# Another integral representation of the Riemann zeta function

#### Random Variable

##### Well-known member
MHB Math Helper
Here is another integral representation of $\zeta(s)$ that is valid for all complex values of $s$.

It's similar to the first one, but a bit harder to derive.

$\displaystyle \zeta(s) = 2 \int_{0}^{\infty} \frac{\sin (s \arctan t)}{(1+t^{2})^{s/2} (e^{2 \pi t} - 1)} \ dt + \frac{1}{2} + \frac{1}{s-1}$

#### Random Variable

##### Well-known member
MHB Math Helper
Maybe I'm jumping the gun, but it doesn't look like anyone is going to attempt this one either.

Again I'm going to add the restriction that $\text{Re}(s) >1$ which can be removed at the end.

$\displaystyle \int_{0}^{\infty} \frac{\sin (s \arctan t)}{(1+t^{2})^{s/2} (e^{2 \pi t} - 1)} \ dt = \int_{0}^{\infty} \text{Im} \ \frac{1}{(1-it)^{s} (e^{2 \pi t}-1)} \ dt$

I'm leaving the imaginary sign inside of the integral since bringing it outside would result in a divergent integral.

$$\displaystyle = \int_{0}^{\infty} \text{Im} \ \frac{1}{(1-it)^{s}} \frac{e^{-\pi t}}{e^{\pi t}-e^{- \pi t}} \ dt = \frac{1}{2} \int_{0}^{\infty} \text{Im} \ \frac{1}{(1-it)^{s}} \Big( \frac{e^{\pi t} + e^{\pi t}}{e^{\pi t} - e^{- \pi t}} -1 \Big) \ dt$$

$$\displaystyle = \frac{1}{2} \int_{0}^{\infty} \text{Im} \ \frac{1}{(1-it)^{s}} \Big( \coth(\pi t) -1 \Big) \ dt = \frac{1}{2} \int_{0}^{\infty} \text{Im} \ \frac{\coth( \pi t) }{(1-it)^{s}} \ dt - \frac{1}{2} \text{Im} \int_{0}^{\infty} \frac{1}{(1-it)^{s}} \ dt$$

$$\displaystyle \int_{0}^{\infty} \frac{1}{(1-it)^{s}} \ dt = \frac{i}{1-s}\frac{1}{ (1-it)^{s-1}} \Big|^{\infty}_{0} = \frac{i}{s-1}$$

$$\int_{0}^{\infty} \text{Im} \ \frac{\coth( \pi t) }{(1-it)^{s}} \ dt = \frac{1}{2} \int_{-\infty}^{\infty} \ \text{Im} \ \frac{\coth(\pi t)}{(1-it)^{s}} \ dt = \frac{1}{2} \text{Im} \ \text{PV} \int_{-\infty}^{\infty} \frac{\coth(\pi t)}{(1-it)^{s}} \ dt$$

Let $$z = 1-it$$

$$= \frac{1}{2} \text{Im} \ \text{PV}\int^{1+ i \infty}_{1- i \infty} i \coth \Big( \pi i (z-1) \Big) \frac{dz}{z^{s}} \ dz = \frac{1}{2} \text{Im} \ \text{PV}\int^{1+ i \infty}_{1- i \infty} \cot \Big( \pi(z-1) \Big) \frac{dz}{z^{s}} = \frac{1}{2} \text{Im} \ \text{PV}\int^{1+ i \infty}_{1- i \infty} \frac{\cot (\pi z)}{z^{s}} \ dz$$

where the contour is a vertical line

Since there is a branch point at the origin, close the contour to the right.

