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I'm having a little trouble figuring this one out. So far I've got $$I = \int_{-\infty}^\infty \frac{e^{2\pi x / 3}}{\cosh{\pi x}}dx$$ Let $$I' = \oint_C \frac{e^{2\pi z / 3}}{\cosh{\pi z}}dz$$

Where the contour $C$ is a rectangle extending from $-R$ to $R$ in the limit of $R \rightarrow \infty$ and of height 1, which gives $$\int_{-R}^R \frac{e^{2\pi x / 3}}{\cosh{\pi x}}dx + \int_{R}^{-R} \frac{e^{2\pi x / 3 + i}}{\cosh{(\pi x + i)}}dx$$ $$=I - \int_{-R}^R \frac{e^{2\pi x / 3 + i}}{\cosh{(\pi x + i)}}dx$$ $$=\text{Res}(z = \frac{i}{z})$$

I'm having trouble solving the problem from here, as I'm not sure how to handle the separation of the denominator $\cosh{(\pi x + i)}$. Any help is appreciated!

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  • $\begingroup$ Try explicitly writing out the $\cosh$ and you should be able to simplify from there. Or, you could look up the handy sum formulas for hyperbolic trig functions $\endgroup$ – Simply Beautiful Art Feb 20 '17 at 1:33
  • $\begingroup$ By the way, you missed the end pieces of your rectangle. The skinny parts at the sides. $\endgroup$ – Simply Beautiful Art Feb 20 '17 at 1:34
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Note that $\cosh(\pi x+i)$ should read $\cosh(\pi x+i\pi)$.

Then note that $\cosh(\pi x\pm i\pi)=\cosh(\pi x)\cos(\pm \pi)+i\sinh(\pi x)\sin(\pm \pi)=-\cosh(\pi x)$.

Therefore, if we take the contour $C$ to be the rectangle with vertices $-R-i$, $R-i$, $R+i$, and $-R+i$, then we can write

$$\begin{align} \oint_C \frac{e^{2\pi z/3}}{\cosh(\pi z)}\,dz&=\int_{-R}^R \frac{e^{-i2\pi/3}e^{2\pi x/3}}{-\cosh(\pi x)}\,dx+\int_{R}^{-R} \frac{e^{i2\pi/3}e^{2\pi x/3}}{-\cosh(\pi x)}\,dx\\\\ &+\int_{-1}^1 \frac{e^{2\pi (R+iy)/3}}{\cosh(R+iy)}\,i\,dy+\int_{1}^{-1}\frac{e^{2\pi (-R+iy)/3}}{\cosh(-R+iy)}\,i\,dy\tag 1 \end{align}$$

Applying the residue theorem to $(1)$ yields

$$\begin{align} \oint_C \frac{e^{2\pi z/3}}{\cosh(\pi z)}\,dz&=2\pi i \text{Res}\left(\frac{e^{2\pi z/3}}{\cosh(\pi z)}, z=\pm i/2\right)\\\\ &=2\pi i \left(\frac{e^{i\pi/3}}{\pi \sinh(i\pi/2)}+\frac{e^{-i\pi/3}}{\pi \sinh(-i\pi/2)}\right)\\\\ &=i2\sqrt 3\tag 2 \end{align}$$

As $R\to \infty$, the third and fourth integrals on the right-hand side of $(1)$ approach $0$. Hence, after letting $R\to \infty$ and setting $(1)$ and $(2)$ equal, we find that

$$\int_{-\infty}^\infty \frac{e^{2\pi x/3}}{\cosh(\pi x)}\,dx=2$$

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