4.2 HW 2

4.2.1 Problems listing

PDF

PDF (letter size)

PDF (legal size)

4.2.2 Problem 2.2.3

Evaluate ${\int }_0^1{e}^{\sqrt{x}}dx$. Show that ${\int }_0^{\mathrm{\infty }}{e}^{-x^4}dx=\mathrm{\Gamma }\left(\frac{5}{4}\right)$

Solution

Let $y=\sqrt{x}$. Therefore$$\begin{array}{rl}\frac{dy}{dx}& =\frac{1}{2}\frac{1}{\sqrt{x}}\\ & =\frac{1}{2}\frac{1}{y}\end{array}$$

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 ${\int }_0^1{e}^y\left(2ydy\right)=2{\int }_0^{1}y{e}^{y}dy$. This integral is evaluated using integration by parts. $$udv={uv|}_0^1-{\int }_0^{1}vdu$$ Let $u=y$ and $dv=e^y$, then $du=dy$ and $v=e^y$. The above becomes$$\begin{array}{rl}2\left({\int }_0^{1}y{e}^{y}dy\right)& =2({uv|}_0^1-{\int }_0^{1}vdu)\\ & =2({y{e}^y|}_0^1-{\int }_0^1{e}^{y}dy)\\ & =2((e^1-0)-{e^y|}_0^1)\\ & =2(e-(e-1))\\ & =2(e-e+1)\\ & =2\end{array}$$

Hence $${\int }_0^1{e}^{\sqrt{x}}dx=2$$ For the second part of the question asking to evaluate ${\int }_0^{\mathrm{\infty }}{e}^{-x^4}dx$, let$$x=y^{\frac{1}{4}}$$ Then $$\frac{dx}{dy}=\frac{1}{4}{y}^{(\frac{1}{4}-1)}$$ When $x=0,y=0$ and when $x=\mathrm{\infty },y=\mathrm{\infty }$. Hence the above integral becomes$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}{e}^{-x^4}dx& ={\int }_0^{\mathrm{\infty }}{e}^{-y}\left(\frac{1}{4}{y}^{(\frac{1}{4}-1)}dy\right)\\ \text{(1)}& & =\frac{1}{4}{\int }_0^{\mathrm{\infty }}{y}^{(\frac{1}{4}-1)}{e}^{-y}dy\end{array}$$

Comparing the above to integral (2.1.39) in the book which says$$\begin{array}{}\text{(2)}& F(n)& ={\int }_0^{\mathrm{\infty }}{y}^n{e}^{-y}dy\text{(3)}& \mathrm{\Gamma }(n)& =F(n-1)\end{array}$$

Then putting $n=\frac{1}{4}$ in (3) gives$$\begin{array}{rl}\mathrm{\Gamma }\left(\frac{1}{4}\right)& =F(\frac{1}{4}-1)\\ & ={\int }_0^{\mathrm{\infty }}{y}^{(\frac{1}{4}-1)}{e}^{-y}dy\end{array}$$

Which is (1). This means that$${\int }_0^{\mathrm{\infty }}{y}^{(\frac{1}{4}-1)}{e}^{-y}dy=\mathrm{\Gamma }\left(\frac{1}{4}\right)$$ Hence $$\begin{array}{}\text{(4)}& \frac{1}{4}{\int }_0^{\mathrm{\infty }}{y}^{(\frac{1}{4}-1)}{e}^{-y}dy=\frac{1}{4}\mathrm{\Gamma }\left(\frac{1}{4}\right)\end{array}$$ To obtain the final form, the following property of Gamma functions is used$$\mathrm{\Gamma }(n+1)=n\mathrm{\Gamma }(n)$$ Which means that when $n=\frac{1}{4}$, the above becomes$$\begin{array}{rl}\mathrm{\Gamma }(\frac{1}{4}+1)& =\frac{1}{4}\mathrm{\Gamma }\left(\frac{1}{4}\right)\\ \mathrm{\Gamma }\left(\frac{5}{4}\right)& =\frac{1}{4}\mathrm{\Gamma }\left(\frac{1}{4}\right)\end{array}$$

