Can this integral be evaluated/approximated?

Question

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.

Answer 1

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).