# 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?

• From MMA: 8 E^(-4 b c) (-( Hypergeometric2F1[4, 2 - (I a)/(2 b), 3 - (I a)/(2 b), -E^(-2 b c)]/(a + 4 I b)) + ( E^(8 b c) Hypergeometric2F1[4, 2 - (I a)/(2 b), 3 - (I a)/(2 b), -E^(2 b c)])/( a + 4 I b) + (-Hypergeometric2F1[4, 2 + (I a)/(2 b), 3 + (I a)/(2 b), -E^(-2 b c)] + E^(8 b c) Hypergeometric2F1[4, 2 + (I a)/(2 b), 3 + (I a)/(2 b), -E^(2 b c)])/(a - 4 I b)) – Mariusz Iwaniuk May 14 '17 at 9:53
• if $\left(\left(e^{-c} \Re\left(i^{\frac{1}{b}}\right)\geq 1\lor e^{-c} \Re\left(i^{\frac{1}{b}}\right)\leq 0\right)\land \left(\Re\left(i^{\frac{1}{b}}\right)=0\lor e^c \Re\left(i^{\frac{1}{b}}\right)=1\lor e^c \Re\left(i^{\frac{1}{b}}\right)\geq 1\lor e^c \Re\left(i^{\frac{1}{b}}\right)\leq 0\right)\right)\lor i^{\frac{1}{b}}\notin \mathbb{R}$ – Mariusz Iwaniuk May 14 '17 at 9:53

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.