Simple looking Integral?

Well, I am thrilled, seriously.

Anyways,

$\displaystyle \int_{0}^{\infty}{\frac{dx}{2cosh(\pi x)(16+{x}^{2})}}$

is our problem and we can write it as -

$\displaystyle \frac{1}{2} \int_{-\infty}^{\infty}{\frac{dx}{2cosh(\pi x)(16+{x}^{2})}}$

We will do just the integral first and then all we would be left with is that half factor.

We know

$\displaystyle \int_{C}{f(z) dz} = \int_{-a}^{a}{f(z) dz} + \int_{arc}{f(z) dz}$

$\displaystyle \int_{C}{f(z) dz} - \int_{arc}{f(z) dz} = \int_{-a}^{a}{f(z) dz}$

where C is the contour described.

Here, we will let C be a semicircle with radius $\displaystyle a = \infty$ and origin as the center in the positive imaginary axis.

Let's compute $\displaystyle \int_{C}{\frac{1}{2cosh(\pi z)(z + 4i)(z - 4i)}}$

We will do the residue theorem.

What are the singularities?

$\displaystyle z - 4i = 0, z = 4i$

$\displaystyle z + 4i = 0, z = -4i$ , Is this a singularity? No! Because $\displaystyle -4i$ is not included in our selected contour.

$\displaystyle cosh(\pi z) = 0, cos(i \pi z) = 0$

$\displaystyle i \pi z = \frac{(2n-1)\pi}{2}, z = \frac{(2n-1)i}{2}$ , where $n \in Z$

(Here we have also found the Weierstrass product and hence the infinite product)

So,

$\displaystyle \displaystyle \int_{C}{\frac{1}{2cosh(\pi z)(z + 4i)(z - 4i)}} = 2\pi i(Res(f(z), 4i) + \sum_{n=1}^{\infty}{Res(f(z), \frac{(2n-1)i}{2})})$

$\displaystyle f(z)$ is of the form $\displaystyle \frac{P(z)}{Q(z)}$ such that singularity is a simple pole so the residue is simply $\displaystyle \frac{P(z)}{Q'(z)}$

Taking derivative tells us that we have to put our value of singularity in

$\displaystyle \frac{1}{4(\frac{(2n-1)i}{2})cosh(\frac{(2n-1)\pi i}{2}) + 2\pi(16 + {\left (\frac{(2n-1)i}{2} \right)}^{2}(sinh(\frac{(2n-1)i}{2}))}$

$\displaystyle Res(f(z), 4i) = \frac{1}{16i}$

$\displaystyle \sum_{n=1}^{\infty}{Res(f(z), \frac{(2n-1)i}{2})} = \sum_{n=1}^{\infty}{\frac{{(-1)}^{n}i}{2\pi (16 - {\left(\frac{2n-1}{2}\right)}^{2})}}$

which can be resolved into

$\displaystyle \sum_{n=1}^{\infty}{\frac{{(-1)}^{n}}{9-2n} + \frac{{(-1)}^{n}}{7+2n}}$

which telescopes to $\displaystyle \frac{152}{105}$

but wait,we have taken out sum constants in that sum resolution i.e. $\displaystyle \frac{i}{8\pi}$ (You can see that)

So, our second residue comes out to be $\displaystyle \frac{19 i}{105\pi}$

Our line integral finally becomes

$\displaystyle \int_{C}{\frac{1}{2cosh(\pi z)(z + 4i)(z - 4i)}} = 2\pi i(\frac{1}{16i} + \frac{19i}{105\pi})$

There is one more integral we need to consider

i.e. $\displaystyle \int_{arc}{\frac{1}{2cosh(\pi z)({z}^{2} + 16)}} \Rightarrow 0, a \Rightarrow \infty$ [We will not give this here since the solution is becoming quite long]

$\displaystyle \int_{-\infty}^{\infty}{\frac{1}{2cosh(\pi z)({z}^{2} + 16)}} = 2\pi i(\frac{1}{16i} + \frac{19i}{105\pi})$

$\displaystyle \int_{0}^{\infty}{\frac{1}{2cosh(\pi z)({z}^{2} + 16)}} = \frac{\pi}{16} - \frac{19}{105}$

check this out.