<s>[INST] zeta-function regularized integrals 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 $\int _0 ^\infty \sqrt{\lambda + \frac{1}{4}} tanh (\pi \sqrt{\lambda})$
 [/INST] 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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