Integration Problem by Aly Ahmed

Consider the integral:

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

Recognising that as $x>0$ :

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

Therefore:

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

Let: $x\ln{3} = z$ . This substitution transforms the integral to:

$I = \frac{1}{(\ln{3})^4}\int_{0}^{\infty} z^3 \left(\sum_{k=1}^{\infty} \mathrm{e}^{-kz} \right)dz$

$I = \frac{1}{(\ln{3})^4}\int_{0}^{\infty} z^3 \left(\mathrm{e}^{-z}+\mathrm{e}^{-2z}+\mathrm{e}^{-3z} + \dots \right)dz$

Now, consider the integral:

$J(k) = \int_{0}^{\infty} z^3 \mathrm{e}^{-kz} \ dz$ Taking $kz = t$ transforms the integral to:

$J(k) = \frac{1}{k^4}\int_{0}^{\infty} t^3 \mathrm{e}^{-t} \ dt$ $\implies J(k) = \frac{1}{k^4} \Gamma(4)$ $J(k) = \frac{6}{k^4}$

Since $\Gamma(n+1) = n!$ . Using the above result transforms $I$ as follows:

$I = \frac{1}{(\ln{3})^4}\left(J(1)+J(2)+J(3)+ \dots \right)$ $\implies I = \frac{6}{(\ln{3})^4} \left(\sum_{k=1}^{\infty} \frac{1}{k^4}\right)$

$\implies I = \frac{6}{(\ln{3})^4} \frac{\pi^4}{90}$ $\implies \boxed{I = \frac{\pi^4}{15(\ln{3})^4}\approx 4.458}$

This is because as per the definition of the Reimann-Zeta function:

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

$\begin{aligned} I & = \int_0^\infty \frac {x^3}{3^x -1} dx & \small \blue{\text{Let }u = x\ln 3 \implies du = \ln 3\ dx} \\ & = \frac 1{\ln^4 3} \int_0^\infty \frac {u^3}{e^u -1} du \\ & = \frac {\blue{\zeta (4)}\red{\Gamma(4)}}{\ln^4 3} & \small \blue{\text{where }\zeta(\cdot) \text{ is the Riemann zeta function}} \\ & = \frac {\blue{\pi^4}\red{3!}}{\blue{90} \ln^4 3} & \small \red{\text{and }\Gamma(\cdot) \text{ is the gamma function}} \\ & \approx \boxed{4.4579} \end{aligned}$

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References:

• Riemann zeta function
• Gamma function