Integral $\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}$ I have stumbled upon the following integral:$$I=\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}=-\frac{\pi}{24}$$
Although I could solve it, I am not quite comfortable with the way I did it.
But first I will show the way. We can substitute $\ln x \rightarrow t\ $ which gives:
$$I=\int_{-\infty}^\infty \frac{t}{\pi^2+t^2}\frac{e^{\frac{t}{2}}}{(1+e^t)^2}dt\overset{t=-x}=\int_{-\infty}^\infty \frac{-x}{\pi^2+x^2}\frac{e^{-\frac{x}{2}}}{(1+e^{-x})^2}dx$$
Also adding the two integral from above and simplify some of it yields:
$$2I= \int_{-\infty}^\infty \frac{x}{\pi^2+x^2}\left(\frac{e^{\frac{x}{2}}}{(1+e^x)^2}-\frac{e^{-\frac{x}{2}}}{(1+e^{-x})^2}\right)dx$$
$$\Rightarrow I=-\frac{1}{4} \int_{-\infty}^\infty \frac{x}{\pi^2+x^2}\frac{\sinh \left(\frac{x}{2}\right)}{\cosh ^2\left(\frac{x}{2}\right)}dx$$
And now a round of IBP gives:
$$I=\frac12 \int_{-\infty}^\infty \left(\frac{x^2-\pi^2}{(x^2+\pi^2)^2}\right)\left(\frac{1}{\cosh \left(\frac{x}{2}\right)}\right)dx$$
Using the Plancherel theorem the integral simplifies to: $$I=\int_0^\infty \left(\sqrt{\frac{\pi}{2}}x\left(-e^{-\pi x}\right)\right)\left(\sqrt{2\pi}\frac{1}{\cosh(\pi x)}\right)dx\overset{\pi x\rightarrow x}=-\frac{1}{\pi}\int_0^\infty \frac{x}{\cosh( x)}e^{- x}dx$$
We also have the following Laplace tranform for:$$f(t)=\frac{t}{\cosh( t)}\rightarrow F(s)=\frac18\left(\psi_1\left(\frac{s+1}{4}\right)-\psi_1\left(\frac{s+3}{4}\right)\right)$$
Where $\displaystyle{\psi_1(z)=\sum_{n=0}^\infty \frac{1}{(z+n)^2}}\,$ is the trigamma function.
$$\Rightarrow I=-\frac{1}{\pi}F(s=1)=-\frac{1}{\pi}\cdot \frac18\left(\psi_1\left(\frac{1}{2}\right)-\psi_1 (1)\right)=-\frac{1}{\pi}\cdot \frac18\left(\frac{\pi^2}{2}-\frac{\pi^2}{6}\right)=-\frac{\pi}{24}$$
Have I done anything wrong, or can it be improved?
I have to admit that I mostly used wolfram when applying Plancherel theorem  and Laplace transform which I'm not comfortable with, but I didn't find an alternative method myself.

For this question I would like to see a different proof that doesn't rely on that theorem.

Probably not needed, but I should mention that my contour integration knowledge is pretty low. Also maybe there is a conexion with this integral, but I didn't find any.
 A: That $\pi^2+\log^2(x)$ makes me think to the integral representation for Gregory coefficients:
$$ \int_{0}^{+\infty}\frac{dx}{(1+x)^n (\pi^2+\log^2 x)} = \frac{1}{n!}\left[\frac{d^n}{dx^n}\frac{z}{\log(1-z)}\right]_{z=0}=[z^n]\frac{z}{\log(1-z)} \tag{1}$$
which can be seen as a consequence of the Lagrange-Buhrmann inversion theorem. We just need to insert a factor $\frac{\log x}{\sqrt{x}}$ in the integrand function appearing in the LHS, so let's go back to the residue theorem.
