How to integrate $\int^1_{-1}\frac{\pi}2 e^{ix}\operatorname{sech}\left(\frac{\pi x}{2}\right)\text{ d}x$?

Question

I have an integral I need to integrate, as follows $$\int^1_{-1}\frac{\pi}2 e^{ix}\operatorname{sech}\left(\frac{\pi x}{2}\right)\text{ d}x$$ This is quite troublesome to do, since I cannot use rectangular contour to contour integrate due to the bounds, and rewriting into exponential form didn't help.

Wolfram alpha spits out a hypergeometric antiderivative, and it seems like the only way to express this in any remotely closed form. Is there a way to derive the hypergeometric expression of the integral? (perhaps through finding the antiderivative or something?)

Thanks

Answer 1

We will prove the integral in question equals

$$\frac{2\pi}{\pi+2i}\left(e^{\pi/2+i}\text{ }_2F_1\left(1, \frac{1}{2}+\frac{i}{\pi}; \frac{3}{2} + \frac{i}{\pi}; -e^{\pi}\right) - e^{-\pi/2-i}\text{ }_2F_1\left(1, \frac{1}{2}+\frac{i}{\pi}; \frac{3}{2} + \frac{i}{\pi}; -e^{-\pi}\right)\right).$$

To do this, we will enforce the Hypergeometric Series of $\displaystyle \frac{1}{1+e^{2x}}$, which equals

$$_2F_1 \left(1, \frac{1}{2}+\frac{i}{\pi}; \frac{3}{2} + \frac{i}{\pi}; -e^{2x}\right) - \frac{2\pi}{3\pi+2i}e^{2x} \text{ }_2F_1 \left(2, \frac{3}{2}+\frac{i}{\pi}; \frac{5}{2} + \frac{i}{\pi}; -e^{2x}\right),$$

and apply the known result

$$\frac{d}{dz} \text{ } _2F_1 (a,b;c;f(z)) = \frac{ab}{c}\text{ } _2F_1 (a+1,b+1;c+1;f(z))f'(z).$$

Proof. Let the integral be $I$. Then using $\displaystyle\frac{\pi}{2}x \mapsto x$ yields

$$ \eqalign{ I =& \int_{-\pi/2}^{\pi/2}e^{2ix/\pi}\operatorname{sech}\left(x\right)dx \cr =& \int_{-\pi/2}^{\pi/2}2e^x e^{2ix/\pi}\left(\frac{1}{1+e^{2x}}\right)dx \cr =& \int_{-\pi/2}^{\pi/2}2e^x e^{2ix/\pi}\bigg(\text{ }_2F_1 \left(1, \frac{1}{2}+\frac{i}{\pi}; \frac{3}{2} + \frac{i}{\pi}; -e^{2x}\right) \cr &- \frac{2\pi}{3\pi+2i}e^{2x} \text{ }_2F_1 \left(2, \frac{3}{2}+\frac{i}{\pi}; \frac{5}{2} + \frac{i}{\pi}; -e^{2x}\right)\bigg)dx \cr =& \int_{-\pi/2}^{\pi/2}2e^{2x}e^{2ix/\pi}\text{ }_2F_1 \left(1, \frac{1}{2}+\frac{i}{\pi}; \frac{3}{2} + \frac{i}{\pi}; -e^{2x}\right)dx \cr &- \frac{4\pi}{3\pi + 2i}e^{3x}e^{2ix/\pi}\text{ }_2F_1 \left(2, \frac{3}{2}+\frac{i}{\pi}; \frac{5}{2} + \frac{i}{\pi}; -e^{2x}\right)dx. } $$ We use integration by parts through $$ \eqalign{ u &= \text{ }_2F_1 \left(1, \frac{1}{2}+\frac{i}{\pi}; \frac{3}{2} + \frac{i}{\pi}; -e^{2x}\right) \cr \implies du &= -\frac{2\pi + 4i}{3\pi + 2i}e^{2x} \text{ }_2F_1 \left(2, \frac{3}{2}+\frac{i}{\pi}; \frac{5}{2} + \frac{i}{\pi}; -e^{2x}\right) } $$ and $$ \eqalign{ dv &= 2e^x e^{2ix/\pi}dx \cr \implies v &= \frac{2\pi}{\pi+2i}e^{x}e^{2ix/\pi}. \cr } $$

