Edit I found the mistake, see my answer below.
I am trying to evaluate the integral
$$\int_0^\infty\frac{x \ln(1+x^2)}{\sinh \pi x}\,dx=\frac{\ln 2}{3} – \log \pi – \frac{1}{2} + 6 \ln A $$
I know this integral was already evaluated in this post utilizing various different techniques. But I am trying to compute yet with another one, but due to a mistake which I cannot spot, I can´t finish.
This is how I proceeded:
First note that
$$
\begin{aligned}
&\frac{1}{\sinh \pi x}=\frac{2}{e^{\pi x}-1}-\frac{2}{e^{2 \pi x}-1}\\
&\frac{1}{\sinh \pi x}=\frac{2}{e^{\pi x}-1}-\frac{2e^{-\pi x}}{e^{ \pi x}-e^{-\pi x}}\\
&\frac{1}{\sinh \pi x}=\frac{2}{e^{\pi x}-1}-\frac{e^{-\pi x}}{\sinh \pi x}\\
&\frac{e^{-\pi x}+1}{\sinh \pi x}=\frac{2}{e^{\pi x}-1}\\
&\frac{1}{\sinh \pi x}=\frac{2}{(e^{\pi x}-1)(e^{-\pi x}+1)}\\
&\frac{1}{\sinh \pi x}=\frac{1}{\sinh \pi x} \qquad \blacksquare\\
\end{aligned}
$$
Then consider the following more general integral, which upon letting $z \to 1$ recovers the goal integral.
$$I(z)=\int_0^\infty\frac{x \ln(z^2+x^2)}{\sinh \pi x}\,dx $$
It can be rewritten as
$$I(z)=2\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{\pi x}-1}\,dx-2\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx \tag{1}$$
Differentiating $(1)$ w.r. to $z$ we obtain
$$I^{\prime}(z)=4\int_0^\infty\frac{zx }{(z^2+x^2)(e^{\pi x}-1)}\,dx-4\int_0^\infty\frac{zx }{(z^2+x^2)(e^{2\pi x}-1)}\,dx \tag{2}$$
Now recall Binet´s Integral representation for the Digamma function
$$\int_0^\infty\frac{x }{(z^2+x^2)(e^{2\pi x}-1)}\,dx=\frac{\log(z)}{2}-\frac{\psi(z)}{2}-\frac{1}{4z} \tag{3}$$
Multiplying $(3)$ by $4 \, z$
$$4\int_0^\infty\frac{zx }{(z^2+x^2)(e^{2\pi x}-1)}\,dx=2z\log(z)-2z\psi(z)-1 \tag{4}$$
Also, making the change of variable $2x \mapsto x$ and multiplying by $4z$ in $(3)$ we get
$$4\int_0^\infty\frac{zx }{(z^2+x^2)(e^{\pi x}-1)}\,dx=2z\log\left(\frac{z}{2}\right)-2z\psi\left(\frac{z}{2}\right)-2 \tag{5}$$
Plugging the R.H.S. of $(4)$ and $(5)$ back in $(2)$ we get
$$\begin{aligned}
I^{\prime}(z)&=2z\log\left(\frac{z}{2}\right)-2z\psi\left(\frac{z}{2}\right)-2-2z\log(z)+2z\psi(z)+1\\
&=2\left(z\ln z-z \ln2 \right)-2z\psi\left(\frac{z}{2}\right)-2z\ln z+2 z \psi(z)-1\\
&=-2 \ln2 z -2z\psi\left(\frac{z}{2}\right)+2 z \psi(z)-1\\
\end{aligned}$$
Now integrate from $0 \,\, \text{to} \,\, z$
$$\begin{aligned}
I(z)&=-2 \ln2 \int_0^zx\,dx -2\int_0^z x\psi\left(\frac{x}{2}\right)\,dx+2 \int_0^z z \psi(x)\,dx-\int_0^z \,dx\\
&=-\ln2 z^2 -2\int_0^z x\psi\left(\frac{x}{2}\right)\,dx+2 \int_0^z x \psi(x)\,dx-z
\end{aligned}$$
Letting $z \to 1$
$$I(1)= -\ln2 -8\int_0^{1/2} x\psi\left(x\right)\,dx+2 \int_0^1 x \psi(x)\,dx-\ln e \tag{6}$$
The two integrals involving the Digamma function I computed as follows
$$
\begin{aligned}
\int_0^1 x \psi(x)\,dx&=-\int_0^1 \ln \Gamma(x) dx\\
&=-\frac12 \ln 2 \pi
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 x \psi\left(\frac{x}{2}\right)\,dx&=4\int_0^{1/2} x \psi(x)\,dx \qquad \left(\frac x2 \to x\right)\\
&=4\left(x\ln \Gamma(x) \Big|_0^{1/2}-\int_0^{1/2} \ln \Gamma(x) dx \right)\\
&= \ln \pi-4\left(\frac32 \ln A+\frac{5}{24}\ln 2+ \frac14 \ln \pi\right)\\
&=-\ln A^6-\frac56 \ln 2
\end{aligned}
$$
where in the last one I used the result
$$\int_0^{1/2} \ln \Gamma(x) dx=\frac32 \ln A+\frac{5}{24}\ln 2+ \frac14 \ln \pi$$
Plugging back the values of the integrals in $(6)$ I got
$$\begin{aligned}
I(1)&=-\ln2 +2\left(\ln A^6+\frac56 \ln 2 \right)- \ln 2 -\ln\pi-\ln e\\
&=-2\ln2 +2\ln A^6+\frac53 \ln 2 -\ln\pi -\ln e\\
\end{aligned}$$
Which does not match with the right answer.