Using this in (4) shows that$$\frac{1}{4}{\int }_0^{\mathrm{\infty }}{y}^{(\frac{1}{4}-1)}{e}^{-y}dy=\mathrm{\Gamma }\left(\frac{5}{4}\right)$$ Which implies$${\int }_0^{\mathrm{\infty }}{e}^{-x^4}dx=\mathrm{\Gamma }\left(\frac{5}{4}\right)$$ Which is what we are asked to show.

4.2.3 Problem 2.2.10 (or part a of problem 2)

Figure 4.3:Problem statment

Solution

Let $$I(a)={\int }_0^1\frac{t^a-1}{\ln t}dt$$ Where $a=1$ for the specific integral in this problem. The above is the parametrized general form. Taking derivative w.r.t $a$ gives $$\begin{array}{rl}\frac{dI(a)}{da}& =\frac{d}{da}\left({\int }_0^1\frac{t^a-1}{\ln t}dt\right)\\ & ={\int }_0^1\frac{d}{da}\left(\frac{t^a-1}{\ln t}\right)dt\\ \text{(1)}& & ={\int }_0^1\frac{1}{\ln t}\frac{d}{da}(t^a-1)dt\end{array}$$

But $$\begin{array}{rl}\frac{d}{da}(t^a-1)& =\frac{d}{da}(e^{a\ln t}-1)\\ \text{(2)}& & =\ln (t)\left(e^{a\ln t}\right)\end{array}$$

Substituting (2) into (1) gives$$\begin{array}{rl}\frac{dI(a)}{da}& ={\int }_0^1\frac{1}{\ln t}(\ln (t)\left(e^{a\ln t}\right))dt\\ & ={\int }_0^1{e}^{a\ln t}dt\\ & ={\int }_0^1{t}^{a}dt\\ & ={\frac{t^{a+1}}{a+1}|}_0^1\\ \text{(3)}& & =\frac{1}{1+a}a\ne -1\end{array}$$

Integrating the above is used to $I(a)$ gives$$\begin{array}{rl}I(a)& ={\int }_0^a\frac{1}{1+\tau }d\tau \\ & ={\ln (1+\tau )|}_0^a\\ & =\ln (1+a)-\ln (1)\\ & =\ln (1+a)a\ne -1\end{array}$$

When $a=1$ the above becomes$$\begin{array}{rl}I(1)& ={\int }_0^1\frac{t-1}{\ln t}dt\\ & =\ln (1+1)\\ & =\ln (2)\end{array}$$

Hence$${\int }_0^1\frac{t-1}{\ln t}dt=\ln (2)$$

4.2.4 Problem 2.2.11 (or part b of problem 2)

Figure 4.4:Problem statment

Solution

4.2.4.1 part (1)

$$I={\int }_0^{\mathrm{\infty }}{e}^{-ax}\sin kxdx$$ Taking derivative w.r.t $a$ gives$$\begin{array}{rl}\frac{dI}{da}& =\frac{d}{da}({\int }_0^{\mathrm{\infty }}{e}^{-ax}\sin kxdx)\\ & ={\int }_0^{\mathrm{\infty }}\frac{d}{da}(e^{-ax}\sin kx)dx\\ & ={\int }_0^{\mathrm{\infty }}-x{e}^{-ax}\sin kxdx\\ & =-{\int }_0^{\mathrm{\infty }}x{e}^{-ax}\sin kxdx\end{array}$$

Which is the integral the problem is asking to find. Therefore, since $I$ is also given as $\frac{k}{a^2+k^2}$ then$$\begin{array}{rl}-{\int }_0^{\mathrm{\infty }}x{e}^{-ax}\sin kxdx& =\frac{d}{da}\left(\frac{k}{a^2+k^2}\right)\\ & =k\frac{d}{da}\left(\frac{1}{a^2+k^2}\right)\\ & =k(-1){(a^2+k^2)}^{-2}\left(2a\right)\\ & =-\frac{2ak}{{(a^2+k^2)}^2}\end{array}$$

