A tale of two integrals

For the first integral:

$\large \displaystyle I = \int_0^{\infty} \frac{xe^{-\pi \sqrt{x}}}{e^{2\pi\sqrt{x}} - 1} dx$

Let $x = \frac{u^2}{\pi^2}$ , $dx = \frac{2u}{\pi^2} du$ :

$\large \displaystyle I = \frac{2}{\pi^4} \int_0^{\infty} \frac{u^3 e^{-u}}{e^{2u}-1}du$

Multiplying above and below by $e^u$ :

$\large \displaystyle I = \frac{2}{\pi^4} \int_0^{\infty} \frac{u^3}{e^{3u}-e^u}du$

Expanding in partial fractions in $e^u$ :

$\large \displaystyle I = \frac{2}{\pi^4} \left [ \frac12 \int_0^{\infty} \frac{u^3}{e^u-1} du + \frac12 \int_0^{\infty} \frac{u^3}{e^u+1} du - \int_0^{\infty} u^3e^{-u} du \right ]$

Recall that:

$\large \displaystyle\int_0^{\infty} \frac{u^{s-1}}{e^u-1} du = \Gamma(s) \zeta(s), \int_0^{\infty} \frac{u^{s-1}}{e^u+1} du = \Gamma(s) \eta(s), \int_0^{\infty} u^{s-1}e^{-u} du = \Gamma(s) = (s-1)!$

Where $\zeta(s)$ is the Riemann Zeta Function , $\eta(s)$ is the Dirichlet Eta Function and $\Gamma(s)$ is the Gamma Function . So:

$\large \displaystyle I = \frac{2}{\pi^4} \left [ \frac12 \zeta(4)\Gamma(4) + \frac12 \eta(4) \Gamma(4) - \Gamma(4) \right ]$

$\large \displaystyle I = \frac{2}{\pi^4} \left [ \frac12 \frac{\pi^4}{90} 3! + \frac12 \frac{7\pi^4}{720} 3!- 3! \right ]$

$\large \displaystyle I = \frac{2}{\pi^4} \left [\frac{\pi^4}{30} + \frac{7\pi^4}{240}- 6 \right ]$

$\large \displaystyle I = \frac{2}{\pi^4} \left [\frac{\pi^4}{16} - 6 \right ]$

$\color{#20A900} \boxed{ \large \displaystyle I = \frac18 - \frac{12}{\pi^4} }$

So:

$\color{#3D99F6} \boxed{\large \displaystyle a = 8 }$

We already have our answer, but, for the second integral, just make $x = \frac{u^3}{\pi^3}$ , $dx = \frac{3u^2}{\pi^3} du$ :

$\large \displaystyle J = \frac{3}{\pi^4} \int_0^{\infty} \frac{u^3 e^{-u}}{e^{u}-1}du$

Multiplying above and below by $e^u$ :

$\large \displaystyle J = \frac{3}{\pi^4} \int_0^{\infty} \frac{u^3}{e^{2u}-e^u}du$

Expanding in partial fractions in $e^u$ :

$\large \displaystyle J = \frac{3}{\pi^4} \left [ \int_0^{\infty} \frac{u^3}{e^{u}-1}du - \int_0^{\infty} u^3 e^{-u} du \right ]$

$\large \displaystyle J = \frac{3}{\pi^4} \left [ \zeta(4) \Gamma(4) - \Gamma(4) \right ]$

$\large \displaystyle J = \frac{3}{\pi^4} \left [ \frac{\pi^4}{15} - 6 \right ]$

$\color{#20A900} \boxed{ \large \displaystyle J = \frac15 - \frac{18}{\pi^4} }$

Likewise:

$\color{#3D99F6} \boxed{\large \displaystyle a = 8 }$