How to check analytically that the following integral equations has this solution?

Question

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

Answer 1

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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

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

Answer 2

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 $$ \chi(s) = \frac{1}{2\pi}\int_0^\infty \frac{1}{ik-1}\frac{1}{2}k^4e^{-k}(e^{iks}-e^{-iks})dk =-\frac{1}{2\pi}\int_0^\infty \frac{ik^4}{1-ik}e^{-k}\sin (ks)dk. $$