Can someone point me where is my mistake(s)?
Best Answer
As I mentioned in my question, when integrating back from $0$ to $z$ the L.H.S. is not our goal integral, nevertheless, we can still evaluate the integral by this method. Here is how to do so:
First, lets integrate back the first integral, namely:
$$ \begin{aligned} &\int_0^z\,\int_0^\infty\frac{2wx }{(w^2+x^2)(e^{2\pi x}-1)}\,dx\,dw=\int_0^zw\log(w)\,dw-\int_0^zw\psi(w)\,dw-\frac12\int_0^zdw\\ &\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx-2\int_0^\infty\frac{x \ln(x)}{e^{2\pi x}-1}\,dx=\frac{z^2 \ln z}{2}-\frac{z^2}{4}-z \ln \Gamma(z)+\int_0^z\ln \Gamma(w)dw-\frac{z}{2}\\ &\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx=2\int_0^\infty\frac{x \ln(x)}{e^{2\pi x}-1}\,dx+\frac{z^2 \ln z}{2}-\frac{z^2}{4}-z \ln \Gamma(z)+\int_0^z\ln \Gamma(w)dw-\frac{z}{2}\\ &\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx=\zeta^{\prime}(-1)+\frac{z^2 \ln z}{2}-\frac{z^2}{4}-z \ln \Gamma(z)+\int_0^z\ln \Gamma(w)dw-\frac{z}{2}\\ \end{aligned} $$
Taking the limit $z \to 1$
$$\int_0^\infty\frac{x \ln(1+x^2)}{e^{2\pi x}-1}\,dx=\zeta^{\prime}(-1)-\frac{3}{4}+\frac12 \ln 2 \pi \qquad \blacksquare$$
Integrating back the second integral:
$$\begin{aligned} &\int_0^z\,\int_0^\infty\frac{2wx }{(w^2+x^2)(e^{\pi x}-1)}\,dx\,dw=\int_0^zw\log\left(\frac{w}{2}\right)\,dw-\int_0^z w\psi\left(\frac{w}{2}\right)\,dw-\int_0^z dw\\ &\int_0^\infty\frac{x \ln(z^2+x^2) }{(e^{\pi x}-1)}\,dx=2\int_0^\infty\frac{x \ln(x) }{(e^{\pi x}-1)}\,dx+\frac{z^2\ln z}{2}-\frac{z^2}{4}-\frac{z^2 \ln 2 }{2}-z-4\int_0^{z/2} w\psi\left(w\right)\,dw\\ &\int_0^\infty\frac{x \ln(z^2+x^2) }{(e^{\pi x}-1)}\,dx=2\int_0^\infty\frac{x \ln(x) }{(e^{\pi x}-1)}\,dx+\frac{z^2\ln z}{2}-\frac{z^2}{4}-\frac{z^2 \ln 2 }{2}-z-4\left(\frac{z}{2} \ln\Gamma\left(\frac{z}{2}\right) -\int_0^{z/2} \ln\Gamma\left(w\right)\,dw \right)\\ \end{aligned} $$
Letting $z \to 1$
$$ \begin{aligned} \int_0^\infty\frac{x \ln(1+x^2) }{(e^{\pi x}-1)}\,dx&=2\left(\frac16+\frac{\ln 2}{6}-2 \ln A \right)-\frac{1}{4}-\frac{ \ln 2 }{2}-1-4\left(\frac{1}{2} \ln\Gamma\left(\frac{1}{2}\right) -\int_0^{1/2} \ln\Gamma\left(w\right)\,dw \right)\\ &=\frac13-4 \ln A -\frac{5}{4}-\frac{ \ln 2 }{6}-4\left(\frac{1}{4} \ln \pi -\left( \frac32\ln A+\frac{5}{24}\ln 2+\frac14 \ln \pi\right)\right)\\ &=\frac13-4 \ln A -\frac{5}{4}-\frac{ \ln 2 }{6}+6\ln A+\frac56\ln 2\\ &=2\ln A -\frac{11}{12}+\frac{ 2 }{3}\ln 2\\ &=-2\zeta^{\prime}(-1)+\frac{ 2 }{3}\ln 2-\frac34 \qquad \blacksquare\\ \end{aligned} $$
