Evaluate $\int_{-\infty}^\infty e^{x/2}\operatorname{sech}(x)\,dx$ without Residue Calculus

Question

I want to integrate: $$\int_{-\infty}^\infty e^{x/2}\operatorname{sech}(x)\,dx.$$

I've trying to test some of my integration skills by attempting some integrals that are standard with residues, but for this one, I must be making a very silly mistake because after checking things with a calculator it doesn't align with what should be right.

I started with integration by parts letting: $u=e^{x/2} => du=\frac{1}{2}e^{x/2}\,dx$ and $dv=\operatorname{sech}(x)\,dx => v=2\arctan(e^x)$ or $-2\arctan(e^{-x})$

I split the integral of $\operatorname{sech}(x)$ into two integrals so it would converge for the limits at infinity (this seemed to work for other integrals I did). After plugging everything:$$-2e^{x/2}\arctan(e^{-x})|_{-\infty}^{\infty}-\int_0^\infty e^{x/2}v\,dx =$$ $$-\int_{-\infty}^0 e^{x/2}\arctan(e^x)\,dx+\int_0^{\infty}e^{x/2}\arctan(e^{-x})\,dx. $$

This, however, is not correct, and I've been staring at it for too long. It would be much appreciated if some help could be given!

Answer 1

METHODOLOGY $1$:

We can proceed by folding the integral, expressing the denominator of the integrand as a geometric series, interchanging the order of integration and summation, and converting the resulting series into a simple integral.

So, here we go ...

$$\begin{align} \int_{-\infty}^\infty e^{x/2}\text{sech}(x)\,dx&=2\int_0^\infty \frac{e^{x/2}+e^{-x/2}}{e^x+e^{-x}}\,dx\\\\ &=2\int_0^\infty \frac{e^{-x/2}+e^{-3x/2}}{1+e^{-2x}}\,dx\\\\ &=2\sum_{n=0}^\infty (-1)^n\int_0^\infty \left (e^{-(2n+1/2)x}+e^{-(2n+3/2)x}\right)\,dx\\\\ &=2\sum_{n=0}^\infty (-1)^n\left(\frac{1}{2n+1/2}+\frac{1}{2n+3/2}\right)\\\\ &=4\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1}+\frac{1}{4n+3}\right)\\\\ &=4\sum_{n=0}^\infty (-1)^n \left(\int_0^1 x^{4n}\,dx+\int_0^1 x^{4n+2}\,dx\right)\\\\ &=4\int_0^1 \left(\frac{1+x^2}{1+x^4}\right)\,dx\\\\ &=\sqrt2 \pi \end{align}$$

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METHODOLOGY $2$:

An alternative approach is to begin with the substitution $x\mapsto\log(x^2)$. Then, we have

$$\begin{align} \int_{-\infty}^\infty e^{x/2}\text{sech}(x)\,dx&=2\int_0^\infty \frac{e^{x/2}+e^{-x/2}}{e^x+e^{-x}}\,dx\\\\ &=4\int_1^\infty \frac{x^2+1}{x^4+1}\,dx\\\\ &\overbrace{=}^{x\mapsto 1/x}4\int_0^1 \left(\frac{1+x^2}{1+x^4}\right)\,dx\\\\ &=\sqrt2 \pi \end{align}$$