Riemann Will Be Happy

$\int\limits_{0}^{\infty}\dfrac{x\sqrt{x}}{e^x-1}dx$

$=\int\limits_{0}^{\infty}\dfrac{x\sqrt{x}}{e^x(1-\frac{1}{e^x})}dx$

$=\int\limits_{0}^{\infty}\dfrac{x\sqrt{x}}{e^x}\Bigg(1 + \dfrac{1}{e^x} + \dfrac{1}{(e^x)^2} + \dots\Bigg) dx ^*$

$=\int\limits_{0}^{\infty}x\sqrt{x}\Bigg(\dfrac{1}{e^x} + \dfrac{1}{(e^x)^2} + \dfrac{1}{(e^x)^3} + \dots\Bigg) dx$

$=\int\limits_{0}^{\infty}x\sqrt{x}\sum\limits_{n=0}^{\infty}\dfrac{1}{e^{nx}} dx$

$=\sum\limits_{n=0}^{\infty}\int\limits_{0}^{\infty}\dfrac{x^{\frac{3}{2}}}{e^{nx}} dx$

$\text{Substituting }nx = t$

$=\sum\limits_{n=0}^{\infty}\int\limits_{0}^{\infty}\dfrac{(\frac{t}{n})^{\frac{3}{2}}}{e^{t}} \dfrac{dt}{n}$

$=\sum\limits_{n=0}^{\infty}\dfrac{1}{n^{\frac{5}{2}}}\int\limits_{0}^{\infty}t^{\frac{3}{2}}e^{-t} dt$

$=\sum\limits_{n=0}^{\infty}\dfrac{1}{n^{\frac{5}{2}}}\Gamma\Bigg({\dfrac{5}{2}}\Bigg) ^{**}$

$=\zeta\Bigg(\dfrac{5}{2}\Bigg)\Gamma\Bigg({\dfrac{5}{2}}\Bigg)$

$\Gamma\Bigg({\dfrac{5}{2}}\Bigg) = \dfrac{3}{2}\cdot\dfrac{1}{2}\cdot\sqrt{\pi}$

$\therefore A = 3, B = 4, C = 5, D = 2, A+B+C+D=\boxed{14}$

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$^* 1 + x + x^2 + \dots = \dfrac{1}{1-x} \text{ for }|x| < 1, \because x > 0$ in the question, $\dfrac{1}{e^x} < 1$ .

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$^{**}$ Refer to the defintion of the Gamma function .

The relationship between gamma function $\Gamma(s)$ and Riemann zeta function $\zeta(s)$ is given by

$\begin{aligned} \int_0^\infty \frac {x^{s-1}}{e^x-1} dx & = \Gamma (s) \zeta (s) \\ \implies \int_0^\infty \frac {x\sqrt x}{e^x-1} dx & = {\color{#3D99F6}\Gamma \left(\frac 52\right)} \zeta \left(\frac 52\right) & \small \color{#3D99F6} \text{From } \Gamma (1+s) = s \Gamma (s) \\ & = {\color{#3D99F6}\frac 32 \Gamma \left(\frac 32\right)} \zeta \left(\frac 52\right) \\ & = {\color{#3D99F6}\frac 12 \cdot \frac 32 \Gamma \left(\frac 12\right)} \zeta \left(\frac 52\right) & \small \color{#3D99F6} \text{and } \Gamma \left(\frac 12\right) = \sqrt \pi \\ & = \frac 34 \sqrt \pi \zeta \left(\frac 52\right) \end{aligned}$

Therefore, $A+B+C+D= 3+4+5+2 = \boxed{14}$ .