Evaluate $\int_{-\infty}^{\infty}\frac{\sinh\left(y\sqrt{\alpha^2-\omega^2}\right)}{\sinh\left(H\sqrt{\alpha^2-\omega^2}\right)}e^{i\alpha x}d\alpha$

Question

I need to evaluate the Fourier inverse integral

$\displaystyle \int_{-\infty}^{\infty}\frac{\sinh\left(y\sqrt{\alpha^2-\omega^2}\right)}{\sinh\left(H\sqrt{\alpha^2-\omega^2}\right)}e^{i\alpha x}d\alpha \tag*{}$

which arose while solving a PDE.

Here, $H>0,x\in\mathbb{R},y\in[0,H]$. The domain of $\omega$ was not given in the original problem, but I am going to assume $\omega>0$ for simplicity.

The problem asks us to introduce proper branch cuts for the square root function before evaluating the integral.

For reference, this is the original question. My attempt up until I get the integral is also shown in the link.

My Attempt

The branch points of the square root functions are at $\alpha =\pm\omega$. So, I considered the following branch cuts and contours.

(Here $\omega$ has an absolute value, but you can ignore it and assume $|\omega|=\omega$)

We, firstly, need to find the poles of the integrand in the upper half-plane. Those are given by the equation

$H\sqrt{\alpha^2-\omega^2}=n\pi i\tag*{}$

Solving this, we obtain

$\displaystyle \alpha =\pm\sqrt{\omega^2-\frac{n^2\pi^2}{H^2}}\tag*{} $

where $n=0,1,2,\cdots$.

The problem is that some of those poles are on the branch cuts depending on the parameters. I have been told this is not permissible, so I am not sure how to proceed.

Edit: The statement that "some of those poles are on the branch cuts" is not correct. It is on the real axis but between $[-\omega,\omega]$.

Edit2: Tables of Fourier Transforms and Fourier Transforms of Distributions by Fritz Oberhettinger states(p37,7.48) that if $a<b$, we have

$\displaystyle \int_{0}^{\infty} \frac{\sinh{(a\sqrt{k^2+x^2}})}{\sinh{(b\sqrt{k^2+x^2}})}\cos{xy}dx = -\pi b^{-1} \sum_{n=0}^{\infty}(-1)^nc_n\sin{(ac_n)}v_n^{-1}e^{-yv_n}$

where $c_n=n\pi/b$, $v_n = (k^2+c_n^2)^{1/2}$. I would guess our integral would have a similar form.

Answer 1

Even though the branch cut $(-\infty,0)$ of the square-root transfers to $\sinh(\sqrt{z})$, the integral actually does not possess any branch cut. The only branch-cut that could possibly occur, would be $(-\omega,\omega)$. However, the phase-jump cancels out in ratio: if for some $\Re(\alpha) \in (-\omega,\omega)$, $\alpha$ goes from the upper half-plane into the lower half-plane, $\sqrt{\alpha^2-\omega^2}$ changes from being of the form $\pm i\cdot r$ to $\mp i\cdot r$, for some real number $r$, but $$\sinh(\pm i r)=\pm i \sin(r) \, .$$ Hence, no jump i.e. no branch-cut.

Since you already know how to obtain the result from the residue theorem, here is a proof that shows the vanishing of the semicircle. However, you do not need to use a semi-circle. You can also take a box of side-length $R$, which turns out to be a bit easier.

Writing $\alpha=r+is$, the contour for $x>0$ is $$C_R=\{r=R,s\in(0,R)\} \cup\{r\in(R,-R),s=R\} \cup \{r=-R,s=(R,0)\} \, ,$$ from which we'll find $$\left|\int_{C_R}\frac{\sinh\left(y\sqrt{\alpha^2-\omega^2}\right)}{\sinh\left(H\sqrt{\alpha^2-\omega^2}\right)} \, e^{i\alpha x} \, d\alpha \right| \leq \int_{0}^R \left|\frac{\sinh\left(y\sqrt{(R+is)^2-\omega^2}\right)}{\sinh\left(H\sqrt{(R+is)^2-\omega^2}\right)} \right| \, e^{-xs} \, ds \\ +\int_{-R}^{R} \left|\frac{\sinh\left(y\sqrt{(r+iR)^2-\omega^2}\right)}{\sinh\left(H\sqrt{(r+iR)^2-\omega^2}\right)} \right| \, e^{-xR} \, dr + \int_{0}^R \left|\frac{\sinh\left(y\sqrt{(-R+is)^2-\omega^2}\right)}{\sinh\left(H\sqrt{(-R+is)^2-\omega^2}\right)} \right| \, e^{-xs} \, ds \\ = I_1 + I_2 + I_3 \, .$$ Now we notice:

1. The first and third integral are the same, since the arguments of the absolute value function are just complex-conjugate to each other.
2. As $|R+is|$ is large, i.e. $$\sqrt{(R+is)^2-\omega^2} = R + is + O\left( \frac{1}{|R+is|} \right)$$ we can remove the square-root. The error made will be small compared to the main term, since $$\sinh(R+is+\epsilon) = \sinh(R+is) \left\{ \cosh(\epsilon) + \coth(R+is) \sinh(\epsilon) \right\} \\ = \sinh(R+is) \left\{1+O(\epsilon)\right\}$$ where $\epsilon=O\left(|R+is|^{-1}\right)$ is small.
3. The following identity holds for all $x,y \in \mathbb{R}$ $$\left| \sinh\left(x+iy\right) \right|^2 = \sinh^2(x) + \sin^2(y) \, .$$
4. In estimating the contribution along the box, we need to avoid the poles of $1/\sinh$ on the imaginary axis at $in\pi$. As $R\rightarrow \infty$, we can assume $HR=(n+1/2)\pi$ as $n \rightarrow \infty$, for integers $n$.

Using these observations, we can estimate as follows: $$I_1 \leq \int_0^R \sqrt{\frac{\sinh^2(yR)+\sin^2(ys)}{\sinh^2(HR)+\sin^2(Hs)}} \, e^{-xs} \, {\rm d}s \leq \int_0^R {\frac{\cosh(yR)}{\sinh(HR)}} \, e^{-xs} \, {\rm d}s \leq \frac{\cosh(yR)}{\sinh(HR)} \frac{1}{x} \\ I_2 \leq \int_{-R}^R \sqrt{\frac{\sinh^2(yr)+\sin^2(yR)}{\sinh^2(Hr)+\sin^2(HR)}} \, e^{-xR} \, {\rm d}r \leq \int_{-R}^R {\frac{\cosh(yr)}{\cosh(Hr)}} \, e^{-xR} \, {\rm d}r \leq 2R e^{-xR} \, .$$