Welcome to our community

Be a part of something great, join today!

An integral representation of the Riemann zeta function

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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
Jan 31, 2012
253
I'm basically going to copy and paste the solution I posted on another forum.



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


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


[tex]\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[/tex]


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


Then [tex]\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)][/tex]


[tex]\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}}[/tex]

[tex]= \frac{2^{1-s}}{\pi} \frac{(-1)^{n}}{i} \frac{1}{(n+1)^{s}}[/tex]


[tex]\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)[/tex]



So [tex]\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)[/tex]


[tex]\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[/tex]



Notice that you get [tex]\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}[/tex]

And [tex]\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}[/tex]