Then $$\int_{0}^{\infty} \text{Im} \ \frac{\coth( \pi t) }{(1-it)^{s}} \ dt = \frac{1}{2} \text{Im} \ \Big( \pi i \text{Res} \Big[ \frac{\cot(\pi z)}{z^{s}} , 1 \Big] + 2 \pi i \sum_{n=2}^{\infty} \text{Res} \Big[\frac{\cot (\pi z)}{z^{s}}, n \Big] \Big)$$

$$=\frac{1}{2} \text{Im} \ \Big( \pi i \Big( \frac{1}{\pi} \Big) + 2 \pi i \sum_{n=2}^{\infty} \frac{1}{\pi n^{s}} \Bigg) = \frac{1}{2} \text{Im} \ \Big( i + 2 i \zeta(s) - 2i \Big) = \zeta(s) - \frac{1}{2}$$

Putting things together we get

$$\displaystyle \int_{0}^{\infty} \frac{\sin (s \arctan t)}{(1+t^{2})^{s/2} (e^{2 \pi t} - 1)} \ dt = \frac{1}{2} \Big( \zeta(s) - \frac{1}{2} \Big) - \frac{1}{2} \text{Im} \ \frac{i}{s-1} = \frac{\zeta(s)}{2} - \frac{1}{4} + \frac{1}{2(1-s)}$$

$$\displaystyle \implies \zeta(s)= 2 \int_{0}^{\infty} \frac{\sin (s \arctan t)}{(1+t^{2})^{s/2} (e^{2 \pi t} - 1)} \ dt+ \frac{1}{2} + \frac{1}{s-1}$$

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#### MarkFL

Staff member
Maybe I'm jumping the gun, but it doesn't look like anyone is going to attempt this one either...
We do recommend that those posting challenges give our members at least a week to respond. This allows those who may only have time to visit us once a week to have a chance to answer our posted challenge questions, all of which we greatly appreciate! #### Random Variable

##### Well-known member
MHB Math Helper
I've waited at least a week in the past (or until I was asked to post a solution/evaluation). The last one went unanswered for about 4 weeks.

I actually wanted to see the complete evaluation myself since I had only scribbled some stuff down on scratch paper.

#### MarkFL

Staff member
...I actually wanted to see the complete evaluation myself since I had only scribbled some stuff down on scratch paper.
Haha...I know what you mean there...most times when I post a challenge question, that's all I have in the way of a solution, and it is from these scratchings that I then construct a (hopefully) coherent solution to post. However, I did feel it was incumbent on me to point out our http://mathhelpboards.com/challenge...nswering-challenging-problem-puzzle-3875.html, especially since you specifically stated that you may be "jumping the gun." #### Random Variable

##### Well-known member
MHB Math Helper
We should check that we get what we expect for $s=0$ and $s=-1$.

$\displaystyle \zeta(0) = 0 + \frac{1}{2} - 1 = - \frac{1}{2}$

$\displaystyle \zeta(-1) = -2 \int_{0}^{\infty} \frac{t}{e^{2 \pi t} -1} \ dt + \frac{1}{2} - \frac{1}{2} = - \frac{1}{2 \pi^{2}} \int_{0}^{\infty} \frac{u}{e^{u}-1} \ du = -\frac{1}{2 \pi^{2}} \Gamma (2) \zeta(2)= - \frac{1}{12}$

If you want to be really adventurous, you could check that $\displaystyle \zeta'(0) = - \frac{\ln (2 \pi)}{2}$.

#### Random Variable

##### Well-known member
MHB Math Helper
Actually it's not that adventurous to find $\zeta' (0)$ from the representation.

Differentiating inside of the integral we get

$$\zeta'(0) = 2 \int_{0}^{\infty} \frac{\arctan t}{e^{2 \pi t} -1} \ dt - 1$$

Binet's integral formula, from which the asymptotic expansion of the Gamma function can be derived, states that

$$2 \int_{0}^{\infty} \frac{\arctan \left( \frac{x}{z} \right)}{e^{2 \pi x} -1} \ dx = \ln \Gamma(z) - \left( z- \frac{1}{2} \right) \ln z + z - \frac{\ln (2 \pi)}{2}$$

So $\displaystyle \zeta'(0) = 1 - \frac{\ln(2 \pi)}{2} - 1 = - \frac{\ln (2 \pi)}{2}$