We the conclude
$$ \begin{aligned} &\int_{0}^{\infty} \frac{x \ln \left(1+x^{2}\right)}{e^{2 \pi x}-1} d x=\zeta^{\prime}(-1)+\ln \sqrt{2 \pi}-\frac{3}{4} \\ & \\ &\int_{0}^{\infty} \frac{x \ln \left(1+x^{2}\right)}{e^{\pi x}-1} d x=-2 \zeta^{\prime}(-1)+\frac{2}{3} \ln 2-\frac{3}{4} \end{aligned} $$
Therefore our integral equals
$$ \begin{aligned} \int_0^\infty\frac{x \ln(1+x^2)}{\sinh \pi x}\,dx&=2\left(-2 \zeta^{\prime}(-1)+\frac{2}{3} \ln 2-\frac{3}{4}-\zeta^{\prime}(-1)-\ln \sqrt{2 \pi}+\frac{3}{4}\right)\\ &=2\left(-3 \zeta^{\prime}(-1)+\frac{2}{3}\ln 2-\frac12 \ln 2-\frac12 \ln \pi\right)\\ &=-6 \zeta^{\prime}(-1)+\frac{4}{3}\ln 2- \ln 2- \ln \pi\\ &=-6 \zeta^{\prime}(-1)+\frac{\ln 2}{3}- \ln \pi\\ &=-6 \left(\frac{1}{12}-\ln A \right)+\frac{\ln 2}{3}- \ln \pi\\ &=-\frac{\ln e}{2}+\ln A^6 +\frac{\ln 2}{3}- \ln \pi\\ &=\ln \, \frac{\sqrt[3]{2}\, A^6}{\pi \sqrt{e}} \qquad \blacksquare \end{aligned} $$
Appendix
Consider the Integral
$$ \begin{aligned} I(s)&=\int_0^\infty \frac{x^{s-1}}{e^{cx}-1}\,dx\\ &=\frac{1}{c^s}\int_0^\infty \frac{x^{s-1}}{e^{x}-1}\,dx\\ &=\frac{1}{c^s}\int_0^\infty \frac{e^{-x}x^{s-1}}{1-e^{-x}}\,dx\\ &=\frac{1}{c^s}\sum_{k=1}^\infty \, \int_0^\infty x^{s-1}e^{-kx}\,dx\\ &=\frac{1}{c^s}\sum_{k=1}^\infty \frac{1}{k^s} \, \int_0^\infty x^{s-1}e^{-x}\,dx\\ &=\frac{1}{c^s}\sum_{k=1}^\infty \frac{1}{k^s} \, \int_0^\infty x^{s-1}e^{-x}\,dx\\ &=\frac{\Gamma(s)\zeta(s)}{c^s} \end{aligned} $$
Differentiating both sides w.r. to s
$$ \begin{aligned} \int_0^\infty \frac{x^{s-1} \ln x}{e^{cx}-1}\,dx&=\frac{d}{ds}\left(\frac{\Gamma(s)\zeta(s)}{c^s} \right)\\ &=\frac{c^s\left(\Gamma^{\prime}(s)\zeta(s)+\Gamma(s)\zeta^{\prime}(s) \right)-c^s\ln c \left( \Gamma(s)\zeta(s) \right)}{(c^s)^2}\\ &=\frac{\left(\Gamma(s)\psi(s)\zeta(s)+\Gamma(s)\zeta^{\prime}(s) \right)-\ln c \left( \Gamma(s)\zeta(s) \right)}{c^s}\\ \end{aligned} $$
Setting $s=2$
$$ \begin{aligned} \int_0^\infty \frac{x \ln x}{e^{cx}-1}\,dx&=\frac{\left(\Gamma(2)\psi(2)\zeta(2)+\Gamma(2)\zeta^{\prime}(2) \right)-\ln c \left( \Gamma(2)\zeta(2) \right)}{c^2}\\ &=\frac{\left((1-\gamma)\zeta(2)+\zeta(2)\left(\ln 2\pi+\gamma-12 \ln A\right) \right)-\zeta(2)\ln c }{c^2}\\ &=\zeta(2)\left(\frac{1-\ln c+\ln 2\pi-12 \ln A }{c^2}\right)\\ \end{aligned} $$
Setting $c=\pi$
