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
Best Answer
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.