Therefore$${\int }_0^{\mathrm{\infty }}x{e}^{-ax}\sin kxdx=\frac{2ak}{{(a^2+k^2)}^2}$$

4.2.4.2 part (2)

$$I={\int }_0^{\mathrm{\infty }}{e}^{-ax}\sin kxdx$$ Taking derivative w.r.t. $k$ gives$$\begin{array}{rl}\frac{dI}{dk}& =\frac{d}{dk}({\int }_0^{\mathrm{\infty }}{e}^{-ax}\sin kxdx)\\ & ={\int }_0^{\mathrm{\infty }}\frac{d}{dk}(e^{-ax}\sin kx)dx\\ & ={\int }_0^{\mathrm{\infty }}{e}^{-ax}\frac{d}{dk}(\sin kx)dx\\ & ={\int }_0^{\mathrm{\infty }}x{e}^{-ax}\cos kxdx\end{array}$$

Which is the integral the problem is asking to find. Therefore, since $I$ is also given as $\frac{k}{a^2+k^2}$ then$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}x{e}^{-ax}\cos kxdx& =\frac{d}{dk}\left(\frac{k}{a^2+k^2}\right)\\ & =\frac{(a^2+k^2)-k\left(2k\right)}{{(a^2+k^2)}^2}\\ & =\frac{a^2+k^2-2k^2}{{(a^2+k^2)}^2}\\ & =\frac{a^2-k^2}{{(a^2+k^2)}^2}\end{array}$$

Hence$${\int }_0^{\mathrm{\infty }}x{e}^{-ax}\cos kxdx=\frac{a^2-k^2}{{(a^2+k^2)}^2}$$

4.2.5 Problem 3

Figure 4.5:Problem statment

Solution

4.2.5.1 Part (a)

$$P(x,\beta )=A{e}^{-\alpha x^2+\beta x^3}$$ Expanding around $\beta =0$ by fixing $x$, gives$$\begin{array}{}\text{(1)}& P(x,\beta )=P(x,0)+\beta {\frac{\partial P}{\partial \beta }|}_{\beta =0}+\frac{{\beta }^2}{2!}{\frac{{\partial }^{2}P}{\partial {\beta }^2}|}_{\beta =0}+\cdots \end{array}$$ But $$\begin{array}{}\text{(2)}& P(x,0)=A{e}^{-\alpha x^2}\end{array}$$ And$$\begin{array}{}\text{(3)}& \frac{\partial P}{\partial \beta }=A{x}^3{e}^{-\alpha x^2+\beta x^3}\end{array}$$ No need to take more derivatives since the problem is asking for first order of $\beta$. Substituting (2,3) into (1) gives$$\begin{array}{rl}P(x,\beta )& =A{e}^{-\alpha x^2}+\beta {A{x}^3{e}^{-\alpha x^2+\beta x^3}|}_{\beta =0}+\cdots \\ \text{(4)}& & =A{e}^{-\alpha x^2}+\beta A{x}^3{e}^{-\alpha x^2}+\cdots \end{array}$$

Using the above in the definition ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}P(x)dx=1$ gives$$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(A{e}^{-\alpha x^2}+\beta A{x}^3{e}^{-\alpha x^2})dx& =1\\ \text{(5)}& A({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\alpha x^2}dx+\beta {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^3{e}^{-\alpha x^2}dx)& =1\end{array}$$