$$\int_0^\infty \frac{x \ln x}{e^{\pi x}-1}\,dx=\frac16+\frac{\ln 2}{6}-2 \ln A \qquad \blacksquare$$
For the LogGamma integral between $0$ and $1/2$
Recall Kummer´s fourier expansion for LogGamma $0<x<1$
$$\ln\left(\Gamma(x)\right)=\frac{\ln\left(2 \pi\right)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos\left(2 \pi k x\right) + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\sin\left(2 \pi k x \right) \tag{A.1}$$
Integrating we obtain
$$\int_0^x\ln\left(\Gamma(u)\right)du=\frac{x \ln\left(2 \pi\right)}{2}+\frac{1}{4 \pi}\sum_{k=1}^{\infty}\frac{\sin\left(2 \pi k x \right)}{k^2} +\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\cos \left(2 \pi k x \right)}{k^2}+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right)}{k^2}-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right) \cos\left(2 \pi k x \right)}{k^2} \tag{A.2}$$
Letting $x=\frac12$ in $(A.2)$ we obtain
$$\begin{aligned} \int_0^{1/2}\ln\left(\Gamma(x)\right)dx&=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{2 \pi^2}\zeta(2)+\frac{\gamma}{2 \pi^2}\eta(2)+\frac{\ln(2\pi)}{2 \pi^2}\zeta(2)+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k\right)}{k^2}+\frac{\ln \left( 2 \pi \right)}{2 \pi^2}\eta(2)-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k \right) \left(-1 \right)^{k}}{k^2}\\ &=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{12 }+\frac{\gamma}{24}+\frac{\ln(2\pi)}{12 }-\frac{1}{2 \pi^2}\zeta^{\prime}(2)+\frac{\ln \left( 2 \pi \right)}{24}-\frac{1}{2 \pi^2}\eta^{\prime}(2)\\ &=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{1}{2 \pi^2} \cdot \frac{\pi^2}{6}\left(\ln(2 \pi)+\gamma-12 \ln A \right)-\frac{1}{2 \pi^2} \left(\frac{\pi^2 \ln(2 )}{12} +\frac{\pi^2}{12}\left(\ln(2 \pi)+\gamma-12 \ln A\right)\right)\\ &=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{\ln(2 \pi)}{12}-\frac{\gamma}{12}+\ln A-\frac{\ln (2)}{24}-\frac{\ln(2 \pi)}{24}-\frac{\gamma}{24}+\frac{\ln A}{2}\\ &=\left(\frac{1}{8}-\frac{1}{12}-\frac{1}{24} \right)\gamma+\left(\frac38-\frac{1}{12}-\frac{1}{24} \right)\ln(2 \pi)-\frac{\ln(2)}{24}+\ln A+\frac{\ln A}{2}\\ &=\frac{3}{2}\ln A+ \frac{6}{24}\ln (2 \pi)-\frac{\ln (2)}{24}\\ &=\frac{3}{2}\ln A+\frac{5}{24}\ln 2+\frac{1}{4}\ln \pi \qquad \blacksquare \end{aligned}$$
We used that
$\zeta^{\prime}(2)=\frac{\pi^2}{6}\left(\ln 2 \pi + \gamma -12 \ln A \right)$
And that
$\eta^{\prime}(s)=2^{1-s}\ln (2) \zeta(s)+(1-2^{1-s})\zeta^{\prime}(s)$