ) -\frac {\sinh \pi }{\pi }-2\frac {\sinh \pi }{\pi }\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left (nx\right ) +n\sin \left (nx\right ) \right ) \right ) \nonumber \\ & =\frac {\sinh \pi }{\pi }\left (\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left (nx\right ) -n\sin \left (nx\right ) \right ) -\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left ( nx\right ) +n\sin \left (nx\right ) \right ) \right ) \nonumber \\ & =\frac {\sinh \pi }{\pi }\left (\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left (nx\right ) -n\sin \left (nx\right ) -\cos \left (nx\right ) -n\sin \left (nx\right ) \right ) \right ) \nonumber \\ & =\frac {\sinh \pi }{\pi }\left (\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (-n\sin \left (nx\right ) -n\sin \left (nx\right ) \right ) \right ) \nonumber \\ & =\frac {\sinh \pi }{\pi }\left (\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (-2n\sin \left (nx\right ) \right ) \right ) \nonumber \end {align}
Hence\begin {equation} \sinh x\sim \frac {2\sinh \pi }{\pi }\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n+1}}{1+n^{2}}n\sin \left (nx\right ) \tag {6} \end {equation} Similarly for \[ \cosh x=\frac {e^{x}+e^{-x}}{2}\] Then substituting (5) in the RHS of the above gives\begin {align} \cosh x & \sim \frac {1}{2}\left (\frac {\sinh \pi }{\pi }\left (1+2\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left ( nx\right ) -n\sin \left (nx\right ) \right ) \right ) +\frac {\sinh \pi }{\pi }\left (1+2\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left ( \cos \left (n\left (-x\right ) \right ) -n\sin \left (n\left (-x\right ) \right ) \right ) \right ) \right ) \nonumber \\ & =\frac {1}{2}\left (\frac {\sinh \pi }{\pi }\left (1+2\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left (nx\right ) -n\sin \left (nx\right ) \right ) \right ) +\frac {\sinh \pi }{\pi }\left ( 1+2\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left ( \cos \left (nx\right ) +n\sin \left (nx\right ) \right ) \right ) \right ) \nonumber \\ & =\frac {1}{2}\left (\frac {\sinh \pi }{\pi }+2\frac {\sinh \pi }{\pi }\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left ( nx\right ) -n\sin \left (nx\right ) \right ) +\left (\frac {\sinh \pi }{\pi }+2\frac {\sinh \pi }{\pi }\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left (nx\right ) +n\sin \left (nx\right ) \right ) \right ) \right ) \nonumber \\ & =\frac {1}{2}\left (\frac {\sinh \pi }{\pi }+2\frac {\sinh \pi }{\pi }\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left ( nx\right ) -n\sin \left (nx\right ) \right ) +\frac {\sinh \pi }{\pi }+2\frac {\sinh \pi }{\pi }\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left (nx\right ) +n\sin \left (nx\right ) \right ) \right ) \nonumber \\ & =\frac {1}{2}\left (2\frac {\sinh \pi }{\pi }+2\frac {\sinh \pi }{\pi }\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left ( nx\right ) -n\sin \left (nx\right ) +\cos \left (nx\right ) +n\sin \left ( nx\right ) \right ) \right ) \nonumber \\ & =\frac {1}{2}\left (2\frac {\sinh \pi }{\pi }+2\frac {\sinh \pi }{\pi }\sum _{n=1}^{\infty }\frac {\left (-1\right ) ^{n}}{1+n^{2}}\left (\cos \left ( nx\right ) +\cos \left (nx\right ) \right ) \right ) \nonumber \\ & =\frac {\sinh \pi }{\pi }+\frac {\sinh \pi }{\pi }\sum _{n=1}^{\infty }\frac {\left ( -1\right ) ^{n}}{1+n^{2}}2\cos \left (nx\right ) \nonumber \end {align}
Hence\begin {equation} \cosh x\sim \frac {\sinh \pi }{\pi }\left (1+2\sum _{n=1}^{\infty }\frac {\left ( -1\right ) ^{n}}{1+n^{2}}\cos \left (nx\right ) \right ) \tag {7} \end {equation}
Perform appropriate integration to show the following results regarding the Dirac delta function
(1) \(\delta \left (ax\right ) =\frac {\delta \relax (x) }{\left \vert a\right \vert }\) where \(a\) is real number. (2) \(\delta \left (f\left ( x\right ) \right ) =\sum _{i}\frac {\delta \left (x-x_{i}\right ) }{\left \vert \frac {df}{dx}\right \vert _{i}}\) where \(x_{i}\) satisfies \(f\left ( x_{i}\right ) =0\) (3) \(\frac {d}{dx}\delta \left (x-x^{\prime }\right ) =\delta \left (x-x^{\prime }\right ) \frac {d}{dx^{\prime }}\)
Solution
Using the integral definition of delta function given by\begin {equation} \delta \relax (x) =\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{ikx}dk \tag {1} \end {equation} Then\[ \delta \left (ax\right ) =\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{ikax}dk \] Case \(a>0\). Let \(u=ak\). Then \(du=adk\). The above becomes\[ \delta \left (ax\right ) =\frac {1}{a}\left (\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{iux}du\right ) \] But \(\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{iux}du=\delta \relax (x) \) by definition. Hence the above becomes\begin {equation} \delta \left (ax\right ) =\frac {1}{a}\delta \relax (x) \tag {2} \end {equation} Case \(a<0\). Let \(u=ak\). Then \(du=adk\). When \(k=\infty ,u=-\infty \) and when \(k=-\infty ,u=+\infty \). The integral becomes\begin {align} \delta \left (ax\right ) & =\frac {1}{a}\left (\frac {1}{2\pi }\int _{\infty }^{-\infty }e^{iux}du\right ) \nonumber \\ & =\frac {1}{-a}\left (\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{iux}du\right ) \nonumber \\ & =\frac {1}{-a}\delta \relax (x) \tag {3} \end {align}
Combining (2,3) gives\[ \delta \left (ax\right ) =\frac {1}{\left \vert a\right \vert }\delta \left ( x\right ) \]
Using
