Interesting integral - think out of the box

Evaluate $\int_0^{\infty} \frac{x^3}{e^x - 1} \; dx $

Answer:

$\frac{1}{1 - x} = 1 + x + x^2 + \cdot \cdot \cdot + x^n + \cdot \cdot \cdot = \sum_{k=0}^{\infty} x^k $, when $|x| \lt 1 $

$\frac{1}{e^x - 1}$

$= \frac{e^{-x}}{1 - e^{-x}}$

$= e^{-x} \sum_{k=0}^{\infty} (e^{-x})^k $, since $e^{-x} \lt 1 $ when $ x\gt 0 $

$= \sum_{k=1}^{\infty} (e^{-x})^k $

$= \sum_{k=1}^{\infty} e^{-kx} $

So,

$\int_0^{\infty} \frac{x^3}{e^x - 1} \; dx $

$= \int_0^{\infty} x^3 \sum_{k=1}^{\infty} e^{-kx}k $

$= \sum_{k=1}^{\infty} \int_0^{\infty} x^3 e^{-kx} $

$= \sum_{k=1}^{\infty} \left. \frac{e^{-kx}}{k^4} \left( k^3x^3 + 3k^2x^2 + 6kx + 6 \right) \right|_0^{\infty} $

$= \sum_{k=1}^{\infty} \frac{6}{k^4} $

$= 6 \sum_{k=1}^{\infty} \frac{1}{k^4} $

$= 6 \cdot \frac{\pi^4}{90} = \frac{\pi^4}{15} $

We used the fact that $\sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{\pi^4}{90}$, an advance result that follows easily from a theorem called Parseval's Theorem, which basically says that if the Fourier series of $f(x)$ is given by

$f(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty} a_k \cos (kx) + \sum_{k=1}^{\infty} b_k \sin (kx) $,

where $a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \; dx $

$a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos (kx) \; dx \;$ and

$b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(kx)\; dx $

then we have the Parseval's Identity

$ \frac{1}{\pi} \int_{-\pi}^{\pi} (f(x))^2 \; dx = \frac{a_0^2}{2} + \sum_{k=1}^{\infty} a_k^2 + b_k^2 $

Consider $f(x) = x^2$, then we have

$ \frac{1}{\pi} \int_{-\pi}^{\pi} x^4 \; dx $

$= \frac{1}{\pi} \left. \frac{x^5}{5} \right|_{-\pi}^{\pi} = \frac{2\pi^4}{5} $

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \; dx$

$= \frac{1}{\pi} \left. \frac{x^3}{3} \right|_{-\pi}^{\pi} = \frac{2\pi^2}{3} $


$a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos (nk) \; dx$

$= \frac{1}{\pi} \left. \frac{2 k x cos(k x)+(-2+k^2 x^2) sin(k x)}{k^3} \right|_{-\pi}^{\pi} $

$= \frac{\pm 4}{k^2} $

$b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \sin (nk) \; dx$

$= \frac{1}{\pi} \left. \frac{(2-k^2 x^2) cos(k x)+2 k x sin(k x)}{k^3} \right|_{-\pi}^{\pi} $

$=0$

So, we have

$ \frac{1}{\pi} \int_{-\pi}^{\pi} (f(x))^2 = \frac{2\pi^4}{5} $ and

$a_0^2 = \frac{4\pi^4}{9}$, $a_k^2 = \frac{16}{k^4}$ and $b_k^2 = 0$

Substituting these values in the Parseval's Identity, we have

$\frac{2\pi^4}{5} = \frac{4\pi^4}{18} + \sum_{k=1}^{\infty} \frac{16}{k^4} $

or, $\sum_{k=1}^{\infty} \frac{16}{k^4} = \frac{2\pi^4}{5} - \frac{4\pi^4}{18} $

$ \sum_{k=1}^{\infty} \frac{16}{k^4} = \frac{16\pi^4}{90} $

$ \sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{\pi^4}{90} $