Evaluate \(\int _{0}^{1}e^{\sqrt {x}}dx\). Show that \(\int _{0}^{\infty }e^{-x^{4}}dx=\Gamma \left (\frac {5}{4}\right ) \)
Solution
Let \(y=\sqrt {x}\). Therefore\begin {align*} \frac {dy}{dx} & =\frac {1}{2}\frac {1}{\sqrt {x}}\\ & =\frac {1}{2}\frac {1}{y} \end {align*}
And\[ dx=2ydy \] When \(x=0\), \(y=0\) and when \(x=1\),\(y=1\). Substituting this back into \(\int _{0}^{1}e^{\sqrt {x}}dx\) gives \(\int _{0}^{1}e^{y}\left (2ydy\right ) =2\int _{0}^{1}ye^{y}dy\). This integral is evaluated using integration by parts. \[ udv=\left . uv\right \vert _{0}^{1}-\int _{0}^{1}vdu \] Let \(u=y\) and \(dv=e^{y}\), then \(du=dy\) and \(v=e^{y}\). The above becomes\begin {align*} 2\left (\int _{0}^{1}ye^{y}dy\right ) & =2\left (\left . uv\right \vert _{0}^{1}-\int _{0}^{1}vdu\right ) \\ & =2\left (\left . ye^{y}\right \vert _{0}^{1}-\int _{0}^{1}e^{y}dy\right ) \\ & =2\left (\left (e^{1}-0\right ) -\left . e^{y}\right \vert _{0}^{1}\right ) \\ & =2\left (e-\left (e-1\right ) \right ) \\ & =2\left (e-e+1\right ) \\ & =2 \end {align*}
Hence \[ \int _{0}^{1}e^{\sqrt {x}}dx=2 \] For the second part of the question asking to evaluate \(\int _{0}^{\infty }e^{-x^{4}}dx\), let\[ x=y^{\frac {1}{4}}\] Then \[ \frac {dx}{dy}=\frac {1}{4}y^{\left (\frac {1}{4}-1\right ) }\] When \(x=0,y=0\) and when \(x=\infty ,y=\infty \). Hence the above integral becomes\begin {align} \int _{0}^{\infty }e^{-x^{4}}dx & =\int _{0}^{\infty }e^{-y}\left (\frac {1}{4}y^{\left (\frac {1}{4}-1\right ) }dy\right ) \nonumber \\ & =\frac {1}{4}\int _{0}^{\infty }y^{\left (\frac {1}{4}-1\right ) }e^{-y}dy\tag {1} \end {align}
Comparing the above to integral (2.1.39) in the book which says\begin {align} F\relax (n) & =\int _{0}^{\infty }y^{n}e^{-y}dy\tag {2}\\ \Gamma \relax (n) & =F\left (n-1\right ) \tag {3} \end {align}
Then putting \(n=\frac {1}{4}\) in (3) gives\begin {align*} \Gamma \left (\frac {1}{4}\right ) & =F\left (\frac {1}{4}-1\right ) \\ & =\int _{0}^{\infty }y^{\left (\frac {1}{4}-1\right ) }e^{-y}dy \end {align*}
Which is (1). This means that\[ \int _{0}^{\infty }y^{\left (\frac {1}{4}-1\right ) }e^{-y}dy=\Gamma \left ( \frac {1}{4}\right ) \] Hence \begin {equation} \frac {1}{4}\int _{0}^{\infty }y^{\left (\frac {1}{4}-1\right ) }e^{-y}dy=\frac {1}{4}\Gamma \left (\frac {1}{4}\right ) \tag {4} \end {equation} To obtain the final form, the following property of Gamma functions is used\[ \Gamma \left (n+1\right ) =n\Gamma \relax (n) \] Which means that when \(n=\frac {1}{4}\), the above becomes\begin {align*} \Gamma \left (\frac {1}{4}+1\right ) & =\frac {1}{4}\Gamma \left (\frac {1}{4}\right ) \\ \Gamma \left (\frac {5}{4}\right ) & =\frac {1}{4}\Gamma \left (\frac {1}{4}\right ) \end {align*}
Using this in (4) shows that\[ \frac {1}{4}\int _{0}^{\infty }y^{\left (\frac {1}{4}-1\right ) }e^{-y}dy=\Gamma \left (\frac {5}{4}\right ) \] Which implies\[ \int _{0}^{\infty }e^{-x^{4}}dx=\Gamma \left (\frac {5}{4}\right ) \] Which is what we are asked to show.