\begin {equation} \int _{-\infty }^{\infty }\delta \left (f\relax (x) \right ) dx=\sum _{i}\int _{x_{i}-\varepsilon }^{x_{i}+\varepsilon }\delta \left (f\left ( x\right ) \right ) dx \tag {1} \end {equation} Where in the RHS, the sum is over the roots of \(f\relax (x) \), where \(f\left (x_{i}\right ) =0\) where \(x_{i}\) is root of \(f\relax (x) \) since \(\delta \relax (u) \) is nonzero only when its argument is zero, which is at the roots of \(f\relax (x) \,\). Now, expanding \(f\left ( x\right ) \) near each one of its roots using Taylor series\[ f\relax (x) =f\left (x_{i}\right ) +\left (x-x_{i}\right ) f^{\prime }\left (x_{i}\right ) +O\left (x^{2}\right ) \] But \(f\left (x_{i}\right ) =0\) since \(x_{i}\) is root, and keeping only linear terms, then (1) now becomes\[ \int _{-\infty }^{\infty }\delta \left (f\relax (x) \right ) dx=\sum _{i}\int _{x_{i}-\varepsilon }^{x_{i}+\varepsilon }\delta \left (\left ( x-x_{i}\right ) f^{\prime }\left (x_{i}\right ) \right ) dx \] But from part (1), we found that \(\delta \left (a\left (x-x_{i}\right ) \right ) =\frac {1}{\left \vert a\right \vert }\delta \left (x-x_{i}\right ) \), where now \(a=f^{\prime }\left (x_{i}\right ) \). Using this relation in the above gives\[ \int _{-\infty }^{\infty }\delta \left (f\relax (x) \right ) dx=\int _{-\infty }^{\infty }\sum _{i}\frac {1}{\left \vert f^{\prime }\left ( x_{i}\right ) \right \vert }\delta \left (x-x_{i}\right ) \] Therefore the integrands on each side is the same. This implies\[ \delta \left (f\relax (x) \right ) =\sum _{i}\frac {\delta \left ( x-x_{i}\right ) }{\left \vert f^{\prime }\left (x_{i}\right ) \right \vert }\]
Starting from property of delta function which is \[ \int \delta \left (x-x^{\prime }\right ) f(x^{\prime })dx^{\prime }=f(x) \] Taking derivative of both sides w.r.t. \(x\) gives\begin {align*} \frac {d}{dx}\int \delta \left (x-x^{\prime }\right ) f(x^{\prime })dx^{\prime } & =\frac {d}{dx}f(x)\\ \int \frac {d\delta \left (x-x^{\prime }\right ) }{dx}f(x^{\prime })dx^{\prime } & =\frac {d}{dx}f(x) \end {align*}
Integration by part. Let \(\frac {d\delta \left (x-x^{\prime }\right ) }{dx}=dv,u=f(x^{\prime })\), then \(v=\left (x-x^{\prime }\right ) ,du=\frac {d}{dx^{\prime }}f\left (x^{\prime }\right ) \). The above becomes\[ \int \delta \left (x-x^{\prime }\right ) \frac {d}{dx^{\prime }}f(x^{\prime })dx^{\prime }=\frac {d}{dx}f(x) \] Therefore\[ \int \frac {d\delta \left (x-x^{\prime }\right ) }{dx}f(x^{\prime })dx^{\prime }=\int \delta \left (x-x^{\prime }\right ) \frac {d}{dx^{\prime }}f(x^{\prime })dx^{\prime }\] or\[ \frac {d\delta \left (x-x^{\prime }\right ) }{dx}f(x^{\prime })=\delta \left ( x-x^{\prime }\right ) \frac {d}{dx^{\prime }}f(x^{\prime }) \] or\[ \frac {d\delta \left (x-x^{\prime }\right ) }{dx}=\delta \left (x-x^{\prime }\right ) \frac {d}{dx^{\prime }}\]
