Evaluate $\int_{-\infty}^{\infty}\sin(ax)\operatorname{sech}^4(b(x-c))dx$, for $a,b,c>0$ $\newcommand{\sech}{\operatorname{sech}}$
I am trying to compute the following integral:
$$\int_{-\infty}^{\infty}\sin(ax)\sech^4(b(x-c))dx; a,b,c>0$$
I cannot seem to tackle this integral. I have tried integration by parts, but this leads nowhere since we need to evaluate $\sin(ax)$ at the infinities. How should I be approaching this integral?
 A: $\newcommand{\sech}{\operatorname{sech}}$
$\newcommand{\res}{\operatorname{Res}}$
Put $x=z/b+c$ and use angle-addition:
$$
\int_{-\infty}^{\infty} \sin(ax)\sech ^4(b(x-c))\,dx
$$
$$
=\frac{1}{b}\int_{-\infty}^{\infty} \sin(a(z/b+c))\sech ^4(z)\,dz
$$
$$
=\frac{\sin(ac)}{b}\int_{-\infty}^{\infty} \cos(\frac{a}{b}z)\sech ^4(z)\,dz+\frac{\cos(ac)}{b}\int_{-\infty}^{\infty} \sin(\frac{a}{b}z)\sech ^4(z)\,dz
$$Now $\sech^4$ is improperly integrable (in fact, rapidly decaying) and even, so the second integral vanishes by symmetry. Note that
$$
\int_{-\infty}^{\infty} \cos(\frac{a}{b}z)\sech ^4(z)\,dz=\Re\left(\int_{-\infty}^{\infty} \exp(\frac{a}{b} i z)\sech ^4(z)\,dz\right)
$$ We will use the Residue Theorem. Let $f(z)$ denote the integrand. A pole of order four occurs at $z=\pi i/2$, so it makes sense to take our contour as the rectangle $\gamma$ (with sub-contours $\gamma_1,\gamma_2,\gamma_3,\gamma_4$) whose path is $$\gamma:-R\longrightarrow R \longrightarrow R+\pi i \longrightarrow -R+\pi i \longrightarrow -R.$$
The details are left to the reader (or to Mathematica or some such), but we have:
$$
\int_{\gamma_3}f(z)\,dz = -\exp(-\frac{a}{b}\pi) \int _{\gamma_1}f(z)\,dz;\\
\lim_{R\to \infty} \int_{\gamma_2}f(z)\,dz=\lim_{R\to \infty} \int_{\gamma_4}f(z)\,dz=0
$$
Computing the residue is tedious, but the end result is:
$$
\res(f(z),\pi i /2)
 = \frac{1}{6}  \lim_{z\to \pi i/2} \frac{d^3}{dz^3}\frac{\exp(\frac{a}{b} iz)(z-\pi i/2)^4}{\sech^4(z)}
$$
$$
= -\frac{i a \left(a^2+4 b^2\right) \exp(-\frac{a}{b}\pi/2)}{6 b^3}
$$ Then in the limit, we have
$$
\lim_{R\to\infty}\int _{\gamma_1}f(z)\,dz = \Re\left((1-\exp(-\frac{a}{b}\pi))^{-1}\cdot 2\pi i \cdot \left(\frac{-i a \left(a^2+4 b^2\right) \exp(-\frac{a}{b}\pi/2)}{6 b^3}\right)\right)
$$
$$=\frac{\pi  a \left(a^2+4 b^2\right) \text{csch}\left(\frac{\pi  a}{2 b}\right)}{6 b^3}
$$
So in summary,
$$
\int_{-\infty}^{\infty} \sin(ax)\sech ^4(b(x-c))\,dx =  \frac{\pi  a \left(a^2+4 b^2\right) \text{csch}\left(\frac{\pi  a}{2
   b}\right) \sin (a c)}{6 b^4}
$$

Note: originally, I took the contour to be the semicircle enclosing the upper half plane, with poles at $t=\pi i(k+1/2)$, $k\in\mathbb{N}^0$. Their sum is a geometric series but showing the integral vanished along the circular arc was slightly tricky because the integrand is large, of size $ \cosh(\frac{a}{b}R)\sec^4(R)$, near $\theta=\pi/2$. If one were careful with the approximation by bounding the size of this region (the integrand rapidly decreases elsewhere), you could do the problem the same way.
