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$f(x)=\sum\limits_{k = 2 }^ \infty e^{-kx} \ln(k) $

$\int\limits_0^{\infty}\int\limits_x^{\infty}\, f(\gamma)\, d\gamma dx=\sum\limits_{k = 2 }^ \infty \frac{1}{k^2} \ln(k) $

$\int\limits_0^{\infty}\int\limits_x^{\infty} f(\gamma)\, d\gamma dx=\sum\limits_{k=2}^ \infty\ln(k^{\frac{1}{k^2}})=\ln(\prod\limits_{k=2}^{\infty}k^{\frac{1}{k^2}}) $

$\prod\limits_{k=2}^{ \infty }k^{\frac{1}{k^2}}=\prod\limits_{k=2}^{ \infty } \sqrt[k^2]{k}=e^{\int\limits_0^{\infty}\int\limits_x^{\infty} f(\gamma) \,d\gamma dx}$

$f(x)=\sum\limits_{k = 2 }^ \infty e^{-kx} \ln(k) $

$f(x)=\sum\limits_{k = 1 }^ \infty e^{-(k+1)x} \ln(k+1) $

$f(x)=e^{-x}\sum\limits_{k = 1 }^ \infty e^{-kx} \ln(k+1) $

$f(x)=e^{-x}\sum\limits_{n = 1 }^ \infty \frac{(-1)^{n+1}}{n} \sum\limits_{k = 1 }^ \infty k^n e^{-kx}$

We know that

$\sum\limits_{k = 1 }^ \infty e^{-kx}= \frac{1}{e^{x}-1} $

$\sum\limits_{k = 1 }^ \infty k^n e^{-kx}= (-1)^n\frac{d^n}{dx^n}(\frac{1}{e^{x}-1}) $

$f(x)=e^{-x}\sum\limits_{n = 1 }^ \infty \frac{(-1)^{n+1}}{n} \sum\limits_{k = 1 }^ \infty k^n e^{-kx} = e^{-x}\sum\limits_{n = 1 }^ \infty \frac{(-1)^{n+1}}{n} (-1)^n\frac{d^n}{dx^n}(\frac{1}{e^{x}-1})$

$f(x)=-e^{-x}\sum\limits_{n = 1 }^ \infty \frac{1}{n} \frac{d^n}{dx^n}(\frac{1}{e^x-1})$

$\int\limits_0^{\infty}\int\limits_x^{\infty} f(\gamma) \,d\gamma dx= -\int\limits_0^{\infty}\int\limits_x^{\infty} e^{-\gamma}\sum\limits_{n = 1 }^ \infty \frac{1}{n} \frac{d^n}{d\gamma^n}(\frac{1}{e^{\gamma}-1})\, d\gamma dx$

I have lost my way after that.

Is it possible to find a closed form in my way? or I need to follow a different way. I need your mathematical sense. Thanks a lot for answers and advice.

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Try using Euler-Maclaurin formula en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula. –  Jon Jul 25 '12 at 15:06
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FWIW: $$-\zeta^\prime(s)=\sum_{k=1}^\infty \frac{\log\,k}{k^s}$$ Now, let $s=2$ and exponentiate your series to turn it into a product... –  J. M. Jul 25 '12 at 15:54
    
It's funny how this question is also on the front page with this question. Now, I see that both questions are yours... –  J. M. Jul 25 '12 at 15:57
    
@unklerhaukus: JFYI instead of \limits one can also just enclosed the expressions in double dollar signs $$...$$ instead of single dollar signs $...$ to force display style instead of inline style. We ask that the maths in titles be written in inline style, but the body text can have displayed formulae. –  Willie Wong Jan 11 '13 at 13:43
    
@WillieWong that works for the products , but not for the integrals –  Elements in Space Jan 11 '13 at 13:58

2 Answers 2

up vote 7 down vote accepted

Yes $\boxed{\displaystyle e^{-\zeta'(2)}}$ I think.

To prove it start with : $$\zeta(2-x)=\sum_{k=1}^\infty \frac {k^x}{k^2}$$

and compute the derivative!

The trick is that the derivation will create a $\ln(k)$ term at the numerator. At the end take the limit as $x\to 0$.

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@anon: you are right (and it seems you didn't see my last 4 micro-edits...:-)) –  Raymond Manzoni Jul 25 '12 at 14:48
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No, I've been kicking back and watching them roll in one after the other :) Still, $s\to2$ makes more sense. –  anon Jul 25 '12 at 14:49
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@RaymondManzoni : It is very nice way. I can see now I proved that $\int_0^{\infty}\int_x^{\infty} e^{-\gamma}\sum\limits_{n = 1 }^ \infty \frac{1}{n} \frac{d^n}{d\gamma^n}(\frac{1}{e^{\gamma}-1})d\gamma dx=\zeta'(2)$. Thanks a lot. –  Mathlover Jul 25 '12 at 14:52
    
@Mathlover: Nice! Note that $\zeta'(2)$ doesn't have a simpler closed form (I think...) and in fact it is used to define the Glaisher-Kinkelin constant so that you found a new expression for this famous constant, felicitations! –  Raymond Manzoni Jul 25 '12 at 14:58
2  
For completeness: $$\zeta'(2)=(\gamma+\log\,2\pi-12\log\,A)\zeta(2)$$ where $A$ is the Glaisher-Kinkelin constant, though I believe $A=\exp\left(\frac1{12}-\zeta^\prime(-1)\right)$ is the more traditional definition. –  J. M. Jul 25 '12 at 15:50

Your infinite product equals to : $e^{-{\zeta}'(2)}$.

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