Applications of Switching Sums and Integrals

The standard integral representation of the Riemann zeta function,

$$ \zeta(s+1)=\frac1{\Gamma(s+1)}\int_0^\infty\frac{x^s}{e^x-1}\:dx, \qquad s>0, $$

is obtained this way, using the uniform convergence, one has $$ \begin{align} \int_0^\infty\frac{x^s}{e^x-1}\:dx&=\int_0^\infty\frac{x^se^{-x}}{1-e^{-x}}\:dx \\&=\int_0^\infty x^s\sum_{n=0}^\infty e^{-(n+1)x}\:dx \\&=\sum_{n=0}^\infty\int_0^\infty x^s e^{-(n+1)x}\:dx \\&=\sum_{n=0}^\infty\frac1{(n+1)^{s+1}}\int_0^\infty u^s e^{-u}\:dx \\&=\sum_{n=1}^\infty\frac1{n^{s+1}}\cdot\Gamma(s+1) \\&=\Gamma(s+1)\cdot \zeta(s+1). \end{align} $$ This integral representation yields many consequences concerning the Riemann zeta function.

First example

For $|a|<1$

$$\int^{2\pi}_0 \frac{1- a \cos(x)}{1+a^2-2a \cos(x)}dx = \sum_{k=0}^{\infty} a^{k} \int^{2\pi}_0 \cos(kx)\,dx = 2\pi $$

Where we used that

$$\sum_{k=0}^{\infty} a^{k} \cos(kx) = \frac{1- a \cos x}{1-2a \cos x + a^{2}} , \ \ |a|<1$$

Second example

$$ \int_{0}^{\infty}\frac{e^{-t}\cosh(a\sqrt{t})}{\sqrt{t}}dt$$

We know that we can expand $\cosh$ using power series

$$\cosh(a\sqrt{t}) = \sum_{n=0}^{\infty}\frac{a^{2n}\cdot t^n}{(2n)!}$$

Substituting back in the integral we have

$$\int_{0}^{\infty}e^{-t}\sum_{n=0}^{\infty}\frac{a^{2n}\cdot t^n}{(2n)!\, \sqrt{t}}dt$$

Now since the series is always positive we can swap the integral and the series

$$\sum_{n=0}^{\infty}\frac{a^{2n}}{(2n)!}\left[ \int_{0}^{\infty}e^{-t} t^{n-\frac{1}{2}}dt\right] $$

Hence we have by using the gamma function

$$\sum^{\infty}_{n=0} \frac{a^{2n}\,\Gamma\left(\frac{1}{2}+n\right)}{(2n!)}$$

Using LDF (Legendre Duplication Formula) we get

$$\sum^{\infty}_{n=0}\frac{a^{2n}}{(2n!)}\left({(2n)! \over 4^n n!} \sqrt{\pi}\right)$$

By further simplification

$$\sqrt{\pi}\,\sum^{\infty}_{n=0} \frac{a^{2n} }{4^n\,n!}$$

Using the expansion of the exponential we get

$$\int_{0}^{\infty}\frac{e^{-t}\cosh(a\sqrt{t})}{\sqrt{t}}dt\,=\,\sqrt{\pi}e^{{a^2 \over 4}}$$