Solution
Let \[ I\relax (a) =\int _{0}^{1}\frac {t^{a}-1}{\ln t}dt \] Where \(a=1\) for the specific integral in this problem. The above is the parametrized general form. Taking derivative w.r.t \(a\) gives \begin {align} \frac {dI\relax (a) }{da} & =\frac {d}{da}\left (\int _{0}^{1}\frac {t^{a}-1}{\ln t}dt\right ) \nonumber \\ & =\int _{0}^{1}\frac {d}{da}\left (\frac {t^{a}-1}{\ln t}\right ) dt\nonumber \\ & =\int _{0}^{1}\frac {1}{\ln t}\frac {d}{da}\left (t^{a}-1\right ) dt\tag {1} \end {align}
But \begin {align} \frac {d}{da}\left (t^{a}-1\right ) & =\frac {d}{da}\left (e^{a\ln t}-1\right ) \nonumber \\ & =\ln \relax (t) \left (e^{a\ln t}\right ) \tag {2} \end {align}
Substituting (2) into (1) gives\begin {align} \frac {dI\relax (a) }{da} & =\int _{0}^{1}\frac {1}{\ln t}\left ( \ln \relax (t) \left (e^{a\ln t}\right ) \right ) dt\nonumber \\ & =\int _{0}^{1}e^{a\ln t}dt\nonumber \\ & =\int _{0}^{1}t^{a}dt\nonumber \\ & =\left . \frac {t^{a+1}}{a+1}\right \vert _{0}^{1}\nonumber \\ & =\frac {1}{1+a}\qquad a\neq -1\tag {3} \end {align}
Integrating the above is used to \(I\relax (a) \) gives\begin {align*} I\relax (a) & =\int _{0}^{a}\frac {1}{1+\tau }d\tau \\ & =\left . \ln \left (1+\tau \right ) \right \vert _{0}^{a}\\ & =\ln \left (1+a\right ) -\ln \relax (1) \\ & =\ln \left (1+a\right ) \qquad a\neq -1 \end {align*}
When \(a=1\) the above becomes\begin {align*} I\relax (1) & =\int _{0}^{1}\frac {t-1}{\ln t}dt\\ & =\ln \left (1+1\right ) \\ & =\ln \relax (2) \end {align*}
Hence\[ \int _{0}^{1}\frac {t-1}{\ln t}dt=\ln \relax (2) \]
Solution
\[ I=\int _{0}^{\infty }e^{-ax}\sin kxdx \] Taking derivative w.r.t \(a\) gives\begin {align*} \frac {dI}{da} & =\frac {d}{da}\left (\int _{0}^{\infty }e^{-ax}\sin kxdx\right ) \\ & =\int _{0}^{\infty }\frac {d}{da}\left (e^{-ax}\sin kx\right ) dx\\ & =\int _{0}^{\infty }-xe^{-ax}\sin kxdx\\ & =-\int _{0}^{\infty }xe^{-ax}\sin kxdx \end {align*}
Which is the integral the problem is asking to find. Therefore, since \(I\) is also given as \(\frac {k}{a^{2}+k^{2}}\) then\begin {align*} -\int _{0}^{\infty }xe^{-ax}\sin kxdx & =\frac {d}{da}\left (\frac {k}{a^{2}+k^{2}}\right ) \\ & =k\frac {d}{da}\left (\frac {1}{a^{2}+k^{2}}\right ) \\ & =k\left (-1\right ) \left (a^{2}+k^{2}\right ) ^{-2}\left (2a\right ) \\ & =-\frac {2ak}{\left (a^{2}+k^{2}\right ) ^{2}} \end {align*}
Therefore\[ \int _{0}^{\infty }xe^{-ax}\sin kxdx=\frac {2ak}{\left (a^{2}+k^{2}\right ) ^{2}}\]
\[ I=\int _{0}^{\infty }e^{-ax}\sin kxdx \] Taking derivative w.r.t. \(k\) gives\begin {align*} \frac {dI}{dk} & =\frac {d}{dk}\left (\int _{0}^{\infty }e^{-ax}\sin kxdx\right ) \\ & =\int _{0}^{\infty }\frac {d}{dk}\left (e^{-ax}\sin kx\right ) dx\\ & =\int _{0}^{\infty }e^{-ax}\frac {d}{dk}\left (\sin kx\right ) dx\\ & =\int _{0}^{\infty }xe^{-ax}\cos kxdx \end {align*}