$$ \int_{0}^{+\infty}\frac{\log(x)\,dx}{\sqrt{x}(1+x)^2(\pi^2+\log^2 x)}=\int_{-\infty}^{+\infty}\frac{t e^{-t/2}}{(2\cosh\frac{t}{2})^2 (\pi^2+t^2)}\,dt$$
equals
$$ -\frac{1}{4}\int_{\mathbb{R}}\frac{t\sinh\frac{t}{2}}{(t^2+\pi^2)\cosh^2\frac{t}{2}}\,dt=-\int_{\mathbb{R}}\frac{t\sinh t}{(4t^2+\pi^2)\cosh^2 t}\,dt. $$
The meromorphic function $\frac{\sinh t}{\cosh^2 t}=-\frac{d}{dt}\left(\frac{1}{\cosh t}\right)$ only has double poles with residue zero, hence all the mass of the last integral comes from the singularity at $\frac{\pi i}{2}$ and from the behaviour at infinity. The residue theorem grants
$$ \frac{1}{\cosh x}=\sum_{n\geq 0}(-1)^n \frac{\pi(2n+1)}{\frac{\pi^2}{4}(2n+1)^2+x^2} $$
and
$$ \frac{\sinh x}{\cosh^2 x} = \sum_{n\geq 0}(-1)^n \frac{2\pi(2n+1)x}{(\frac{\pi^2}{4}(2n+1)^2+x^2)^2}.$$
Since 
$$ \int_{\mathbb{R}}\frac{2\pi(2n+1)x^2}{(\frac{\pi^2}{4}(2n+1)^2+x^2)^2 (\pi^2+4x^2)}\,dx = \frac{1}{2\pi(n+1)^2}$$
our integral equals $-\frac{1}{2\pi}\eta(2)=\color{red}{-\frac{\pi}{24}}$ by the dominated convergence theorem, allowing to switch $\int_{\mathbb{R}}$ and $\sum_{n\geq 0}$. $\frac{1}{24}$ is also the coefficient of $z^3$ in $\frac{z}{\log(1-z)}$, but so far I have not found a direct way to relate the original integral to the $n=3$ instance of $(1)$.
A: This is not a full answer, but one does not need the full Laplace transform of $x/\cosh x$, as the integral can be done by expanding $1/\cosh x$ into a geometric series
$$\begin{aligned} I = \int_{0}^{\infty}\frac{xe^{-x}}{\cosh x}\,\mathrm{d}x &= 2\int_{0}^{\infty}\frac{xe^{-x}}{e^{x}}\frac{\mathrm{d}x}{1+e^{-2x}} = 2\sum_{n=0}^{\infty}(-1)^{n}\int_{0}^{\infty}xe^{-(2+2n)x}\,\mathrm{d}x \\
&= 2\sum_{n=0}^{\infty}\frac{(-1)^{n}}{4(1+n)^{2}}\int_{0}^{\infty}ue^{-u}\,\mathrm{d}u = \frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(1+n)^{2}} \\ &= \frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{2}} = \frac{\eta(2)}{2} = \frac{\pi^{2}}{24}\end{aligned}$$
where $\eta(s)$ is the Dirichlet eta function.
A: To show that $$(-1)^{n-1} \int_{0}^{\infty}\frac{ \mathrm dx}{ (\pi^{2}+\ln^{2} x)(1+x)^{n}} $$ is an integral representation of the Gregory coefficients, some textbooks integrate the function $\frac{1}{(\ln z - i\pi)(1+z)^{n}} $ around a keyhole contour.  We can do something similar here.
Let's integrate the function $$f(z)= \frac{1}{(\ln z - i \pi) (1+z)^{2}} \frac{1}{\sqrt{z}}$$ around a keyhole contour where the branch cut is along the positive real axis.
Integrating around the contour, we get  $$ \int_{0}^{\infty}\frac{1}{(\ln x - i \pi) (1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} + \int_{\infty}^{0} \frac{1}{(\ln x + 2 \pi i - i \pi) (1+x)^{2}} \frac{\mathrm dx}{\sqrt{e^{2 \pi i}x}}  = 2 \pi i \operatorname{Res} \left[f(z), -1 \right] $$
The left side of the above equation is
$$ \int_{0}^{\infty}\frac{1}{(\ln x - i \pi) (1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} + \int_{0}^{\infty} \frac{1}{(\ln x + i \pi) (1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}}  $$
$$= \int_{0}^{\infty} \frac{2 \ln x}{\left(\ln^{2}(x)+\pi^{2}\right)(1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} $$
Since $f(z)$ has pole at $z=e^{i \pi}=-1$ of order three, calculating the residue of $f(z)$ at $z=-1$ is a bit tedious.
