# An integral representation of the Riemann zeta function

#### Random Variable

##### Well-known member
MHB Math Helper
Show that $\displaystyle \zeta(s) = \frac{2^{s-1}}{1-2^{1-s}} \int_{0}^{\infty} \frac{\cos (s \arctan t)}{(1+t^{2})^{s/2} \cosh \left( \frac{\pi t}{2} \right)} \ dt$

The cool thing about this representation is that it is valid for all complex values of $s$ excluding $s=1$.

This integral is similar to another integral I recently came across, so I knew immediately how to approach it.

#### Random Variable

##### Well-known member
MHB Math Helper
I'm basically going to copy and paste the solution I posted on another forum.

$$\int_{0}^{\infty} \frac{\cos (s \arctan t)}{(1+t^{2})^{s/2} \cosh \left( \frac{\pi t}{2} \right)} \ dt$$

Let's add the restriction that $$\text{Re}(s)>1$$. This can be removed at the end by analytic continuation.

$$\int_{0}^{\infty} \frac{\cos (s \arctan t)}{(1+t^{2})^{s/2} \cosh \left( \frac{\pi t}{2} \right)} \ dt = \frac{1}{2} \ \text{Re} \int_{-\infty}^{\infty} \frac{1}{(1-it)^{s} \cosh \left( \frac{\pi t}{2} \right)} \ dt$$

Let $$f(z) = \frac{1}{(1-iz)^{s} \cosh \left( \frac{\pi z}{2} \right)}$$ and integrate around a rectangle with vertices at $$z=N, z=N + 2Ni, z=-N + 2Ni$$ and $$z=N$$.

Then $$\frac{1}{2} \ \text{Re} \int_{-\infty}^{\infty} \frac{1}{(1-it)^{s} \cosh \left( \frac{\pi t}{2} \right)} \ dt = \text{Re} \ \pi i \sum_{n=0}^{\infty} \text{Res} [f(z),i(2n+1)]$$

$$\text{Res} [f(z),i(2n+1)] = \lim_{z \to i(2n+1)} \frac{1}{-is(1-iz)^{s-1} \cosh \left( \frac{\pi z}{2} \right) + (1-iz)^{s} \frac{\pi}{2} \sinh \left( \frac{\pi z}{2} \right)} = \frac{2}{\pi} \frac{(-1)^{n}}{i} \frac{1}{(2+2n)^{s}}$$

$$= \frac{2^{1-s}}{\pi} \frac{(-1)^{n}}{i} \frac{1}{(n+1)^{s}}$$

$$\text{Re} \ \pi i \sum_{n=1}^{\infty} \text{Res} [f(z),i(2n+1)] = 2^{1-s} \eta(s) = 2^{1-s} \Big( 1 - 2^{1-s} \Big) \zeta(s)$$

So $$\displaystyle \int_{0}^{\infty} \frac{\cos (s \arctan t)}{(1+t^{2})^{s/2} \cosh \left( \frac{\pi t}{2} \right)} \ dt = 2^{1-s} \Big( 1 - 2^{1-s} \Big) \zeta(s)$$

$$\implies \zeta(s) = \frac{2^{s-1}}{1-2^{1-s}} \int_{0}^{\infty} \frac{\cos (s \arctan t)}{(1+t^{2})^{s/2} \cosh \left( \frac{\pi t}{2} \right)} \ dt$$

Notice that you get $$\zeta(0) = -\frac{1}{2} \int_{0}^{\infty} \frac{1}{\cosh \left( \frac{\pi t}{2} \right)} \ dt = -\frac{1}{2} (1) = -\frac{1}{2}$$

And $$\zeta(-1) = -\frac{1}{12} \int_{0}^{\infty} \frac{\cos (\arctan t)}{(1+t^{2})^{-1/2} \cosh \left( \frac{\pi t}{2} \right)} \ dt = - \frac{1}{12} \int_{0}^{\infty} \frac{1}{\cosh \left( \frac{\pi t}{2} \right)} \ dt = - \frac{1}{12}$$