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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:
$$I=int_{-infty}^infty frac{t}{pi^2+t^2}frac{e^{frac{t}{2}}}{(1+e^t)^2}dtoverset{t=-x}=int_{-infty}^infty frac{-x}{pi^2+x^2}frac{e^{-frac{x}{2}}}{(1+e^{-x})^2}dx$$
Also adding the two integral from above and simplify some of it yields:
$$2I= int_{-infty}^infty frac{x}{pi^2+x^2}left(frac{e^{frac{x}{2}}}{(1+e^x)^2}-frac{e^{-frac{x}{2}}}{(1+e^{-x})^2}right)dx$$
$$Rightarrow I=-frac{1}{4} int_{-infty}^infty frac{x}{pi^2+x^2}frac{sinh left(frac{x}{2}right)}{cosh ^2left(frac{x}{2}right)}dx$$
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: $$I=int_0^infty left(sqrt{frac{pi}{2}}xleft(-e^{-pi x}right)right)left(sqrt{2pi}frac{1}{cosh(pi x)}right)dxoverset{pi xrightarrow x}=-frac{1}{pi}int_0^infty frac{x}{cosh( x)}e^{- x}dx$$
We also have the following Laplace tranform for:$$f(t)=frac{t}{cosh( t)}rightarrow F(s)=frac18left(psi_1left(frac{s+1}{4}right)-psi_1left(frac{s+3}{4}right)right)$$
Where $displaystyle{psi_1(z)=sum_{n=0}^infty frac{1}{(z+n)^2}},$ is the trigamma function.
$$Rightarrow I=-frac{1}{pi}F(s=1)=-frac{1}{pi}cdot frac18left(psi_1left(frac{1}{2}right)-psi_1 (1)right)=-frac{1}{pi}cdot frac18left(frac{pi^2}{2}-frac{pi^2}{6}right)=-frac{pi}{24}$$
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.

From the identity

$$Imint_0^infty e^{-(pi-it)x},dx=frac t{pi^2+t^2}$$

we see that it suffices to compute the imaginary part of the integral

$$int_0^infty dxint_{-infty}^infty dt;
frac{e^{alpha t}}{(1+e^t)^2}e^{-pi x}$$

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

$$int_0^infty pileft(frac12-ixright)frac{e^{-pi x}}{cosh(pi x)},dx.$$

Taking the imaginary part we see that the problem boils down to compute the integral

$$int_0^infty frac{x e^{-pi x}}{cosh(pi x)},dx
=2int_0^infty frac{x}{1+e^{2pi x}},dx$$

which, after the substitution $v=2pi x$, reduces to the integral representation of the eta function $eta(2)$. Also notice that taking the real part we obtain the evaluation

$$int_0^inftyfrac1{(pi^2+log^2 x)(1+x)^2} frac{dx}{sqrt x}=
frac{log2}{2pi}.$$

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{3log (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

$$int_0^infty frac{sqrt xln x}{(pi^2+ln^2 x)(1+x)^3};dx=0.$$

This, however, should not be surprising due to the symmetry $xmapsto1/x$.

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.

$$ int_{0}^{+infty}frac{log(x),dx}{sqrt{x}(1+x)^2(pi^2+log^2 x)}=int_{-infty}^{+infty}frac{t e^{-t/2}}{(2coshfrac{t}{2})^2 (pi^2+t^2)},dt$$
equals
$$ -frac{1}{4}int_{mathbb{R}}frac{tsinhfrac{t}{2}}{(t^2+pi^2)cosh^2frac{t}{2}},dt=-int_{mathbb{R}}frac{tsinh t}{(4t^2+pi^2)cosh^2 t},dt. $$
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
$$ frac{1}{cosh x}=sum_{ngeq 0}(-1)^n frac{pi(2n+1)}{frac{pi^2}{4}(2n+1)^2+x^2} $$
and
$$ frac{sinh x}{cosh^2 x} = sum_{ngeq 0}(-1)^n frac{2pi(2n+1)x}{(frac{pi^2}{4}(2n+1)^2+x^2)^2}.$$
Since
$$ int_{mathbb{R}}frac{2pi(2n+1)x^2}{(frac{pi^2}{4}(2n+1)^2+x^2)^2 (pi^2+4x^2)},dx = frac{1}{2pi(n+1)^2}$$
our integral equals $-frac{1}{2pi}eta(2)=color{red}{-frac{pi}{24}}$ by the dominated convergence theorem, allowing to switch $int_{mathbb{R}}$ and $sum_{ngeq 0}$. $frac{1}{24}$ 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 $(1)$.

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

$$begin{aligned} I = int_{0}^{infty}frac{xe^{-x}}{cosh x},mathrm{d}x &= 2int_{0}^{infty}frac{xe^{-x}}{e^{x}}frac{mathrm{d}x}{1+e^{-2x}} = 2sum_{n=0}^{infty}(-1)^{n}int_{0}^{infty}xe^{-(2+2n)x},mathrm{d}x \
&= 2sum_{n=0}^{infty}frac{(-1)^{n}}{4(1+n)^{2}}int_{0}^{infty}ue^{-u},mathrm{d}u = frac{1}{2}sum_{n=0}^{infty}frac{(-1)^{n}}{(1+n)^{2}} \ &= frac{1}{2}sum_{n=1}^{infty}frac{(-1)^{n-1}}{n^{2}} = frac{eta(2)}{2} = frac{pi^{2}}{24}end{aligned}$$

where $eta(s)$ is the Dirichlet eta function.