Tricky Integrals - 1

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

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

$\int_{0}^{\infty} \sum_{n=0}^{\infty} x^3 e^{-x}{e^{-nx}}dx$

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

$\text{Let} , u = nx$

$\sum_{n=1}^{\infty} (\frac{1}{n^4} \int_{0}^{\infty} u^3 e^{-u} du)$

$(\sum_{n=1}^{\infty}\frac{1}{n^4}) (\int_{0}^{\infty} u^3 e^{-u} du)$

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

$\int_{0}^{\infty} u^3 e^{-u} du = 3!$

$\therefore \int_{0}^{\infty} \frac{x^3}{e^x - 1} dx = \frac{\pi^4}{15}$

$a = 4, b=15$

$a+b = 19$

$\text{From the relation : } \Gamma{(n)}\cdot \zeta{(n)} = \displaystyle\int_{0}^{\infty} \dfrac{t^{n-1}}{e^{t}-1} dt \\ \implies \Gamma{(4)}\cdot \zeta{(4)} = \displaystyle\int_{0}^{\infty} \dfrac{t^{3}}{e^{t}-1} dt \implies 3! \times \dfrac{\pi^{4}}{90} \\ \implies \boxed{ \dfrac{\pi^{\color{#3D99F6}{4}}}{\color{#D61F06}{15}} } \\ a+b=\boxed{19}$

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Proof of the Relation provided by @Vatsalya Tandon