复杂的对数积分（三）
\[\Large\displaystyle \int_0^\infty \frac{\ln \left(1+\dfrac{\pi^2}{4x} \right)}{e^{\sqrt{x}}-1}\mathrm{d}x
\]
\(\Large\mathbf{Solution:}\)
Step 1 - Split
Let \(\displaystyle I=\int_0^\infty \frac{\ln \left(1+\dfrac{\pi^2}{4x} \right)}{e^{\sqrt{x}}-1}\mathrm{d}x\). Now put \(x\mapsto x^2\)
\[\begin{align*} I &=2\int_0^\infty \frac{x\ln\left( 1+\dfrac{\pi^2}{4x^2} \right)}{e^x-1}\mathrm{d}x\\&= 2\int_0^\infty \frac{x \ln \left( 4x^2+\pi^2\right)-x\ln (4)-2x\ln(x)}{e^x-1}\mathrm{d}x \\ &= 2\underbrace{\int_0^\infty \frac{x \ln \left( 4x^2+\pi^2\right)}{e^x-1}\mathrm{d}x}_{=I_1}-4\ln(2) \underbrace{\int_0^\infty \frac{x}{e^x-1}\mathrm{d}x}_{=I_2}-4 \underbrace{\int_0^\infty \frac{x\ln(x)}{e^x-1}\mathrm{d}x}_{=I_3} \tag{1} \end{align*}
\]
Step 2 - Evaluation of \(I_3\)
Note that for \(\Re (z)>1\), we have
\[\begin{align*} \int_0^\infty \frac{x^{z-1}}{e^x-1}\mathrm{d}x = \Gamma(z)\zeta(z) \tag{2}\end{align*}
\]
Differentiate both sides with respect to \(z\).
\[\begin{align*} \int_0^\infty \frac{x^{z-1}\ln(x)}{e^x-1}\mathrm{d}x= \psi_0(z)\Gamma(z) \zeta(z)+\Gamma(z)\zeta '(z) \tag{3}\end{align*}
\]
Put \(z=2\) to obtain
\[\begin{align*}\int_0^\infty \frac{x \ln(x)}{e^x-1}\mathrm{d}x &= \frac{\pi^2}{6}(1-\gamma)-\frac{\pi^2}{6} \left(12 \ln\mathbf{A}-\gamma-\ln(2\pi) \right) \\&= \frac{\pi^2}{6} \left( 1+\ln(2\pi)-12 \ln\mathbf{A} \right) \tag{4}\end{align*}
\]
Step 3 - Evaluation of \(I_2\)
\(I_2\) can be evaluated easily by means of equation (2).
\[\begin{align*}
\int_0^\infty \frac{x}{e^x-1}\mathrm{d}x = \frac{\pi^2}{6} \tag{5}
\end{align*}\]
Step 4 - Evaluation of \(I_1\)
From here begins the dirty job. We require equation (4) of Adamchik's paper.
\[\begin{align*} \int_0^\infty \frac{x \ln(x^2+z^2)}{e^{2\pi x}-1}\mathrm{d}x &= \zeta'(-1,z)-\frac{z^2}{2}\ln z+\frac{z^2}{4}+\frac{z}{2}\ln z \\
&~~~-2z \int_0^\infty \frac{\arctan \left( \dfrac{x}{z}\right)}{e^{2\pi x}-1}\mathrm{d}x \quad \Re(z)>0 \end{align*}\]
Now, from Binet's second formula we have
\[\begin{align*} \int_0^\infty \frac{x \ln(x^2+z^2)}{e^{2\pi x}-1}\mathrm{d}x &= \zeta'(-1,z)-\frac{z^2}{2}\ln z+\frac{z^2}{4}+\frac{z}{2}\ln z \\ &\quad -z \left\{\ln \Gamma(z) -\left(z-\frac{1}{2} \right)\ln z+z -\frac{1}{2}\ln(2\pi) \right\} \tag{6} \end{align*}
\]
Set \(x \mapsto \dfrac{x}{2\pi}\) and put \(z=\dfrac{1}{4}\).