For each energy eigenstate of a particle of mass \(m\) in the infinitely-deep potential well between \(x=0\) and \(L\), find the probability distribution of the possible results when the particle momentum is measured.
Solution
The goal is to determine \(\left \vert \langle \phi _{p}|\psi \rangle \right \vert ^{2}\) which will give the probability of measuring momentum \(p\). But \begin {align} \langle \phi _{p}|\psi _{n}\rangle & =\int _{0}^{\infty }\langle \phi _{p}|x\rangle \langle x|\psi _{n}\rangle dx\nonumber \\ & =\int _{0}^{\infty }\langle x|\phi _{p}\rangle ^{\ast }\langle x|\psi _{n}\rangle dx \tag {1} \end {align}
But \(\langle x|\phi _{p}\rangle =\phi _{p}\relax (x) \) and \(\langle x|\psi _{n}\rangle =\psi _{n}\relax (x) \). From lecture notes, \begin {align*} \phi _{p}\relax (x) & =\frac {1}{\sqrt {2\pi \hbar }}e^{i\frac {px}{\hbar }}\\ \psi _{n}\relax (x) & =\left \{ \begin {array} [c]{ccc}\sqrt {\frac {2}{L}}\sin \frac {n\pi x}{L} & & 0<x<L\\ 0 & & \text {otherwise}\end {array} \right \} \end {align*}
For \(n=1,2,3,\cdots \). Substituting the above in (1) gives\begin {align} \langle \phi _{p}|\psi _{n}\rangle & =\int _{0}^{\infty }\phi _{p}^{\ast }\left ( x\right ) \psi _{n}\relax (x) dx\nonumber \\ & =\int _{0}^{L}\frac {1}{\sqrt {2\pi \hbar }}e^{-i\frac {px}{\hbar }}\sqrt {\frac {2}{L}}\sin \frac {n\pi x}{L}dx\nonumber \\ & =\frac {1}{\sqrt {\pi \hbar L}}\int _{0}^{L}e^{-i\frac {px}{\hbar }}\sin \left ( \frac {n\pi x}{L}\right ) dx \tag {2} \end {align}
To evaluate \(I=\int _{0}^{L}e^{-i\frac {px}{\hbar }}\sin \left (\frac {n\pi x}{L}\right ) dx\) we do Integration by parts twice. Let \(u=\sin \left ( \frac {n\pi x}{L}\right ) ,dv=e^{-i\frac {px}{\hbar }}\) then \(du=\frac {n\pi }{L}\cos \left (\frac {n\pi x}{L}\right ) dx\) and \(v=\hbar \frac {e^{-i\frac {px}{\hbar }}}{-ip}\). Hence\begin {align*} I & =\left [ uv\right ] _{0}^{L}-\int _{0}^{L}vdu\\ & =\left [ \sin \left (\frac {n\pi x}{L}\right ) \hbar \frac {e^{-i\frac {px}{\hbar }}}{-ip}\right ] _{0}^{L}-\int _{0}^{L}\frac {n\pi }{L}\cos \left ( \frac {n\pi x}{L}\right ) \hbar \frac {e^{-i\frac {px}{\hbar }}}{-ip}dx\\ & =\left [ \sin \left (\frac {n\pi L}{L}\right ) \hbar \frac {e^{-i\frac {pL}{\hbar }}}{-ip}-0\right ] +\frac {\hbar n\pi }{ipL}\int _{0}^{L}\cos \left ( \frac {n\pi x}{L}\right ) e^{-i\frac {px}{\hbar }}dx \end {align*}
