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Referring to a previous question, i want help with the integral :

$$\int_{0}^{\infty}\frac{\log(1\pm ix)^{2}}{\left(\frac{t}{2}\log(1 \pm ix) \right )^{2}-\pi ^{2}n^{2}}e^{-2\pi mx}dx$$ Where $n,m$ are positive integers, and $t$ is a real variable. I have tried repeated integration by parts, but it becomes very confusing after a couple of steps.

Hints: $$\int\frac{\log(1+ix)^{2}}{\left(\frac{t}{2}\log(1 + ix) \right )^{2}-\pi ^{2}n^{2}}dx=\frac{4\pi in}{t}\left(e^{-2\pi n/t}\text{Ei}\left(\log(1+ix)+\frac{2\pi n}{t}\right)-e^{2\pi n/t}\text{Ei}\left(\log(1+ix)-\frac{2\pi n}{t}\right) \right )+\frac{4x-4i}{t^{2}}$$

$$\int \text{Ei}\left(\log(1+ix)\pm\frac{2\pi n}{t}\right)dx=(x-i)\text{Ei}\left(\log(1+ix)\pm\frac{2\pi n}{t}\right)+ie^{\mp 2\pi n/t}\text{Ei}\left(2\log(1+ix)\pm\frac{4\pi n}{t}\right)$$

EDIT

We notice that:

$$\frac{\log(1+ ix)^{2}}{\left(\frac{t}{2}\log(1 + ix) \right )^{2}-\pi ^{2}n^{2}}=\frac{4\log(1+ix)}{t^{2}}\int_{0}^{\infty}\sinh\left(\frac{2\pi ny}{t} \right )(1+ix)^{-y}dy$$

So we need to evaluate : $$f(y,m)=\int_{0}^{\infty}\log(1+ix)(1+ix)^{-y}e^{-2\pi m x}dx$$ And :

$$\frac{4}{t^{2}}\int_{0}^{\infty}f(y,m)\sinh\left(\frac{2\pi ny}{t} \right )dy$$

But : $$f(y,m)=-\frac{d}{dy}\left[\int_{0}^{\infty}(1+ix)^{-y}e^{-2\pi m x}dx\right]$$

And the problem reduces to this last integral !

EDIT 2 $$(1+ix)^{-y}=\sum_{k=0}^{\infty}i^{k}x^{k}\frac{\Gamma(1-y)}{k!\Gamma(1-k-y)}$$ Thus : $$\int_{0}^{\infty}(1+ix)^{-y}e^{-2\pi mx}dx=\sum_{k=0}^{\infty}i^{k}\frac{\Gamma(1-y)}{\Gamma(1-k-y)}\frac{1}{(2\pi m)^{k+1}}$$ $$=i(-2\pi i m)^{y-1}\Gamma(1-y,-2\pi i m )$$

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  • $\begingroup$ What is your final question? Or have you already solved everything yourself? Then maybe you should consider separating your solution from the question and posting it as an answer $\endgroup$ – Yuriy S Mar 16 '16 at 12:03

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