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I gather that the following two identities about $\xi(3)$ hold via some notion of zeta-function regularized integrals.

$\xi(3) = \frac{(2\pi)^3}{3}\int _0 ^\infty d\lambda \frac{\sqrt{\lambda} }{1 + e^{2 \pi \sqrt{\lambda} } } = - \frac{4 \pi^3 }{3} \int _0^\infty d\lambda \sqrt{\lambda}\text{ } tanh (\pi \sqrt{\lambda}) $

This comes up in the context of Quantum Field Theory but I haven't been able to locate any QFT resource either which proves these.

  • I would like to know the proof of the above.

  • I would like to know if there is any generalization of these to $\xi(n)$

  • I am hoping that there is generalization of the second identity to cases like $\int _0 ^\infty \sqrt{\lambda + \frac{1}{4}} tanh (\pi \sqrt{\lambda})$

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up vote 3 down vote accepted

We have $$\tanh(x) = \dfrac{1 - e^{-2x}}{1 + e^{-2x}} = (1-e^{-2x}) \sum_{k=0}^{\infty}(-1)^k e^{-2kx} = 1 + 2 \sum_{k=1}^{\infty}(-1)^ke^{-2kx}$$ Now we have $$\sqrt{x} \tanh(\sqrt{x}) = \sqrt{x} + 2 \sum_{k=1}^{\infty}(-1)^k \sqrt{x}e^{-2k\sqrt{x}}$$ Now throwing away the divergent part, i.e., $\sqrt{x}$, as every good QFT person does, we get $$\text{Regularized}\left(\int_0^{\infty}\sqrt{x} \tanh(\sqrt{x}) \right) = 2 \sum_{k=1}^{\infty}(-1)^k \int_0^{\infty}\sqrt{x}e^{-2k\sqrt{x}} \tag{$\star$}$$ Now note that$$\int_0^{\infty}\sqrt{x}e^{-2k\sqrt{x}}dx = \dfrac1{2k^3}$$which is obtained by setting $\sqrt{x}=t$ and integrating by parts. Plugging it back into $\star$ gives us $$\text{Regularized}\left(\int_0^{\infty}\sqrt{x} \tanh(\sqrt{x}) \right) = \sum_{k=1}^{\infty}(-1)^k \dfrac1{k^3} = -\dfrac34 \zeta(3)$$ I will let you fix the constant that scale during the integration process.

Note that we landed up with $\zeta(3)$, since the integral was of the form $\sqrt{x} \tanh(\sqrt{x})$. If we were to start with the integral of the form $x^{1/n} \tanh(x^{1/n})$ and mimic the process above, we will get $\zeta(n+1)$.

Added on OP's request We have $$\zeta(3) = 1 + \dfrac1{2^3} + \dfrac1{3^3} + \dfrac1{4^3} + \cdots$$ Note that $$\dfrac1{2^3} + \dfrac1{4^3} + \dfrac1{6^3} + \cdots = \dfrac1{2^3}\left( 1 + \dfrac1{2^3} + \dfrac1{3^3} + \dfrac1{4^3} + \cdots\right) = \dfrac{\zeta(3)}8$$ Hence, $$1 + \dfrac1{3^3} + \dfrac1{5^3} + \cdots = \zeta(3) - \dfrac{\zeta(3)}8 = \dfrac78 \zeta(3)$$ Therefor, the sum $$\sum_{k=1}^{\infty}(-1)^k \dfrac1{k^3} = -\dfrac1{1^3} + \dfrac1{2^3} - \dfrac1{3^3} + \dfrac1{4^3} \mp = -\dfrac78 \zeta(3) + \dfrac18 \zeta(3) = -\dfrac34 \zeta(3)$$

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I am a bit confused about what you are saying - (1) Why is $\xi(3) = - \frac{4}{3} \sum_{k=1}^\infty (-1)^k \frac{1}{k^3}$ ? (...that doesn't naively seem to be the usual definition of the Riemann zeta function...) (2) Are you proving the first integration equality also somehow? –  user6818 Dec 5 '13 at 22:12
    
@user6818 What is the first integration equality you are talking about? –  user11000 Dec 5 '13 at 22:23
    
Thanks for the efforts. Now I remember that there is this general statement that for $Re(q)>0$ one can write, $\xi(q) = \frac{\pi^q}{2^{1-2q}(2^q - 2) \Gamma(q)} \int _0 ^\infty dx \frac{ e^{-2\pi\sqrt{x}} x^{\frac{q}{2} -1 } }{1 + e^{-2\pi\sqrt{x}} }$ - now for $q=3$ this seems to match the first integral equality I wrote down - and now if I power-series expand the denominator of this integrand and integrate term-by-term I get, $\xi(q) = \frac{2^q}{2-2^q}\sum_{s=1}^{\infty} \frac{ (-1)^s }{s^q } $ - which matches your expression for $\xi(3)$ –  user6818 Dec 5 '13 at 22:53
    
- I guess a similar proof you have given for $\xi(3)$ will also give this above. –  user6818 Dec 5 '13 at 22:54
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