But at $z=-1$,  $\ln(z) - i\pi$ has the Taylor series expansion $$\ln(z) - i \pi = - (z+1)-\frac{(z+1)^{2}}{2!}- \frac{(z+1)^{3}}{3!} + O \left((z+1)^{4} \right) $$
Using polynomial long division, one can then show that Laurent series expansion of $\frac{1}{\ln z - i \pi}$ about $z=-1 $ is $$\frac{1}{\ln z - i \pi} = - \frac{1}{z+1} + \frac{1}{2} + \frac{z+1}{12} + O \left((z+1)^{2} \right)  $$
So near $z=-1$, $$f(z) = \frac{1}{\sqrt{z}} \left(- \frac{1}{(z+1)^{3}} + \frac{1}{2(z+1)^{2}} + \frac{1}{12(z+1)} +O(1)\right), $$
from which we get $$ \begin{align} \small \operatorname{Res}\left[f(z), -1 \right] &= \small -\operatorname{Res}\left[\frac{1}{\sqrt{z}} \frac{1}{(z+1)^{3}}, -1 \right] + \frac{1}{2} \,  \operatorname{Res}\left[\frac{1}{\sqrt{z}} \frac{1}{(z+1)^{2}}, -1 \right] + \frac{1}{12} \, \operatorname{Res}\left[\frac{1}{\sqrt{z}} \frac{1}{z+1}, -1 \right] \\ &=- \frac{1}{2!}  \frac{3/4}{(e^{i \pi})^{5/2}} - \frac{1}{2} \frac{1/2}{(e^{i \pi)^{3/2}}} + \frac{1}{12} \frac{1}{(e^{i \pi})^{1/2}}  \\ &= - \frac{3}{8i} + \frac{1}{4i} + \frac{1}{12i} = -\frac{1}{24i} \end{align}$$
(Alternatively, we could have also expanded $\frac{1}{\sqrt{z}}$ at $z=-1$ to get the first few terms of the Laurent series expansion of $f(z)$ about  $z=-1$.)
Therefore, $$\int_{0}^{\infty} \frac{\ln x}{(\pi^{2}+\ln^{2} x)(1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} = \frac{2 \pi i}{2}  \left(-\frac{1}{24i} \right) = -\frac{\pi}{24}$$

If $a>0$ and $a \ne 1$, then the same approach shows that $$\int_{0}^{\infty} \frac{\ln x}{(\pi^{2}+ \ln^{2}x)(a+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} = \pi \left(\frac{2 + \ln a}{2a^{3/2} \ln^{2}a}- \frac{1}{(a-1)^{2}} \right) $$
Applying L'Hôpital's rule 4 times shows that the limit of the right side of the above equation as $a \to 1$ is $-\frac{\pi}{24}$.
A: From the identity
$$\Im\int_0^\infty e^{-(\pi-it)x}\,dx=\frac t{\pi^2+t^2}$$
we see that it suffices to compute the imaginary part of the integral
$$\int_0^\infty dx\int_{-\infty}^\infty dt\;
\frac{e^{\alpha t}}{(1+e^t)^2}e^{-\pi x}$$
where $\alpha=1/2+ix$. Now, the integral with respect to $t$ is easy by means of the substitution $u=e^t$ and using the beta function. We thus get
$$\int_0^\infty \pi\left(\frac12-ix\right)\frac{e^{-\pi x}}{\cosh(\pi x)}\,dx.$$
Taking the imaginary part we see that the problem boils down to compute the integral
$$\int_0^\infty \frac{x e^{-\pi x}}{\cosh(\pi x)}\,dx
=2\int_0^\infty \frac{x}{1+e^{2\pi x}}\,dx$$
which, after the substitution $v=2\pi x$, reduces to the integral representation of the eta function $\eta(2)$. Also notice that taking the real part we obtain the evaluation 
$$\int_0^\infty\frac1{(\pi^2+\log^2 x)(1+x)^2} \frac{dx}{\sqrt x}=
\frac{\log2}{2\pi}.$$
This method generalises to other integrals such as
\begin{align*}
\int_0^\infty \frac{1}{(\pi^2+\ln^2 x)(1+x)^3} \frac{dx}{\sqrt x}
&=\frac{3\log (2)}{8 \pi }-\frac{3 \zeta (3)}{16 \pi ^3}\\
\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^3} \frac{dx}{\sqrt x}
&=-\frac{\pi }{24}\\
\int_0^\infty \frac{1}{(\pi^2+\ln^2 x)(1+x)^4} \frac{dx}{\sqrt x}
&=\frac{5 \log (2)}{16 \pi }-\frac{9 \zeta (3)}{32 \pi ^3}\\
\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^4} \frac{dx}{\sqrt x}
&=-\frac{223 \pi }{5760}\\
\int_0^\infty \frac{1}{(\pi^2+\ln^2 x)(1+x)^5} \frac{dx}{\sqrt x}
&=-\frac{43 \zeta (3)}{128 \pi ^3}+\frac{15 \zeta (5)}{256 \pi ^5}+\frac{35 \log (2)}{128
   \pi }\\
\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^5} \frac{dx}{\sqrt x}
&=-\frac{103 \pi }{2880}\\
\end{align*}
Incidentally, since the integrals $\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^k} \frac{dx}{\sqrt x}$ both yield the same value for $k=2,3$, we also deduce
$$\int_0^\infty \frac{\sqrt x\ln x}{(\pi^2+\ln^2 x)(1+x)^3}\;dx=0.$$
This, however, should not be surprising due to the symmetry $x\mapsto1/x$. 
