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I've been trying to evaluate this integral without much success: $\displaystyle \int_{-\infty}^\infty dx\, e^{iax} \frac{1- e^{-c\sinh^2 bx}}{\sinh^2 bx}$

I've tried contour integration. There are no poles, but I can't find a suitable contour such that all contributions other than that from the real line disappear. If anyone can help with evaluating/ approximating this in the regime where $c$ is large, that would be greatly appreciated.

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  • $\begingroup$ Any reason to believe that there is a closed form? $\endgroup$ – tired Jan 12 '16 at 10:08
  • $\begingroup$ Not really. But it's the definite integral that I'm after. Any help in obtaining an approximate answer would also be very useful. this is what I can do: I can obtain a contour integral of $ \displaystyle \int_{-\infty}^\infty dz\, e^{iax} \frac{1- e^{-c\sinh^2 bz}}{\sinh^2 bz}$ $\endgroup$ – Nirmalya Kajuri Jan 12 '16 at 10:19
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    $\begingroup$ then i would suggest that u also add the "asymtptotics" and or the "approximation" tag $\endgroup$ – tired Jan 12 '16 at 10:21
  • $\begingroup$ ..with the rectangular contour :$ (-\infty, -i\pi/4)$ to $(\infty,-i\pi/4)$ to $(\infty, i\pi/4)$ to $(-\infty, -i\pi/4)$ to $(-\infty, -i\pi/4)$. This is zero due to the absence of poles, and the contributions at $\pm \infty$ vanish. Also the integrals parallel to the real line are proportional. So I get : $ \displaystyle \int dz\, e^{iax} \frac{1- e^{-c\sinh^2 bz}}{\sinh^2 bz} = 0$ along the lines z=$(-\infty, \pm i\pi/4)$ to $(\infty, \pm i\pi/4)$. @tired: OK I will do that, thanks for the suggestion. $\endgroup$ – Nirmalya Kajuri Jan 12 '16 at 10:28
  • $\begingroup$ Furthermore u could scale away one parameter. that makes things easier :) $\endgroup$ – tired Jan 12 '16 at 10:31
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Let's begin by converting this to a simpler double integral:

$$\begin{align}b \int_{-\infty}^{\infty} dx \, e^{i a x/b} \int_0^c du \, e^{-u \sinh^2{x}} &= b \int_0^c du \int_{-\infty}^{\infty} dx \, e^{i a x/b} \, e^{-u \sinh^2{x}}\\ &=b \int_0^c du \, e^{u/2}\,\int_0^{\infty} dx \, \cos{\left (\frac{a x}{2 b} \right )} e^{-(u/2) \cosh{x}} \\ &= 2 b\int_0^{c/2} du \, e^{u} \, K_{i \frac{a}{2 b}} \left (u \right ) \end{align}$$

I turned to Mathematica for this integral, and it is a little ugly but could be worse as an exact representation of the integral:

$$-b \operatorname{Re}{\left [2^{i a/b} c^{1-i a/(2 b)} \Gamma \left (-1-i \frac{a}{2 b} \right ) \, _2F_2 \left (\begin{array} \\ \frac12+i \frac{a}{2 b} & 1+i \frac{a}{2 b}\\2+i \frac{a}{2 b}&1+i \frac{a}{b} \end{array} ; c\right ) \right ]} $$

I did a numerical verification of this using $a=2$, $b=1$, and plotted against $c \in [0,1]$ and got indistinguishable plots.

Of course, we can debate whether this is truly a useful closed form. Within Mathematica, the answer is clear - it sure is. Mathematica can compute the closed form in a small fraction of the time it took to compute a numerical approximation to the integral (and it had a hard time doing so).

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    $\begingroup$ The nice thing is that ur integral is welled suited for an asymptotic analysis , because the asymptotics of bessel functions are well known :) (+1) $\endgroup$ – tired Jan 12 '16 at 12:53
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    $\begingroup$ Of course, it would be nice if the OP could specify what approximation (s)he wants. Is $a$ large? Is $c$? Or are they small? Or is it $b$ that is restricted? Hard to do asymptotic/approximations when we have no idea about the domains of our inputs. $\endgroup$ – Ron Gordon Jan 12 '16 at 13:03
  • $\begingroup$ I was going to suggest something similar: Differentiating under the integral sign with respect to c. $\endgroup$ – Lucian Jan 12 '16 at 14:03
  • $\begingroup$ Are you quite sure about the integration over x? because mathematica is not giving me anything on that one. Sorry, I should have specified the domain for approximation. $c$ is large. $\endgroup$ – Nirmalya Kajuri Jan 12 '16 at 14:04
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    $\begingroup$ @NirmalyaKajuri: I am quite sure about that one. Here's a ref: dlmf.nist.gov/10.32#i (#10.32.9). For this, use the fact that $\cos{z} = \cosh{(i z)}$. Besides, I checked the numerics before posting. $\endgroup$ – Ron Gordon Jan 12 '16 at 14:08

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