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We have the integral :

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

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 was stuck, the internal complex $\log$ makes it confusing to do so !

EDIT

The integral is equivalent to : $$\frac{1}{2\pi n}\int_{0}^{\infty}\frac{e^{2\pi i nx}}{(1+x)\left(\frac{4\pi^{2}}{s^{2}}+\log(1+x) \right )}dx$$ Setting $y=\log(1+x)$, it's also equivalent to : $$\frac{1}{2\pi n}\int_{0}^{\infty}\frac{\exp[{2\pi i n \left(e^{y}-1 \right )]}}{\left(\frac{4\pi^{2}}{s^{2}}+y \right )}dy$$

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    $\begingroup$ I am not sure if it helps, but you can check that the integral is equal to $$ \frac{1}{2n\pi} \int_{0}^{\infty} \frac{e^{2\pi i n x}}{(1+x)(\alpha + \log(1+x))} \, dx $$ where $\alpha = 4\pi^2 / s^2$. $\endgroup$ – Sangchul Lee Feb 28 '15 at 4:56
  • $\begingroup$ @sos440 can you show in steps how you arrived at this equality? Now we can split the integral into a real and an imaginary part! $\endgroup$ – Mohammad Al Jamal Feb 28 '15 at 15:26
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    $\begingroup$ I first made integration by parts, then shifted the contour by 90 degree to obtain that integral. $\endgroup$ – Sangchul Lee Mar 1 '15 at 0:06
  • $\begingroup$ I am unclear on how you "shifted the contour by 90 degree" yet still have real limits on the integral. While you might be able to use contour integration in some cases to rotate the limit back to the real positive infinity, it would depend on the value of $s$, as it might put poles within the area bounded by the contour integral. $\endgroup$ – Glen O Mar 3 '15 at 3:01
  • $\begingroup$ @GlenO, Of course, precise derivation should include control over poles and decay speed along the quarter arc that goes to infinity. In this case, thankfully, all these problems are avoided. $\endgroup$ – Sangchul Lee Mar 3 '15 at 16:23
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I am not able to comment due to lack of reputation, but I did manage to split up the imaginary and real components (which is something you seem interested in):

$$ \int^\infty_0 \frac{1}{2}\log\left(\left(1+\frac{s^2\log\left(x^2+1\right)}{8\pi^2}\right)^2+\frac{s^4\arctan\left(x\right)^2}{16\pi^4}\right)\text{e}^{-2\pi n x}+i\arctan\left(\frac{s^2\arctan\left(x\right)}{4\pi^2\left(1+\frac{s^2\log\left(x^2+1\right)}{8\pi^2}\right)}\right)\text{e}^{-2\pi n x}\,\text{d}x $$

Although the real term will still have some imaginary components after integration (from the log functions).

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  • $\begingroup$ This only works for real $s$. In the question, it states that $s$ is a complex parameter. $\endgroup$ – Glen O Mar 3 '15 at 3:05
  • $\begingroup$ Ah, I didn't notice. I think he updated the question after I posted this. $\endgroup$ – Ron Mar 3 '15 at 5:56
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Starting with your integral in terms of $y$;

change $e^{2\pi in(e^y-1)}$ into $e^{2\pi ine^y}e^{-2\pi in}=e^{2\pi ine^y}$ which simplifies it a bit.

Then change variables with $u=y+\frac{4 \pi^2}{s^2}$

Then change variables with v= ln(u)

You get the integral in the form $\int_b^{\infty}exp[ce^{e^v}] dv$

Are there any theorems about integrals of triple exponentials?

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