Which is the integral the problem is asking to find. Therefore, since \(I\) is also given as \(\frac {k}{a^{2}+k^{2}}\) then\begin {align*} \int _{0}^{\infty }xe^{-ax}\cos kxdx & =\frac {d}{dk}\left (\frac {k}{a^{2}+k^{2}}\right ) \\ & =\frac {\left (a^{2}+k^{2}\right ) -k\left (2k\right ) }{\left ( a^{2}+k^{2}\right ) ^{2}}\\ & =\frac {a^{2}+k^{2}-2k^{2}}{\left (a^{2}+k^{2}\right ) ^{2}}\\ & =\frac {a^{2}-k^{2}}{\left (a^{2}+k^{2}\right ) ^{2}} \end {align*}
Hence\[ \int _{0}^{\infty }xe^{-ax}\cos kxdx=\frac {a^{2}-k^{2}}{\left (a^{2}+k^{2}\right ) ^{2}}\]
Solution
\[ P\left (x,\beta \right ) =Ae^{-\alpha x^{2}+\beta x^{3}}\] Expanding around \(\beta =0\) by fixing \(x\), gives\begin {equation} P\left (x,\beta \right ) =P\left (x,0\right ) +\beta \left . \frac {\partial P}{\partial \beta }\right \vert _{\beta =0}+\frac {\beta ^{2}}{2!}\left . \frac {\partial ^{2}P}{\partial \beta ^{2}}\right \vert _{\beta =0}+\cdots \tag {1} \end {equation} But \begin {equation} P\left (x,0\right ) =Ae^{-\alpha x^{2}}\tag {2} \end {equation} And\begin {equation} \frac {\partial P}{\partial \beta }=Ax^{3}e^{-\alpha x^{2}+\beta x^{3}}\tag {3} \end {equation} No need to take more derivatives since the problem is asking for first order of \(\beta \). Substituting (2,3) into (1) gives\begin {align} P\left (x,\beta \right ) & =Ae^{-\alpha x^{2}}+\beta \left . Ax^{3}e^{-\alpha x^{2}+\beta x^{3}}\right \vert _{\beta =0}+\cdots \nonumber \\ & =Ae^{-\alpha x^{2}}+\beta Ax^{3}e^{-\alpha x^{2}}+\cdots \tag {4} \end {align}
Using the above in the definition \(\int _{-\infty }^{\infty }P\relax (x) dx=1\) gives\begin {align} \int _{-\infty }^{\infty }\left (Ae^{-\alpha x^{2}}+\beta Ax^{3}e^{-\alpha x^{2}}\right ) dx & =1\nonumber \\ A\left (\int _{-\infty }^{\infty }e^{-\alpha x^{2}}dx+\beta \int _{-\infty }^{\infty }x^{3}e^{-\alpha x^{2}}dx\right ) & =1\tag {5} \end {align}
But \[ \int _{-\infty }^{\infty }x^{3}e^{-\alpha x^{2}}dx=0 \] This is because \(e^{-\alpha x^{2}}\) is an even function over \(\left ( -\infty ,+\infty \right ) \) and \(x^{3}\) is odd. Eq (5) now simplifies to\[ A\int _{-\infty }^{\infty }e^{-\alpha x^{2}}dx=1 \] But \(\int _{-\infty }^{\infty }e^{-\alpha x^{2}}dx=\sqrt {\frac {\pi }{\alpha }}\) (\(\alpha >0\)) because it is standard Gaussian integral. The above now becomes\begin {align*} A\sqrt {\frac {\pi }{\alpha }} & =1\\ A & =\sqrt {\frac {\alpha }{\pi }}\qquad \alpha >0 \end {align*}
\[ \bar {x}=\int _{-\infty }^{\infty }xP\relax (x) dx \] Using Eq. (4) from part (a), the above becomes\begin {align*} \bar {x} & =\int _{-\infty }^{\infty }x\left (Ae^{-\alpha x^{2}}+\beta Ax^{3}e^{-\alpha x^{2}}\right ) dx\\ & =A\int _{-\infty }^{\infty }xe^{-\alpha x^{2}}dx+A\int _{-\infty }^{\infty }\beta x^{4}e^{-\alpha x^{2}}dx \end {align*}
