A calculus problem by Sudipan Mallick

We begin by rearranging: $\int_0^\infty \dfrac{x^3}{e^x-1}~\text d x=\int_0^\infty x^3 \dfrac{1/e^x}{1-1/e^x}~\text d x.$ Now write $\dfrac{1/e^x}{1-1/e^x}=\sum_{n=1}^\infty \left(\dfrac{1}{e^x}\right)^n=\sum_{n=1}^\infty \dfrac{1}{e^{xn}}$ which follows directly from the infinite summation formula of geometric series, to get $\int_0^\infty x^3 \dfrac{1/e^x}{1-1/e^x}~\text d x=\int_0^\infty x^3 \sum_{n=1}^\infty \dfrac{1}{e^{xn}}~\text d x=\sum_{n=1}^\infty \int_0^\infty \dfrac{x^3}{e^{xn}}~\text d x.$ We substitute $z=xn$ so that $\text d z=n~\text d x$ . Putting these yields $\sum_{n=1}^\infty \int_0^\infty \dfrac{x^3}{e^{xn}}~\text d x=\sum_{n=1}^\infty \int_0^\infty \dfrac{z^3}{e^z}\dfrac{1}{n^4}~\text d z=\left(\sum_{n=1}^\infty \dfrac{1}{n^4}\right)\left(\int_0^\infty \dfrac{z^3}{e^z}~\text d z\right).$ By the definitions of Gamma and Zeta functions, it now easily follows that $\left(\sum_{n=1}^\infty \dfrac{1}{n^4}\right)\left(\int_0^\infty \dfrac{z^3}{e^z}~\text d z\right)=\zeta(4)~\Gamma(4)=\dfrac{\pi^4}{90}\times 3!=\boxed{\dfrac{\pi^4}{15}}$ which is approximately $6.5$ . Notice that $x^3$ can be replaced by $x^k$ and the same proof above yields $\zeta(k+1)~\Gamma(k+1)$ .

First we prove the following proposition:

Proposition :

$\int_0^1 x^a \ln^k x\ dx=\frac{(-1)^k\, k!}{(a+1)^{k+1}} \qquad\text{for }\ k=0,1,2,\ldots$

Proof :

Note that $\int_0^1 x^a\ dx=\frac1{a+1}\qquad\text{for }\ \alpha>-1.$ Differentiating equation above $k$ times w.r.t. $a$ we have $\int_0^1 \frac{\partial^k}{\partial a^k}\left(x^a\right)\ dx=\int_0^1 x^a \ln^k x\ dx=\frac{(-1)^k\, k!}{(a+1)^{k+1}}\qquad\text{Q.E.D.}$

Now, we will evaluate the general case of

$\int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx=\int_0^\infty\frac{x^{a-1}\,e^{-bx}}{1-e^{-bx}}\,dx\qquad\text{for }\ a,b>0$

Set $y=e^{-bx}$ , then

$\int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx=\frac{(-1)^{a-1}}{b^a}\int_0^1\frac{\ln^{a-1}y}{1-y}\,dy$

Use a geometric series for $\dfrac{1}{1-y}$ then interchange the integral and summation sign which is justified by the Fubini–Tonelli theorem and apply the previous proposition, we have

$\begin{aligned} \int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx&=\frac{(-1)^{a-1}}{b^a}\int_0^1\sum_{k=0}^\infty y^k\,\ln^{a-1}y\,dy\\ &=\frac{(-1)^{a-1}}{b^a}\sum_{k=0}^\infty\int_0^1 y^k\,\ln^{a-1}y\,dy\\ &=\frac{(-1)^{a-1}}{b^a}\sum_{k=0}^\infty \frac{(-1)^{a-1}\, (a-1)!}{(k+1)^{a}}\\ &=\frac{(a-1)!}{b^a}\sum_{k=1}^\infty \frac{1}{k^{a}}\\ &=\frac{\Gamma(a)\,\zeta(a)}{b^a} \end{aligned}$

where $\Gamma(a)$ is the gamma function and $\zeta(a)$ is the Riemann zeta function .

Hence, by setting $a=4$ and $b=1$ , we have

$\int_0^\infty\frac{x^{3}}{e^{x}-1}\,dx=\Gamma(4)\,\zeta(4)=3!\cdot\frac{\pi^4}{90}=\frac{\pi^4}{15}$

Well this is one of the definations of Riemann Zeta function

http://mathworld.wolfram.com/RiemannZetaFunction.html