# Integral that gives one or minus one

Consider following integral

$$\int_{-\infty }^{\infty } \frac{\text{sech}\left(\frac{\pi (-a+x+1)}{c}\right)}{c} \, dx$$ with $$c$$ real and $$a$$ is a complex number of the form $$e^{i\phi}$$ for real $$\phi$$.

Looks like this integral is either 1 or -1 which is what I got from Mathematica.

How do I determine the range of $$a$$ (or $$c e^{i\phi}$$) for which this integral is positive 1 and range for which it is $$-1$$?

I suspect this is related to $$$$\frac{2 \left(\tan ^{-1}(x)+\cot ^{-1}(x)\right)}{\pi }=\pm 1$$$$ for complex x. But I am not quite sure.

• There should be a typo, since my calculations show that this doesn't even make sense. Commented Feb 23, 2023 at 6:04
• There is no typo. If you take $a=i c$, you get -1 and if you take $a=2ic$, you get 1, for example. Commented Feb 23, 2023 at 6:20
• Okay, I see. try converting the hyperbolic secant into its exponential form and see what happens. Commented Feb 23, 2023 at 6:23

Let $$y=\frac{\pi(x+1)}{c}$$ and $$A=\frac{a\pi}{c}$$. Then the integral becomes $$\int_{-\infty}^{\infty}\frac{2}{\pi} \frac{e^{y-A}}{e^{2(y-A)}+1} dy.$$ If $$A$$ were real, this would be (translating out $$A$$) $$\left|\frac{2}{\pi} \arctan\big(e^y\big)\right|_{-\infty}^{\infty}=1.$$ The effect of $$A$$ not being real is to shift the line of integration in the imaginary direction by $$\mathrm{Im } A$$ to get $$\int_{-\infty+ i\mathrm{Im}A}^{\infty+ i\mathrm{Im}A}\frac{2}{\pi} \frac{e^{y}}{e^{2y}+1} dy.$$ and this will affect the integral by moving across poles of the integrand. I think the residue at each pole is $$2$$ or $$-2$$, so I would expect the value of the integral to alternate between $$\pm 1$$ as the imaginary part of $$\frac{a}{c}$$ increases, changing each time $$\mathrm{Im} A$$ passes a pole, at $$(n+\frac{1}{2})\pi$$.
$$\int_{-\infty }^{\infty } \frac{\text{sech}\left(\frac{\pi \left(e^{i a}+x-1\right)}{c}\right)}{c} \, dx=\frac{2}{\pi } \left(\tan ^{-1}\left(e^{\frac{\pi \left(-1+e^{i a}\right)}{c}}\right)+\cot ^{-1}\left(e^{\frac{\pi \left(-1+e^{i a}\right)}{c}}\right)\right)$$
and now it is pretty clear that it has to be $$\pm 1$$ depending on whether the real part of the argument is positive or negative for the principal values.