But $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^3{e}^{-\alpha x^2}dx=0$$ This is because $e^{-\alpha x^2}$ is an even function over $(-\mathrm{\infty },+\mathrm{\infty })$ and $x^3$ is odd. Eq (5) now simplifies to$$A{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\alpha x^2}dx=1$$ But ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\alpha x^2}dx=\sqrt{\frac{\pi }{\alpha }}$ ($\alpha >0$) because it is standard Gaussian integral. The above now becomes$$\begin{array}{rl}A\sqrt{\frac{\pi }{\alpha }}& =1\\ A& =\sqrt{\frac{\alpha }{\pi }}\alpha >0\end{array}$$

4.2.5.2 Part b

$$\overline{x}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xP(x)dx$$ Using Eq. (4) from part (a), the above becomes$$\begin{array}{rl}\overline{x}& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(A{e}^{-\alpha x^2}+\beta A{x}^3{e}^{-\alpha x^2})dx\\ & =A{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x{e}^{-\alpha x^2}dx+A{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\beta x^4{e}^{-\alpha x^2}dx\end{array}$$

But ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x{e}^{-\alpha x^2}dx=0$ since $e^{-\alpha x^2}$ is an even function over $(-\mathrm{\infty },+\mathrm{\infty })$ and $x$ is an odd function. The above simplifies to$$\begin{array}{}\text{(6)}& \overline{x}=A\beta {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^4{e}^{-\alpha x^2}dx\end{array}$$ To evaluate the above, starting from the standard Gaussian integral given by $$I\left(\alpha \right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\alpha x^2}dx=\sqrt{\frac{\pi }{\alpha }}$$ Taking derivative w.r.t $\alpha$ of both sides of the above results in$$\begin{array}{rl}{I}^{\prime }\left(\alpha \right)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{d}{d\alpha }{e}^{-\alpha x^2}dx=\frac{d}{d\alpha }\sqrt{\frac{\pi }{\alpha }}\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}-x^2{e}^{-\alpha x^2}dx=\sqrt{\pi }(-\frac{1}{2}){\alpha }^{-\frac{3}{2}}\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^2{e}^{-\alpha x^2}dx=\frac{\sqrt{\pi }}{2}{\alpha }^{-\frac{3}{2}}\end{array}$$

Taking one more derivative w.r.t $\alpha$ gives$$\begin{array}{rl}{I}^{\prime \prime }\left(\alpha \right)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{d}{d\alpha }{x}^2{e}^{-\alpha x^2}dx=\frac{d}{d\alpha }\left(\frac{\sqrt{\pi }}{2}{\alpha }^{-\frac{3}{2}}\right)\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}-x^4{e}^{-\alpha x^2}dx=\frac{\sqrt{\pi }}{2}(-\frac{3}{2}{\alpha }^{-\frac{5}{2}})\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^4{e}^{-\alpha x^2}dx=\frac{\sqrt{\pi }}{2}\left(\frac{3}{2}{\alpha }^{-\frac{5}{2}}\right)\end{array}$$

Now the integrand is the one we want. This shows that$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^4{e}^{-\alpha x^2}dx=\frac{3\sqrt{\pi }}{4{\alpha }^{\frac{5}{2}}}$$ Using the above result in (6) gives$$\overline{x}=A\beta \left(\frac{3\sqrt{\pi }}{4{\alpha }^{\frac{5}{2}}}\right)$$ But $A=\sqrt{\frac{\alpha }{\pi }}$ from part(a). Hence the above becomes$$\begin{array}{rl}\overline{x}& =\sqrt{\frac{\alpha }{\pi }}\beta \left(\frac{3\sqrt{\pi }}{4{\alpha }^{\frac{5}{2}}}\right)\\ & ={\alpha }^{\frac{1}{2}}\beta \frac{3}{4{\alpha }^{\frac{5}{2}}}\\ & =\beta \frac{3}{4{\alpha }^{\frac{5}{2}-\frac{1}{2}}}\\ & =\frac{3}{4}\frac{\beta }{{\alpha }^2}\alpha >0\end{array}$$