Then

$$ \eqalign{ I =& \Bigg[\frac{2\pi}{\pi+2i}e^x e^{2ix/\pi} \text{ }_2F_1 \left(1, \frac{1}{2}+\frac{i}{\pi}; \frac{3}{2} + \frac{i}{\pi}; -e^{2x}\right) \Bigg]_{-\pi/2}^{\pi/2} \cr &- \int_{-\pi/2}^{\pi/2}-\frac{2\pi}{\pi+2i}e^x e^{2ix/\pi} e^{2x}\frac{2\pi + 4i}{3\pi + 2i}\text{ }_2F_1\left(2, \frac{3}{2}+\frac{i}{\pi}; \frac{5}{2} + \frac{i}{\pi}; -e^{2x}\right)dx \cr &- \int_{-\pi/2}^{\pi/2} \frac{4\pi}{3\pi+2i}e^{3x}e^{2ix/\pi} \text{ }_2F_1\left(2, \frac{3}{2}+\frac{i}{\pi}; \frac{5}{2} + \frac{i}{\pi}; -e^{2x}\right)dx \cr =& \Bigg[\frac{2\pi}{\pi+2i}e^x e^{2ix/\pi} \text{ }_2F_1 \left(1, \frac{1}{2}+\frac{i}{\pi}; \frac{3}{2} + \frac{i}{\pi}; -e^{2x}\right) \Bigg]_{-\pi/2}^{\pi/2} \cr =& \frac{2\pi}{\pi+2i}\bigg(e^{\pi/2+i}\text{ }_2F_1\left(1, \frac{1}{2}+\frac{i}{\pi}; \frac{3}{2} + \frac{i}{\pi}; -e^{\pi}\right) \cr &- e^{-\pi/2-i}\text{ }_2F_1\left(1, \frac{1}{2}+\frac{i}{\pi}; \frac{3}{2} + \frac{i}{\pi}; -e^{-\pi}\right)\bigg) \cr } $$ as desired! Q.E.D.

Answer 2

Because of the symmetry $$I=\frac{\pi}2\int^{+1}_{-1} e^{ix}\operatorname{sech}\left(\frac{\pi x}{2}\right)\text{ d}x=\pi \int_0^1\cos(x)\operatorname{sech}\left(\frac{\pi x}{2}\right)\text{ d}x$$ I do not think that you can avoid the Gaussian hypergeometric function (even if it is one of the simplest).

Expanding the complex numbers, $$\color{blue}{I=-\frac{2 e \pi }{1+e^2}+\frac{2 e^{\pi /2} \pi (2 \sin (1)+\pi \cos (1))}{4+\pi ^2}\,(H_1+H_2)-}$$ $$\color{blue}{\frac{2 e^{\pi /2} \pi (\pi \sin (1)-2 \cos (1))}{4+\pi ^2}\,(H_1-H_2)\, \color{red}{\large i}}$$ where

$$H_1=\, _2F_1\left(1,\frac{1}{2}-\frac{i}{\pi };\frac{3}{2}-\frac{i}{\pi};-e^{\pi }\right)$$ $$H_2=\, _2F_1\left(1,\frac{1}{2}+\frac{i}{\pi };\frac{3}{2}+\frac{i}{\pi};-e^{\pi }\right)$$

$(H_1+H_2)$ is a real number $$H_1+H_2=0.419817581157418905902747916112\cdots$$ and $(H_1-H_2)$ is an imaginary number $$H_1-H_2=0.287876065036803717328869876227\cdots i$$

None of them is recognized by inverse symbolic alculators even in terms of special functions.