Problem 8

Exercise 2.8 (Complex integral involving a $\cosh$)   Evaluate

$$\int_{-\infty}^\infty \frac{e^{-2\pi i x \xi}}{\cosh(\pi x)}dx$$ (2.46)

Proof. This integral can be evaluated by appealing to the residues of the complexified function at $z=\pm i/2$. For any $R$, enclose these residues in the rectangle with vertices in counter-clockwise order

$$\{R-i, R+i, -R+i, -R-i\}$$ (2.47)

Then the integral over this rectangle is given by either the residue theorem or directly

$$\int_\gamma f(z)dz = I_1 + I_2 + I_3 + I_4$$ (2.48)

$$= \int_{-R}^R f(x-i) + \int_{-1}^1 f(R+iy)dy + \int_{R}^{-R}f(x+i) + \int_1^{-1} f(R+iy)dy$$ (2.49)

The important contributions are given by

$$I_1 = \int_{-R}^R f(x-i) =\int_{-R}^R \frac{e^{-2\pi i(x-i)\xi}}{\cosh(\pi(x-i))} = -e^{-2\pi \xi} \int_{-R}^R\frac{e^{-2\pi i x\xi}}{\cosh(\pi x)}$$ (2.50)

and similarly

$$I_3 = \int_R^{-R} f(x+i) =-\int_{-R}^R \frac{e^{-2\pi i(x+i)\xi}}{\cosh(\pi(x+i))} = e^{2\pi \xi} \int_{-R}^R\frac{e^{-2\pi i x\xi}}{\cosh(\pi x)}$$ (2.51)

As $R\to\infty$, they each converge to a constant multiple of the desired integral.

The other integrals vanish as $R\to\infty$, and this can be seen:

$$\vert I_2\vert = \left\vert\int_{-1}^1 f(R+iy)dy\right\vert \leq \int_{-1}^1 \frac{\vert e^{-2\pi i(R+iy)\xi}\vert}{\vert\cosh(\pi(R+iy))\vert}$$ (2.52)

$$\leq \int_{-1}^1 \frac{e^{2\pi y\xi}}{\vert\cosh(\pi(R+iy))\vert}$$ (2.53)

Given $\epsilon>0$, select by uniform continuity $\delta>0$ such that

$$\left\vert \frac{1}{\vert\cosh(\pi(R+iy))\vert} - \frac{1}{\vert\cosh(\pi R)\vert} \right\vert < \epsilon$$ (2.54)

Also realize that on this interval $e^{2\pi y \xi}\leq e^{2\pi \xi}$. Then break the integral into pieces of size $\delta$.

$$\int_{-1}^1 \frac{e^{2\pi y\xi}}{\vert\cosh(\pi(R+iy))\vert} \leq \left( \sum_{k=0}^n \int_{-1+k\delta}^{-1+(k+1)\delta} + \in... ...^1 \right) e^{2\pi \xi}\left(\frac{1}{\vert\cosh(\pi R)\vert} + \epsilon\right)$$ (2.55)

$$\leq \int_{-1}^1 e^{2\pi \xi}\left(\frac{1}{\vert\cosh(\pi R)\vert} + \epsilon\right)$$ (2.56)

$$= 2e^{2\pi \xi}\left(\frac{1}{\vert\cosh(\pi R)\vert} + \epsilon\right)$$ (2.57)

Send $\epsilon\to 0$ and $R\to 0$ to see the quantity vanish as $R\to\infty$. Similar for the other integral.

The residue theorem says that

$$\lim_{R\to\infty}\left[ I_1 + I_2 + I_3 + I_4 \right] = 2\pi i \operatorname{Res}(f, \pm i/2) =2(e^{\pi\xi}-e^{-\pi\xi})$$ (2.58)

But we know $I_2$ and $I_4$ vanish so we are left with

$$2(e^{\pi\xi}-e^{-\pi\xi}) = \lim_{R\to\infty} I_1 + I_3 = e^{2\pi\xi}I - e^{-2\pi\xi} I = (e^{\pi\xi} - e^{-\pi\xi})(e^{\pi\xi} + e^{-\pi \xi})I$$ (2.59)

This implies

$$I(\xi) = \frac{2}{e^{\pi\xi}+e^{-\pi\xi}}=\frac{1}{\cosh(\pi\xi)}$$ (2.60)

Let us compute these residues directly

$$\lim_{z\to i/2}(z-i/2)f(z) = \lim_{z\to i/2}\frac{(z-i/2)e^{-2\pi i z \xi}}{\cosh(\pi z)} =\... ...2\pi i z\xi} + (z-i/2) \times e^{-2\pi i z \xi} (-2\pi i \xi)}{\pi\sinh(\pi z)}$$ (2.61)

$$= \frac{e^{-2\pi i(i/2)\xi}}{\pi i} = \frac{e^{\pi \xi}}{\pi i}$$ (2.62)

Similarly

$$\lim_{z\to -i/2}(z+i/2)f(z) = \lim_{z\to -i/2}\frac{(z+i/2)e^{-2\pi i z \xi}}{\cosh(\pi z)} =... ...2\pi i z\xi} + (z+i/2) \times e^{-2\pi i z \xi} (-2\pi i \xi)}{\pi\sinh(\pi z)}$$ (2.63)

$$= \frac{e^{-2\pi i(-i/2)\xi}}{\pi i} = \frac{e^{-\pi \xi}}{\pi i}$$ (2.64)

$\qedsymbol$