4.2.6 Problem 4

Figure 4.6:Problem statment

Solution

4.2.6.1 Part a

$$dN=\left(\frac{4\pi V}{h^3{c}^3}\right)\frac{E^2}{1+e^{\frac{E}{kT}}}dE$$ The total energy is therefore$$E_{total}=\int EdN$$ Hence the energy density $\rho$ is$$\begin{array}{rl}\rho & =\frac{1}{V}\int EdN\\ & =\frac{1}{V}{\int }_0^{\mathrm{\infty }}\left(\frac{4\pi V}{h^3{c}^3}\right)\frac{E{E}^2}{1+e^{\frac{E}{kT}}}dE\\ & =\left(\frac{1}{V}\right)\left(\frac{4\pi V}{h^3{c}^3}\right){\int }_0^{\mathrm{\infty }}\frac{E^3}{1+e^{\frac{E}{kT}}}dE\\ \text{(1)}& & =\frac{4\pi }{h^3{c}^3}{\int }_0^{\mathrm{\infty }}\frac{E^3}{1+e^{\frac{E}{kT}}}dE\end{array}$$

$k$ (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 $E=\mathrm{\infty },x=\mathrm{\infty }$. Substituting this into the integral in (1) gives$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}\frac{E^3}{1+e^{\frac{E}{kT}}}dE& ={\int }_0^{\mathrm{\infty }}\frac{{\left(xkT\right)}^3}{1+e^x}\left(kTdx\right)\\ \text{(2)}& & ={\left(kT\right)}^4{\int }_0^{\mathrm{\infty }}\frac{x^3}{1+e^x}dx\end{array}$$

Substituting (2) into (1) gives$$\begin{array}{}\text{(3)}& \rho =\left(\frac{4\pi }{h^3{c}^3}\right){\left(kT\right)}^4{\int }_0^{\mathrm{\infty }}\frac{x^3}{1+e^x}dx\end{array}$$ 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$$\begin{array}{rl}\left(\frac{4\pi }{h^3{c}^3}\right){\left(kT\right)}^4& =\frac{1}{{\left(\left[J\right]\left[T\right]\right)}^3{\left(\frac{\left[L\right]}{\left[T\right]}\right)}^3}{\left(\frac{\left[J\right]}{\left[K\right]}\left[K\right]\right)}^4\\ & =\frac{1}{{\left[J\right]}^3{\left[L\right]}^3}{\left(\left[J\right]\right)}^4\\ & =\frac{\left[J\right]}{{\left[L\right]}^3}\end{array}$$

Which has the correct units of energy density. Let this factor be called $\mathrm{\Phi }=\left(\frac{4\pi }{h^3{c}^3}\right){\left(kT\right)}^4$. Then (3) can be written as$$\rho =\mathrm{\Phi }{\int }_0^{\mathrm{\infty }}\frac{x^3}{1+e^x}dx$$

4.2.6.2 Part b

The dimensionless integral found in part (a) is$$\begin{array}{}\text{(1)}& I={\int }_0^{\mathrm{\infty }}\frac{x^3}{e^x+1}dx\end{array}$$ 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$$\begin{array}{}\text{(2)}& I={\int }_0^{\mathrm{\infty }}\frac{x^3}{e^x-1}dx-2{\int }_0^{\mathrm{\infty }}\frac{x^3}{e^{2x}-1}dx\end{array}$$ The first integral has the standard form ${\int }_0^{\mathrm{\infty }}\frac{x^n}{e^x-1}dx$. Hence$${\int }_0^{\mathrm{\infty }}\frac{x^3}{e^x-1}=(3!)\xi (4)$$ (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$${\int }_0^{\mathrm{\infty }}\frac{\frac{y^3}{8}}{e^y-1}\frac{dy}{2}=\frac{1}{16}{\int }_0^{\mathrm{\infty }}\frac{y^3}{e^y-1}dy$$ We see that ${\int }_0^{\mathrm{\infty }}\frac{y^3}{e^y-1}dy$ now has the same form as the first integral. Hence ${\int }_0^{\mathrm{\infty }}\frac{y^3}{e^y-1}dy=(3!)\xi (4)$. Putting these two results back into (2) gives the final result$$\begin{array}{rl}I& =(3!)\xi (4)-2(\frac{1}{16}(3!)\xi (4))\\ & =(3!)\xi (4)(1-2\left(\frac{1}{16}\right))\\ & =(6)\xi (4)(1-\frac{1}{8})\\ & =(6)\xi (4)\frac{7}{8}\\ & =\frac{21}{4}\xi (4)\end{array}$$