But \(\int _{-\infty }^{\infty }xe^{-\alpha x^{2}}dx=0\) since \(e^{-\alpha x^{2}}\) is an even function over \(\left (-\infty ,+\infty \right ) \) and \(x\) is an odd function. The above simplifies to\begin {equation} \bar {x}=A\beta \int _{-\infty }^{\infty }x^{4}e^{-\alpha x^{2}}dx\tag {6} \end {equation} To evaluate the above, starting from the standard Gaussian integral given by \[ I\left (\alpha \right ) =\int _{-\infty }^{\infty }e^{-\alpha x^{2}}dx=\sqrt {\frac {\pi }{\alpha }}\] Taking derivative w.r.t \(\alpha \) of both sides of the above results in\begin {align*} I^{\prime }\left (\alpha \right ) & =\int _{-\infty }^{\infty }\frac {d}{d\alpha }e^{-\alpha x^{2}}dx=\frac {d}{d\alpha }\sqrt {\frac {\pi }{\alpha }}\\ & =\int _{-\infty }^{\infty }-x^{2}e^{-\alpha x^{2}}dx=\sqrt {\pi }\left ( -\frac {1}{2}\right ) \alpha ^{-\frac {3}{2}}\\ & =\int _{-\infty }^{\infty }x^{2}e^{-\alpha x^{2}}dx=\frac {\sqrt {\pi }}{2}\alpha ^{-\frac {3}{2}} \end {align*}
Taking one more derivative w.r.t \(\alpha \) gives\begin {align*} I^{\prime \prime }\left (\alpha \right ) & =\int _{-\infty }^{\infty }\frac {d}{d\alpha }x^{2}e^{-\alpha x^{2}}dx=\frac {d}{d\alpha }\left (\frac {\sqrt {\pi }}{2}\alpha ^{-\frac {3}{2}}\right ) \\ & =\int _{-\infty }^{\infty }-x^{4}e^{-\alpha x^{2}}dx=\frac {\sqrt {\pi }}{2}\left (-\frac {3}{2}\alpha ^{-\frac {5}{2}}\right ) \\ & =\int _{-\infty }^{\infty }x^{4}e^{-\alpha x^{2}}dx=\frac {\sqrt {\pi }}{2}\left (\frac {3}{2}\alpha ^{-\frac {5}{2}}\right ) \end {align*}
Now the integrand is the one we want. This shows that\[ \int _{-\infty }^{\infty }x^{4}e^{-\alpha x^{2}}dx=\frac {3\sqrt {\pi }}{4\alpha ^{\frac {5}{2}}}\] Using the above result in (6) gives\[ \bar {x}=A\beta \left (\frac {3\sqrt {\pi }}{4\alpha ^{\frac {5}{2}}}\right ) \] But \(A=\sqrt {\frac {\alpha }{\pi }}\) from part(a). Hence the above becomes\begin {align*} \bar {x} & =\sqrt {\frac {\alpha }{\pi }}\beta \left (\frac {3\sqrt {\pi }}{4\alpha ^{\frac {5}{2}}}\right ) \\ & =\alpha ^{\frac {1}{2}}\beta \frac {3}{4\alpha ^{\frac {5}{2}}}\\ & =\beta \frac {3}{4\alpha ^{\frac {5}{2}-\frac {1}{2}}}\\ & =\frac {3}{4}\frac {\beta }{\alpha ^{2}}\qquad \alpha >0 \end {align*}
Solution
\[ dN=\left (\frac {4\pi V}{h^{3}c^{3}}\right ) \frac {E^{2}}{1+e^{\frac {E}{kT}}}dE \] The total energy is therefore\[ E_{total}=\ \int EdN \] Hence the energy density \(\rho \) is\begin {align} \rho & =\frac {1}{V}\int EdN\nonumber \\ & =\frac {1}{V}\int _{0}^{\infty }\left (\frac {4\pi V}{h^{3}c^{3}}\right ) \frac {EE^{2}}{1+e^{\frac {E}{kT}}}dE\nonumber \\ & =\left (\frac {1}{V}\right ) \left (\frac {4\pi V}{h^{3}c^{3}}\right ) \int _{0}^{\infty }\frac {E^{3}}{1+e^{\frac {E}{kT}}}dE\nonumber \\ & =\frac {4\pi }{h^{3}c^{3}}\int _{0}^{\infty }\frac {E^{3}}{1+e^{\frac {E}{kT}}}dE\tag {1} \end {align}
\(k\) (Boltzmann constant) has units of \(\frac {\left [ J\right ] }{\left [ K\right ] }\) where \(J\) is joule and \(K\) is temperature in Kelvin. Hence units of \(\frac {E}{kT}\) is \(\frac {\left [ J\right ] }{\frac {\left [ J\right ] }{\left [ K\right ] }\left [ K\right ] }\) which is dimensionless. Let \[ x=\frac {E}{kT}\] Therefore \(\frac {dx}{dE}=\frac {1}{kT}\,\) When \(E=0,x=0\) and when \(E=\infty ,x=\infty \). Substituting this into the integral in (1) gives\begin {align} \int _{0}^{\infty }\frac {E^{3}}{1+e^{\frac {E}{kT}}}dE & =\int _{0}^{\infty }\frac {\left (xkT\right ) ^{3}}{1+e^{x}}\left (kTdx\right ) \nonumber \\ & =\left (kT\right ) ^{4}\int _{0}^{\infty }\frac {x^{3}}{1+e^{x}}dx\tag {2} \end {align}
