We have the integral :

$$\int_{0}^{\infty}\log\left(1+\frac{s^{2}}{4\pi^{2}} \log^{2}(1+ix)\right ) e^{-2\pi nx}dx$$

Where s is a complex parameter, and n is a positive integer. The integral converges by virtue of the exponential factor. I tried to deform the path of integration such that we avoid the branch cut(s) of the logarithm. But here is where i got stuck, the internal complex log makes it confusing to do so !

A different version of this problem was posted here with no answers.

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    $\begingroup$ What exactly do you want to do? To compute the integral or merely to find an appropriate path? $\endgroup$ – Alex Degtyarev Mar 12 '15 at 21:32
  • $\begingroup$ i want to compute the integral ! $\endgroup$ – mohammad-83 Mar 12 '15 at 22:29
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    $\begingroup$ AFAIK there is no branch cut here. The arguments of both log's have positive imaginary parts. $\endgroup$ – Fan Zheng Mar 13 '15 at 3:39
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    $\begingroup$ In Mathematics Stack Exchange the inner log was not squared. What is the motivation that allow these differences? $\endgroup$ – juan Mar 15 '15 at 18:15
  • $\begingroup$ Originally, i wanted to compute the integral with the inner log squared. mistakingly, i dropped the squared sign, and then quickly realised that we can factor the argument of the outer log, and have two integrals without it being there. $\endgroup$ – mohammad-83 Mar 15 '15 at 18:29

$$I_n(s)=\int_{0}^{\infty}\log\left(1+\frac{s^{2}}{4\pi^{2}} \log^{2}(1+ix)\right ) e^{-2\pi nx}dx$$

a closed-form evaluation of this integral does not look promising, but small and large-$|s|$ asymptotics is doable:

$$I_n(s)=\frac{s^2}{48\pi^3 n}\left(24 i \pi n \; _3F_3(1,1,1;2,2,2;2 i n \pi )-\tfrac{3}{2} [\pi +2 i \log (2 \pi n)] [2 i \log (2 \pi n)+4 i \gamma +\pi ]+6 \gamma ^2+\pi ^2\right)+{\rm order}(|s|^4)$$

  • large $|s|$:

$$I_n(s)=\frac{1}{\pi n}\log(s)+{\rm order}(|s|^0)$$

  • and for large $n$ the integral decays as $1/n^3$,

$$\lim_{n\rightarrow\infty} n^3 I_n(s)=-\frac{s^2}{16\pi^5}$$

  • $\begingroup$ Thanks Carlo ... i am still hoping for an explicit evaluation, regardless of how complicated it looks ! $\endgroup$ – mohammad-83 Mar 16 '15 at 21:45

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