Is Riemann here?

Note that

$\begin{aligned} \zeta(s) & = \frac 1{\Gamma(s)} \int_0^\infty \frac {x^{s-1}}{e^x - 1} dx & \small \blue{\text{where } \zeta (\cdot) \text{ denotes the Riemann zeta function}} \\ \implies \int_0^\infty \frac {x^3}{e^x-1} dx & = \blue{\zeta (4)} \red{\Gamma (4)} & \small \blue{\text{and }\Gamma (\cdot) \text{ denotes the gamma function.}} \\ & = \blue{\frac {\pi^4}{90}} \cdot \red {(4-1)!} \\ & = \frac {\pi^4}{15} \approx \boxed{6.49} \end{aligned}$

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References:

• Riemann zeta function
• Gamma function

$I=\int_0^∞\frac{x^3}{e^x-1} dx$ = $\int_0^∞ x^3( \displaystyle\sum_{n=1}^∞ e^{-nx}dx)$

$I$ = $\displaystyle\sum_{n=1}^∞ \int_{0}^∞ x^3 e^{-nx}dx$

let's take $nx=u$ so $n=du/dx$

$I$ = $\displaystyle\sum_{n=1}^∞ \frac{1}{n} \int_{0}^∞ \frac {u^3}{n^3} e^{-u} du$

$I$ = $\displaystyle\sum_{n=1}^∞ \frac{1}{n^4} Γ(4)$

Note $Γ(s)$ = $\int_{0}^∞ x^{s-1} e^{-x}dx$

And $Γ(s)$ = $(s-1)!$

$I =Γ(4) \displaystyle\sum_{n=1}^∞ \frac{1}{n^4}$

$I = 3! ζ(4)$

$I =6×\frac{π^4}{90}$

$I= \boxed{\frac{π^4}{15}}$

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In general $\int_{0}^∞ \frac{x^n}{e^x-1} = n!ζ(n+1)$

$I = \int_{0}^{\infty} \frac{x^3}{\mathrm{e}^x-1} \ dx$

Multiplying numeritor and denominator by $\mathrm{e}^{-x}$ gives: $I = \int_{0}^{\infty} \frac{\mathrm{e}^{-x}x^3}{1 - \mathrm{e}^{-x}} \ dx$

Expanding the denominator as an infinite geometric series, gives: $I = \int_{0}^{\infty} x^3\left( \mathrm{e}^{-x} + \mathrm{e}^{-2x} + \mathrm{e}^{-3x} + \mathrm{e}^{-4x} + \dots\right) \ dx$ $\because \mathrm{e}^{-x}<1 \ \forall \ x>0$ $I = \sum_{k=1}^{\infty} \left( \int_{0}^{\infty}x^3 \mathrm{e}^{-kx} \ dx\right)$

Let:

$I_2 = \int_{0}^{\infty}x^3 \mathrm{e}^{-kx} \ dx$ Let: $kx = z$ . This transforms the integral to:

$I_2 = \frac{1}{k^4}\int_{0}^{\infty}z^3 \mathrm{e}^{-z} \ dz$ $I_2 = \frac{\Gamma(4)}{k^4}$

$\because \Gamma(4)=3!$ $\implies I_2 = \frac{6}{k^4}$

Using the above result in $I$ gives:

$I = \sum_{k=1}^{\infty} \left( \int_{0}^{\infty}x^3 \mathrm{e}^{-kx} \ dx\right) = \sum_{k=1}^{\infty} \frac{6}{k^4}$

$I = 6\sum_{k=1}^{\infty} \frac{1}{k^4}$ $I = 6 \zeta(4)$ $I = 6\left(\frac{\pi^4}{90}\right)$ $I = \frac{\pi^4}{15}$

$\because \zeta(4) = \frac{\pi^4}{90}$

Where $\zeta(n)$ denotes the Reimann-Zeta function and $\Gamma(n)$ denotes the Gamma function.