Substituting (2) into (1) gives\begin {equation} \rho =\left (\frac {4\pi }{h^{3}c^{3}}\right ) \left (kT\right ) ^{4}\int _{0}^{\infty }\frac {x^{3}}{1+e^{x}}dx\tag {3} \end {equation} Units of \(c\) (speed of light) is \(\frac {\left [ L\right ] }{\left [ T\right ] }\) where \(\left [ L\right ] \) is length in meters and \(\left [ T\right ] \) is time in seconds. Units for Planck constant \(h\) is \(\left [ J\right ] \left [ T\right ] \) (Joule-second). Therefore the factor \(\left ( \frac {4\pi }{h^{3}c^{3}}\right ) \left (kT\right ) ^{4}\) above in (3) in front of the integral has units\begin {align*} \left (\frac {4\pi }{h^{3}c^{3}}\right ) \left (kT\right ) ^{4} & =\frac {1}{\left (\left [ J\right ] \left [ T\right ] \right ) ^{3}\left ( \frac {\left [ L\right ] }{\left [ T\right ] }\right ) ^{3}}\left ( \frac {\left [ J\right ] }{\left [ K\right ] }\left [ K\right ] \right ) ^{4}\\ & =\frac {1}{\left [ J\right ] ^{3}\left [ L\right ] ^{3}}\left (\left [ J\right ] \right ) ^{4}\\ & =\frac {\left [ J\right ] }{\left [ L\right ] ^{3}} \end {align*}
Which has the correct units of energy density. Let this factor be called \(\Phi =\left (\frac {4\pi }{h^{3}c^{3}}\right ) \left (kT\right ) ^{4}\). Then (3) can be written as\[ \rho =\Phi \int _{0}^{\infty }\frac {x^{3}}{1+e^{x}}dx \]
The dimensionless integral found in part (a) is\begin {equation} I=\int _{0}^{\infty }\frac {x^{3}}{e^{x}+1}dx\tag {1} \end {equation} But \[ \frac {1}{e^{x}+1}=\frac {1}{e^{x}-1}-2\frac {1}{e^{2x}-1}\] We did the above, to make the denominator has the form \(e^{x}-1\), which is easier to work with following the lecture notes than working with \(e^{x}+1\). Eq (1) now becomes\begin {equation} I=\int _{0}^{\infty }\frac {x^{3}}{e^{x}-1}dx-2\int _{0}^{\infty }\frac {x^{3}}{e^{2x}-1}dx\tag {2} \end {equation} The first integral has the standard form \(\int _{0}^{\infty }\frac {x^{n}}{e^{x}-1}dx\). Hence\[ \int _{0}^{\infty }\frac {x^{3}}{e^{x}-1}=\left (3!\right ) \xi \relax (4) \] (Derivations of the above is given at the end of this problem). Now we evaluate on the second integral in (2). Let \(y=2x\), then \(\frac {dy}{dx}=2\). The limits do not change. The integral becomes\[ \int _{0}^{\infty }\frac {\frac {y^{3}}{8}}{e^{y}-1}\frac {dy}{2}=\frac {1}{16}\int _{0}^{\infty }\frac {y^{3}}{e^{y}-1}dy \] We see that \(\int _{0}^{\infty }\frac {y^{3}}{e^{y}-1}dy\) now has the same form as the first integral. Hence \(\int _{0}^{\infty }\frac {y^{3}}{e^{y}-1}dy=\left ( 3!\right ) \xi \relax (4) \). Putting these two results back into (2) gives the final result\begin {align*} I & =\left (3!\right ) \xi \relax (4) -2\left (\frac {1}{16}\left ( 3!\right ) \xi \relax (4) \right ) \\ & =\left (3!\right ) \xi \relax (4) \left (1-2\left (\frac {1}{16}\right ) \right ) \\ & =\relax (6) \xi \relax (4) \left (1-\frac {1}{8}\right ) \\ & =\relax (6) \xi \relax (4) \frac {7}{8}\\ & =\frac {21}{4}\xi \relax (4) \end {align*}