\[\begin{align*} &\frac{1}{4\pi^2}\int_0^\infty \frac{x \ln \left( 4x^2+\pi^2\right)}{e^x-1}\mathrm{d}x-\frac{\ln 2+\ln(2\pi)}{12} \\=& \zeta' \left(-1, \frac{1}{4}\right)+\frac{1}{16}\ln(2)+\frac{1}{64}-\frac{1}{4}\ln(2)-\frac{1}{4} \left\{\ln \Gamma \left(\frac{1}{4} \right) -\frac{\ln (2)}{2}+\frac{1}{4} -\frac{1}{2}\ln(2\pi) \right\} \\ &\int_0^\infty \frac{x \ln \left(4x^2 +\pi^2\right)}{e^x-1}\mathrm{d}x \\=& 4\pi^2 \zeta' \left( -1,\frac{1}{4}\right)+\frac{\pi^2}{12}\ln(2)+\frac{5}{6}\pi^2 \ln(2\pi) -\frac{3}{16}\pi^2-\pi^2 \ln \Gamma \left( \frac{1}{4}\right) \tag{7} \end{align*}
\]
Now, from equations (3) and (11) of Adamchik's paper we get
\[\begin{align*} \zeta' \left(-1,\frac{1}{4} \right) &= \zeta'(-1)-\frac{3}{4}\ln \Gamma \left( \frac{1}{4}\right)-\ln \mathrm{G} \left( \frac{1}{4}\right) \\ &= \frac{1}{12}-\ln\mathbf{A}-\frac{3}{4}\ln \Gamma \left( \frac{1}{4}\right)-\left(\frac{3}{32} -\frac{\mathbf{G}}{4\pi}-\frac{3}{4}\ln \Gamma \left( \frac{1}{4}\right)-\frac{9}{8}\ln\mathbf{A}\right) \\ &= \frac{\ln\mathbf{A}}{8}+\frac{\mathbf{G}}{4\pi}-\frac{1}{96} \tag{8} \end{align*}
\]
where \(\mathrm{G}(z)\) denotes the Barnes G Function.
Using this result in (7), we get
\[\begin{align*} &\int_0^\infty \frac{x \ln \left(4x^2 +\pi^2\right)}{e^x-1}\mathrm{d}x \\&= 4\pi^2 \left( \frac{\ln\mathbf{A}}{8}+\frac{\mathbf{G}}{4\pi}-\frac{1}{96}\right)+\frac{\pi^2}{12}\ln(2)+\frac{5}{6}\pi^2 \ln(2\pi) -\frac{3}{16}\pi^2-\pi^2 \ln \Gamma \left( \frac{1}{4}\right) \\ &= \frac{\pi^2}{2}\ln\mathbf{A}+\frac{\pi^2}{12}\ln(2)-\frac{11}{48}\pi^2+\frac{5\pi^2}{6}\ln(2\pi) +\pi \mathbf{G} - \pi^2 \ln \Gamma \left( \frac{1}{4}\right)\tag{9} \end{align*}
\]
Step 5 - Final Answer
Combining everything with the help of equation (1), we get
\[\begin{align*} I &= 2 \left(\frac{\pi^2}{2}\ln\mathbf{A}+\frac{\pi^2}{12}\ln(2)-\frac{11}{48}\pi^2+\frac{5\pi^2}{6}\ln(2\pi) +\pi \mathbf{G} - \pi^2 \ln \Gamma \left( \frac{1}{4}\right)\right) \\&~~~-4\ln(2) \left(\frac{\pi^2}{6} \right)-4 \left( \frac{\pi^2}{6} \left( 1+\ln(2\pi)-12 \ln\mathbf{A} \right)\right) \\ &=\Large\boxed{\displaystyle\color{blue}{\pi^2 \left\{ \ln \left( \frac{\pi \mathbf{A}^9 \sqrt{2}}{\Gamma^2 \left( \dfrac{1}{4}\right)}\right)-\frac{9}{8}\right\}+2\pi \mathbf{G}}} \end{align*}
\]
