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- Jan 31, 2012

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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}$

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- Thread starter
- #1

- Jan 31, 2012

- 253

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}$

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- #2

- Jan 31, 2012

- 253

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.

[tex] \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 [/tex]

[tex] \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 [/tex]

[tex] \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} [/tex]

[tex] \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 [/tex]

Let [tex] z = 1-it [/tex]

[tex] = \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 [/tex]

where the contour is a vertical line

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

Then [tex] \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) [/tex]

[tex] =\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} [/tex]

Putting things together we get

[tex] \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)} [/tex]

[tex] \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} [/tex]

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.

[tex] \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 [/tex]

[tex] \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 [/tex]

[tex] \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} [/tex]

[tex] \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 [/tex]

Let [tex] z = 1-it [/tex]

[tex] = \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 [/tex]

where the contour is a vertical line

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

Then [tex] \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) [/tex]

[tex] =\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} [/tex]

Putting things together we get

[tex] \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)} [/tex]

[tex] \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} [/tex]

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- #3

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!Maybe I'm jumping the gun, but it doesn't look like anyone is going to attempt this one either...

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- Jan 31, 2012

- 253

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

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

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."

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- #6

- Jan 31, 2012

- 253

$ \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} $.

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- #7

- Jan 31, 2012

- 253

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}$