How to check analytically that the following integral equations has this solution? Consider the following integral equation
$$
\chi(s)= \frac{12}{\pi} \frac{s(1-s^2)}{(1+s^2)^4} + \frac{1}{i\pi} \int_0^\infty \chi(t) \ln \left| \frac{s+t}{s-t} \right| \, \mathrm{d} t \, , \quad\quad (s\ge 0) \, , 
$$
which has a solution of the form
$$
\chi(s) = -\frac{1}{2\pi} \int_0^\infty \frac{i q^4}{1-iq} e^{-q} \sin (qs) \, \mathrm{d}q \, .
$$
This solution has been obtained by solving the initial problem differently and can be checked numerically to be correct. 
I was wondering whether analytical treatment is possible in order to confirm that the solution verifies the above integral equation.
Your helps welcome.
Thanks
RF
 A: 
Proof of the formula in the comment

$$
I=-\int_{-1}^{1}\frac{\log(|t|)}{(1+t)^2}\sin\left(a \frac{1-t}{1+t}\right)dt=\frac{\pi}{2 a}\sin(a)
$$

Fractional linear transformation $\frac{1-t}{1+t}=x$
$$
I=\frac{1}{2}\int_{0}^{\infty}\log\big|\frac{1-x}{1+x}\big|\sin\left(a x\right)dx
$$
split 
$$
\frac{1}{2}\int_{0}^{1}\log\big(\frac{1-x}{1+x}\big)\sin\left(a x\right)dx+\frac{1}{2}\int_{1}^{\infty}\log\big(\frac{x-1}{x+1}\big)\sin\left(a x\right)dx
$$ 
Integration by part is valid because the divergent boundary terms from both integrals cancel out 
$$
I=I_1+I_2=\frac{1}{a}\int_{0}^{1_-}\frac{\cos\left(a x\right)}{x^2-1}dx+\frac{1}{a}\int_{1_+}^{\infty}\frac{\cos\left(a x\right)}{x^2-1}dx
$$ 
where $1_{\pm} =\lim_{x\rightarrow 1\pm\epsilon}$
to proceed we perform a partial fraction decompositon
$$
2 a I_1=\int_0^{1-}\frac{\cos(ax)}{1-x}-\int_0^{1}\frac{\cos(ax)}{1+x}=\\
\cos(a)\int_{0_+}^{2a}\frac{\cos(q)}{q}+\sin(a)\int_{0}^{2a}\frac{\sin(q)}{q}
$$
and
$$
2aI_2=\int_{1+}^{\infty}\frac{\cos(ax)}{1-x}-\int_{1}^{\infty}\frac{\cos(ax)}{1+x}=\\
-\cos(a)\int_{0_+}^{2a}\frac{\cos(q)}{q}+\sin(a)\int_{0}^{\infty}\frac{\sin(q)}{q}+\sin(a)\int_{2a}^{\infty}\frac{\sin(q)}{q}
$$
Now using a famous integral, the magic happens
$$
2aI_1+2aI_2=2 \sin(a)\int_0^{\infty}\frac{\sin(q)}{q}=\pi \sin(a)
$$
and

$$
I=I_1+I_2=\pi\frac{\sin(a)}{2
 a}
$$

A: We can also show the solution directly, after some manipulation. First, the function $\chi(s)$ is extended to the whole real axis as an odd function by $\chi(-s)=-\chi(s)$ if $s<0$. Then the logarithmic integral can be written as
$$
-\frac{1}{\pi i}\int_{-\infty}^\infty \chi(t)\ln|s-t|dt
$$
by using the fact that
$$
\frac{1}{\pi i}\int_0^\infty \chi(t)\ln|s+t|dt = 
\frac{1}{\pi i}\int_{-\infty}^0 \chi(-t)\ln|s-t|dt
=-\frac{1}{\pi i}\int_{-\infty}^0 \chi(t)\ln|s-t|dt.
$$ 
The original integral becomes
$$
\chi(s)= \frac{12}{\pi} \frac{s(1-s^2)}{(1+s^2)^4}
-\frac{1}{\pi i}\int_{-\infty}^\infty \chi(t)\ln|s-t|dt.
$$
Taking derivative w.r.t $s$ on both sides
$$
\chi'(s) = \frac{12}{\pi}\frac{5s^4-10s^2+1}{(1+s^2)^5}+iP.V\frac{1}{\pi}\int_{-\infty}^\infty \frac{\chi(t)}{s-t}dt.
$$
The principal integral $P.V\frac{1}{\pi}\int_{-\infty}^\infty \frac{\chi(t)}{s-t}dt$ on the right hand is exactly the Hilbert transform of $\chi$. Now take the Fourier transform ($\hat{f}(k)=\int_{-\infty}^\infty f(x)e^{-ikx}dx$) of both sides, we get
$$
ik\hat{\chi}(k)=\frac{1}{2}k^4e^{-|k|}+\mbox{sign}(k)\hat{\chi}(k)
$$
or
$$
\hat{\chi}(k) = \frac{1}{ik-\mbox{sign}(k)}\frac{1}{2}k^4e^{-|k|}.
$$
Then $\chi$ is given by the inverse Fourier transform
$$
\chi(s) = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{\chi}(k)e^{iks}dk
=\frac{1}{2\pi}\int_0^\infty 
\frac{1}{ik-1}\frac{1}{2}k^4e^{-k}e^{iks}dk
+\frac{1}{2\pi}\int_{-\infty}^0
\frac{1}{ik+1}\frac{1}{2}k^4e^{k}e^{iks}dk.
$$
Finally, convert the variable in the second integral on $(0,\infty)$
$$
\frac{1}{2\pi}\int_{-\infty}^0
\frac{1}{ik+1}\frac{1}{2}k^4e^{k}e^{iks}dk
=\frac{1}{2\pi}\int_0^\infty \frac{1}{-ik+1}\frac{1}{2}k^4e^{-k}e^{-iks}dk,
$$
and we get
$$
\chi(s) = \frac{1}{2\pi}\int_0^\infty 
\frac{1}{ik-1}\frac{1}{2}k^4e^{-k}(e^{iks}-e^{-iks})dk
=-\frac{1}{2\pi}\int_0^\infty \frac{ik^4}{1-ik}e^{-k}\sin (ks)dk.
$$
