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 $\int _0 ^\infty \sqrt{\lambda + \frac{1}{4}} tanh (\pi \sqrt{\lambda})$


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 $$\sum_{k=1}^{\infty}(-1)^k \dfrac1{k^3} = -\dfrac1{1^3} + \dfrac1{2^3} - \dfrac1{3^3} + \dfrac1{4^3} \mp = -\dfrac78 \zeta(3) + \dfrac18 \zeta(3) = -\dfrac34 \zeta(3)$$

  • $\begingroup$ 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? $\endgroup$ – user6818 Dec 5 '13 at 22:12
  • $\begingroup$ @user6818 What is the first integration equality you are talking about? $\endgroup$ – user11000 Dec 5 '13 at 22:23
  • $\begingroup$ 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 $\xi(3)$ $\endgroup$ – user6818 Dec 5 '13 at 22:53
  • $\begingroup$ - I guess a similar proof you have given for $\xi(3)$ will also give this above. $\endgroup$ – user6818 Dec 5 '13 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.