Since \(n\) is integer, then boundary terms are zero. \[ I=\frac {\hbar n\pi }{ipL}\int _{0}^{L}\cos \left (\frac {n\pi x}{L}\right ) e^{-i\frac {px}{\hbar }}dx \] Doing integration by parts one more time. Let \(u=\cos \left (\frac {n\pi x}{L}\right ) ,dv=e^{-i\frac {px}{\hbar }}\) then \(du=-\frac {n\pi }{L}\sin \left ( \frac {n\pi x}{L}\right ) dx\), then the above becomes\begin {align*} I & =\frac {\hbar n\pi }{ipL}\left (\left [ \cos \left (\frac {n\pi x}{L}\right ) \hbar \frac {e^{-i\frac {px}{\hbar }}}{-ip}\right ] _{0}^{L}+\int _{0}^{L}\frac {\hbar e^{-i\frac {px}{\hbar }}}{-ip}\frac {n\pi }{L}\sin \left ( \frac {n\pi x}{L}\right ) dx\right ) \\ & =\frac {\hbar n\pi }{ipL}\left (\frac {\hbar }{-ip}\left [ \cos \left ( n\pi \right ) e^{-i\frac {pL}{\hbar }}-1\right ] -\frac {\hbar n\pi }{ipL}\int _{0}^{L}e^{-i\frac {px}{\hbar }}\sin \left (\frac {n\pi x}{L}\right ) dx\right ) \\ & =\frac {\hbar ^{2}}{-ip}\frac {n\pi }{ipL}\left [ \cos \left (n\pi \right ) e^{-i\frac {pL}{\hbar }}-1\right ] -\frac {n\pi }{ipL}\frac {\hbar ^{2}n\pi }{ipL}\int _{0}^{L}e^{-i\frac {px}{\hbar }}\sin \left (\frac {n\pi x}{L}\right ) dx\\ & =\frac {\hbar ^{2}n\pi }{p^{2}L}\left [ \cos \left (n\pi \right ) e^{-i\frac {pL}{\hbar }}-1\right ] +\frac {\hbar ^{2}n^{2}\pi ^{2}}{p^{2}L^{2}}\int _{0}^{L}e^{-i\frac {px}{\hbar }}\sin \left (\frac {n\pi x}{L}\right ) dx \end {align*}
But \(\int _{0}^{L}e^{-i\frac {px}{\hbar }}\sin \left (\frac {n\pi x}{L}\right ) dx=I\). Therefore the above becomes\[ I=\frac {\hbar ^{2}n\pi }{p^{2}L}\left (\cos \left (n\pi \right ) e^{-i\frac {pL}{\hbar }}-1\right ) +\frac {\hbar ^{2}n^{2}\pi ^{2}}{p^{2}L^{2}}I \] Solving for \(I\)\begin {align*} I-\frac {\hbar ^{2}n^{2}\pi ^{2}}{p^{2}L^{2}}I & =\frac {\hbar ^{2}n\pi }{p^{2}L}\left (\cos \left (n\pi \right ) e^{-i\frac {pL}{\hbar }}-1\right ) \\ I\left (1-\frac {\hbar ^{2}n^{2}\pi ^{2}}{p^{2}L^{2}}\right ) & =\frac {\hbar ^{2}n\pi }{p^{2}L}\left (\cos \left (n\pi \right ) e^{-i\frac {pL}{\hbar }}-1\right ) \\ I & =\frac {\hbar ^{2}n\pi }{p^{2}L}\frac {\left (\cos \left (n\pi \right ) e^{-i\frac {pL}{\hbar }}-1\right ) }{\left (1-\frac {\hbar ^{2}n^{2}\pi ^{2}}{p^{2}L^{2}}\right ) }\\ & =p^{2}L^{2}\frac {\hbar ^{2}n\pi }{p^{2}L}\frac {\left (\cos \left ( n\pi \right ) e^{-i\frac {pL}{\hbar }}-1\right ) }{\left (p^{2}L^{2}-\hbar ^{2}n^{2}\pi ^{2}\right ) }\\ & =\frac {\hbar ^{2}n\pi L}{p^{2}L^{2}-\hbar ^{2}n^{2}\pi ^{2}}\left (\left ( -1\right ) ^{n}e^{-i\frac {pL}{\hbar }}-1\right ) \end {align*}
Substituting the above in (2) gives\begin {align*} \langle \phi _{p}|\psi _{n}\rangle & =\frac {1}{\sqrt {\pi \hbar L}}\left ( \frac {\hbar ^{2}n\pi L}{p^{2}L^{2}-\hbar ^{2}n^{2}\pi ^{2}}\left (\left ( -1\right ) ^{n}e^{-i\frac {pL}{\hbar }}-1\right ) \right ) \\ & =\frac {\hbar ^{2}n\pi L\left (\sqrt {\pi \hbar L}\right ) }{\left (\pi \hbar L\right ) \left (p^{2}L^{2}-\hbar ^{2}n^{2}\pi ^{2}\right ) }\left (\left ( -1\right ) ^{n}e^{-\frac {ipL}{\hbar }}-1\right ) \\ & =\frac {n\hbar \sqrt {\pi \hbar L}}{\left (p^{2}L^{2}-\hbar ^{2}n^{2}\pi ^{2}\right ) }\left (\left (-1\right ) ^{n}e^{-\frac {ipL}{\hbar }}-1\right ) \end {align*}