But from class handout, $\xi (4)=\frac{{\pi }^4}{90}$. Hence$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}\frac{x^3}{e^x+1}dx& =\frac{21}{4}\left(\frac{{\pi }^4}{90}\right)\\ & =\frac{7}{4}\left(\frac{{\pi }^4}{30}\right)\\ & =\frac{7}{120}{\pi }^4\\ & \approx 5.6822\end{array}$$

Using this in the result obtained in part (a) gives the energy density as$$\begin{array}{rl}\rho & =\mathrm{\Phi }{\int }_0^{\mathrm{\infty }}\frac{x^3}{1+e^x}dx\\ & =\left(\frac{7{\pi }^4}{120}\right)\left(\frac{4\pi }{h^3{c}^3}\right){\left(kT\right)}^4\end{array}$$

Derivation of the integral

In the above, we used the result that ${\int }_0^{\mathrm{\infty }}\frac{x^n}{e^x-1}dx=(n!)\xi (n+1)$. For $n=3$ this becomes $(3!)\xi (4)$.

To show how this came above, we start by multiplying the numerator and denominator of the integrand by $e^{-x}$. This gives$$\begin{array}{}\text{(3)}& {\int }_0^{\mathrm{\infty }}\frac{x^n{e}^{-x}}{1-e^{-x}}dx\end{array}$$ Let $y=e^{-x}$ then$$\begin{array}{rl}\frac{e^{-x}}{1-e^{-x}}& =\frac{y}{1-y}\\ & =y(1+y+y^2+y^3+\cdots )\\ & =y+y^2+y^3+\cdots \\ & =\sum _{k=1}^{\mathrm{\infty }}{y}^k\\ & =\sum _{k=1}^{\mathrm{\infty }}{e}^{-kx}\end{array}$$

Using the above sum in Eq (3) gives$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}\frac{x^n{e}^{-x}}{1-e^{-x}}dx& ={\int }_0^{\mathrm{\infty }}{x}^n\sum _{k=1}^{\mathrm{\infty }}{e}^{-kx}dx\\ & =\sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{x}^n{e}^{-kx}dx\end{array}$$

Let $z=kx$. Then $\frac{dz}{dx}=k$. When $x=0,z=0$ and when $x=\mathrm{\infty },z=\mathrm{\infty }$. The above becomes$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}\frac{x^n{e}^{-x}}{1-e^{-x}}dx& =\sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\left(\frac{z}{k}\right)}^n{e}^{-z}\left(\frac{dz}{k}\right)\\ & =\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{n+1}}{\int }_0^{\mathrm{\infty }}{z}^n{e}^{-z}dz\\ & =\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{n+1}}\left({\int }_0^{\mathrm{\infty }}{x}^n{e}^{-x}dx\right)\end{array}$$

But ${\int }_0^{\mathrm{\infty }}{x}^n{e}^{-x}dx=n!$, which can be shown by integration by parts repeatedly $n$ times. The above integral now becomes$${\int }_0^{\mathrm{\infty }}\frac{x^n{e}^{-x}}{1-e^{-x}}dx=(n!)\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{n+1}}$$ The sum $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{n+1}}$ is called the Zeta function $\zeta (n+1)$. When $n=3$ the above result becomes$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}\frac{x^3}{e^x-1}dx& =(3!)\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^4}\\ & =(3!)\zeta (4)\end{array}$$

Which is the result used earlier.

4.2.7 key solution for HW 2

PDF