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I have a Fredholm integral equation of second kind

$$\frac{-1}{2\pi \omega'^2}+\int_{-\infty}^{\infty}\frac{1}{\pi \omega'^2}\frac{\psi(\omega)d\omega}{{\text{sinch}}((\omega-\omega')\pi/2)}=\psi(\omega'),$$

where $\text{sinch}(x)=\sinh(x)/x.$ I've been trying to solve for $\psi$ by putting $\psi(\omega)=\sum_{n=0}^\infty a_n\omega^n$. However the integral $$a_n\int_{-\infty}^{\infty}\frac{1}{\pi \omega'^2}\frac{\omega^nd\omega}{{\text{sinch}}((\omega-\omega')\pi/2)},$$ seems to be cumbersome.

Any thoughts on how to solve these types of integrals would be appreciated. I also failed to find it in any integral table.

AB

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It is sufficient to compute $$ \int_{-\infty}^\infty \frac{w^n dw}{\mathrm{sinch}(\pi w/2)}, $$ since you can shift by $w'$ and expand the numerator in the original integral. Now, this integral converges very well, and there is no harm in adding a factor $\exp(i\epsilon\omega)$ to the integrand, for any $\epsilon>0$. However, now you can deform the contour into the upper half-plane with no harm for the convergence. In this way you can reduce to sum of the residues at the poles $w=2ik$, $k=1,2,\ldots$ $$ \int_{-\infty}^\infty \frac{w^n e^{i\epsilon w} dw}{\mathrm{sinch}(\pi w/2)}=2\pi i\sum_{k=1}^\infty \mathrm{res}(2ik)=-4\pi \sum_{k=1}^\infty(2ik)^nk(-1)^ke^{-2\epsilon k}=2\pi i^n(-\partial_\epsilon)^{n+1}\frac{1}{e^{2\epsilon}+1}. $$ Now, we have for Euler polynomials $$ \frac{2e^{xt}}{e^t+1}=\sum_{k=0}^\infty E_k(x)\frac{t^k}{k!}, $$ so that $$ \frac{1}{e^{2\epsilon}+1}=\sum_{k=0}^\infty 2^{k-1}E_k(0)\frac{\epsilon^k}{k!}, $$ giving for the integral $$ \left.\int_{-\infty}^\infty \frac{w^n e^{i\epsilon w} dw}{\mathrm{sinch}(\pi w/2)}\right\vert_{\epsilon=0}=\left.2\pi i^n(-\partial_\epsilon)^{n+1}\frac{1}{e^{2\epsilon}+1}\right\vert_{\epsilon=0}=-2\pi (-2i)^nE_{n+1}(0). $$ So finally $$ \int_{-\infty}^\infty \frac{w^n dw}{\mathrm{sinch}(\pi w/2)}=-2\pi (-2i)^nE_{n+1}(0). $$ This is always real, since both sides vanish for odd $n$.


Note that the Euler polynomials form an Appell sequence, and so have a fundamental property $$ E_n(x+y)=\sum_{k=0}^n \binom{n}{k}E_k(x)y^{n-k}, $$ so a better statement of the result is that (note the change from $\mathrm{sinch}$ to $\sinh$) $$ \int_{-\infty}^\infty \frac{(w+w')^n dw}{\mathrm{sinh}(\pi w/2)+i0}=(-2i)^{n+1}E_{n}\left(\frac{iw'}{2}\right), $$ where $+i0$ in the denominator tells you that your integration contour should go above the pole at $w=0$ in the complex plane. Curiously, this integral representation of Euler polynomials seems to be absent from Wolfram functions site. However, I guess it should be known.

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    $\begingroup$ To meet the formula of the OP the nominator of the integrand in the very last formula should be $w (w+w')^{n-1}$, which then gives a less neat result. If you want to (in a sense) double check: The very first integral in the answer can also be found in Gradshteyn & Ryshik, formula 3.523 2. $\endgroup$ – Johannes Trost May 9 '15 at 10:57
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    $\begingroup$ @JohannesTrost, Thanks for the reference! Regarding the integral of the OP, it can be easily reduced to the integral computed in the middle of the answer. I just included the neat version because it is neat :) $\endgroup$ – Peter Kravchuk May 9 '15 at 23:12

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