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