But from class handout, \(\xi \relax (4) =\frac {\pi ^{4}}{90}\). Hence\begin {align*} \int _{0}^{\infty }\frac {x^{3}}{e^{x}+1}dx & =\frac {21}{4}\left (\frac {\pi ^{4}}{90}\right ) \\ & =\frac {7}{4}\left (\frac {\pi ^{4}}{30}\right ) \\ & =\frac {7}{120}\pi ^{4}\\ & \approx 5.6822 \end {align*}
Using this in the result obtained in part (a) gives the energy density as\begin {align*} \rho & =\Phi \int _{0}^{\infty }\frac {x^{3}}{1+e^{x}}dx\\ & =\left (\frac {7\pi ^{4}}{120}\right ) \left (\frac {4\pi }{h^{3}c^{3}}\right ) \left (kT\right ) ^{4} \end {align*}
Derivation of the integral
In the above, we used the result that \(\int _{0}^{\infty }\frac {x^{n}}{e^{x}-1}dx=\left (n!\right ) \xi \left (n+1\right ) \). For \(n=3\) this becomes \(\left (3!\right ) \xi \relax (4) \).
To show how this came above, we start by multiplying the numerator and denominator of the integrand by \(e^{-x}\). This gives\begin {equation} \int _{0}^{\infty }\frac {x^{n}e^{-x}}{1-e^{-x}}dx\tag {3} \end {equation} Let \(y=e^{-x}\) then\begin {align*} \frac {e^{-x}}{1-e^{-x}} & =\frac {y}{1-y}\\ & =y\left (1+y+y^{2}+y^{3}+\cdots \right ) \\ & =y+y^{2}+y^{3}+\cdots \\ & =\sum _{k=1}^{\infty }y^{k}\\ & =\sum _{k=1}^{\infty }e^{-kx} \end {align*}
Using the above sum in Eq (3) gives\begin {align*} \int _{0}^{\infty }\frac {x^{n}e^{-x}}{1-e^{-x}}dx & =\int _{0}^{\infty }x^{n}\sum _{k=1}^{\infty }e^{-kx}dx\\ & =\sum _{k=1}^{\infty }\int _{0}^{\infty }x^{n}e^{-kx}dx \end {align*}
Let \(z=kx\). Then \(\frac {dz}{dx}=k\). When \(x=0,z=0\) and when \(x=\infty ,z=\infty \). The above becomes\begin {align*} \int _{0}^{\infty }\frac {x^{n}e^{-x}}{1-e^{-x}}dx & =\sum _{k=1}^{\infty }\int _{0}^{\infty }\left (\frac {z}{k}\right ) ^{n}e^{-z}\left (\frac {dz}{k}\right ) \\ & =\sum _{k=1}^{\infty }\frac {1}{k^{n+1}}\int _{0}^{\infty }z^{n}e^{-z}dz\\ & =\sum _{k=1}^{\infty }\frac {1}{k^{n+1}}\left (\int _{0}^{\infty }x^{n}e^{-x}dx\right ) \end {align*}
But \(\int _{0}^{\infty }x^{n}e^{-x}dx=n!\), which can be shown by integration by parts repeatedly \(n\) times. The above integral now becomes\[ \int _{0}^{\infty }\frac {x^{n}e^{-x}}{1-e^{-x}}dx=\left (n!\right ) \sum _{k=1}^{\infty }\frac {1}{k^{n+1}}\] The sum \(\sum _{k=1}^{\infty }\frac {1}{k^{n+1}}\) is called the Zeta function \(\zeta \left (n+1\right ) \). When \(n=3\) the above result becomes\begin {align*} \int _{0}^{\infty }\frac {x^{3}}{e^{x}-1}dx & =\left (3!\right ) \sum _{k=1}^{\infty }\frac {1}{k^{4}}\\ & =\left (3!\right ) \zeta \relax (4) \end {align*}
Which is the result used earlier.