I've been trying to evaluate this integral without much success: $$ \int_{-\infty}^{\infty}\mathrm{e}^{\mathrm{i}ax}\,\, \frac{1 - \mathrm{e}^{-c\sinh^{\,2}\,\left(bx\right)}}{\sinh^{2}\left(bx\right)}\,\,\mathrm{d}x $$ 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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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).

| cite | improve this answer | |
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.