Convoluted Lorentzian and Fourier Transformation To describe a measurement, I have to calculate the convolution of three functions $f,g,h$:


*

*$f(x)=\frac{1}{(W/2)^2+x^2} \, , W>0$

*$g(x)=e^{\beta x} \frac{(\beta x-2)e^{\beta x}+\beta x+2}{(e^{\beta x}-1)^3} \, ,\beta >0$

*$h(x)=\begin{cases}
  (1 - x^2)^{3/2} &,\,   |x| < 1 \\
  0 &,\, |x| \geq 1
  \end{cases}$  


So the goal is to calculate $s=f*g*h$. I tried using Fourier transformation:


*

*$(\mathcal F f)(k)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) e^{ikx} dx = \frac{\sqrt{2\pi}}{W}e^{-|k|\frac{W}{2}}$

*$(\mathcal F g)(k)=\sqrt{\frac{\pi^3}{2}}\frac{1}{\beta^3} \left(\frac{k}{\mathrm{sinh}^2(\pi k/\beta)}\right)^2$

*$(\mathcal F h)(k)=\frac{3}{2}\sqrt{2\pi}\frac{J_2(k)}{k^2} \, , \text{ where } J_2 \text{ is a Bessel function of first kind} $


Using the convolution theorem leads to
$$s(x)=c \int_{-\infty}^\infty e^{-|k|\frac{W}{2}}J_2(k)\frac{e^{-ikx}}{\mathrm{sinh}^2(\pi k /\beta)} dk \, ,$$
where $c$ contains the prefactors independent of $k$ or $x$. Is it possible to analytically solve this integral (I thought about the residue theorem, but the integrand is not holomorphic due to $|k|$)?

Let's consider $k \ge 0$ first:
$$\int_0^{\infty} dk \, J_2(k) \, \frac{e^{-a k}}{\sinh^2{b k}}$$
where $a = (W/2) - i x$ and $b = \pi/\beta$.  We may rewrite as
$$\int_0^{\infty} dk \, J_2(k) \, e^{-(a+2 b) k} \left (1-e^{-2 b k}\right)^{-2}$$
We may write out the full Taylor expansion of the quantity in parentheses to get
$$\sum_{m=0}^{\infty} (m+1) \int_0^{\infty} dk \, J_2(k) \, e^{-[a+2 (m+1) b] k} $$
Now use the Laplace transform:
$$\int_0^{\infty} dx \, J_2(t x) \, e^{-p x} = \frac{t^2}{\sqrt{p^2+t^2} (p+\sqrt{p^2+t^2})^2}$$
to get that
$$\int_0^{\infty} dk \, J_2(k) \, \frac{e^{-a k}}{\sinh^2{b k}} = \sum_{m=1}^{\infty} \frac{m}{\sqrt{1+[a+2 m b]^2} \left [a+2 m b + \sqrt{1+[a+2 m b]^2} \right ]^2} $$ 
Considering $k < 0$, we should get the same thing, with the sign of $x$ reversed, i.e., take the conjugate of $a$.  Thus, the final result will be twice the real part of the above sum.  That's about as good as I can get it for now.