Let \(k_{n}=\frac {n\hbar \sqrt {\pi \hbar L}}{\left (p^{2}L^{2}-\hbar ^{2}n^{2}\pi ^{2}\right ) }\), then \begin {align*} \langle \phi _{p}|\psi _{n}\rangle & =k_{n}\left (\left (-1\right ) ^{n}e^{-\frac {ipL}{\hbar }}-1\right ) \\ & =\left (-1\right ) ^{n}k_{n}e^{-\frac {ipL}{\hbar }}-k_{n}\\ & =\left (-1\right ) ^{n}k_{n}\left (\cos \frac {pL}{\hbar }-i\sin \frac {pL}{\hbar }\right ) -k_{n}\\ & =\left (-1\right ) ^{n}k_{n}\cos \frac {pL}{\hbar }-i\left (-1\right ) ^{n}k_{n}\sin \frac {pL}{\hbar }-k_{n}\\ & =\left (\left (-1\right ) ^{n}k_{n}\cos \frac {pL}{\hbar }-k_{n}\right ) -i\left (\left (-1\right ) ^{n}k_{n}\sin \frac {pL}{\hbar }\right ) \end {align*}
Hence \[ \left \vert \langle \phi _{p}|\psi _{n}\rangle \right \vert =\sqrt {\left (\left ( -1\right ) ^{n}k_{n}\cos \frac {pL}{\hbar }-k_{n}\right ) ^{2}+\left (\left ( -1\right ) ^{n}k_{n}\sin \frac {pL}{\hbar }\right ) ^{2}}\] And\begin {align*} \left \vert \langle \phi _{p}|\psi _{n}\rangle \right \vert ^{2} & =\left ( \left (-1\right ) ^{n}k_{n}\cos \frac {pL}{\hbar }-k_{n}\right ) ^{2}+\left ( \left (-1\right ) ^{n}k_{n}\sin \frac {pL}{\hbar }\right ) ^{2}\\ & =\left (-1\right ) ^{2n}k_{n}^{2}\cos ^{2}\frac {pL}{\hbar }+k_{n}^{2}-2k_{n}^{2}\left (-1\right ) ^{n}\cos \frac {pL}{\hbar }+\left (-1\right ) ^{2n}k_{n}^{2}\sin ^{2}\frac {pL}{\hbar }\\ & =\left (-1\right ) ^{2n}k_{n}^{2}\left (\cos ^{2}\frac {pL}{\hbar }+\sin ^{2}\frac {pL}{\hbar }\right ) +k_{n}^{2}-2k_{n}^{2}\left (-1\right ) ^{n}\cos \frac {pL}{\hbar }\\ & =\left (-1\right ) ^{2n}k_{n}^{2}+k_{n}^{2}-2k_{n}^{2}\left (-1\right ) ^{n}\cos \frac {pL}{\hbar }\\ & =k_{n}^{2}\left (1+\left (-1\right ) ^{2n}-2\left (-1\right ) ^{n}\cos \frac {pL}{\hbar }\right ) \end {align*}
But \(k_{n}=\frac {n\hbar \sqrt {\pi \hbar L}}{\left (p^{2}L^{2}-\hbar ^{2}n^{2}\pi ^{2}\right ) }\), therefore the above becomes\begin {align*} \left \vert \langle \phi _{p}|\psi _{n}\rangle \right \vert ^{2} & =\left ( \frac {n\hbar \sqrt {\pi \hbar L}}{\left (p^{2}L^{2}-\hbar ^{2}n^{2}\pi ^{2}\right ) }\right ) ^{2}\left (1+\left (-1\right ) ^{2n}-2\left ( -1\right ) ^{n}\cos \frac {pL}{\hbar }\right ) \\ & =\frac {n^{2}\hbar ^{3}\pi L}{\left (p^{2}L^{2}-\hbar ^{2}n^{2}\pi ^{2}\right ) ^{2}}\left (1+\left (-1\right ) ^{2n}-2\left (-1\right ) ^{n}\cos \left (\frac {pL}{\hbar }\right ) \right ) \end {align*}
The above gives the probability of measurement of \(p\), where \(n=1,2,3,\cdots \). For illustration, the following two tables are generated to see how the probability of measuring say \(p=1\) and \(p=2\) changes as function of \(n\). To generate this, \(L\) is taken as \(1\) and \(\hbar =1\) for simplicity.
