MHB An integral representation of the Riemann zeta function

polygamma
Messages
227
Reaction score
0
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.
 
Mathematics news on Phys.org
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)} \ dtLet'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)} \ dtLet 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}
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top