Ok here's my approach on the second integral, I hope some of you can confirm or correct it :)
We define the following contour: Set the cut for the log to be the positive real axis. We start at the origin, in fact at a circle of radius epsilon around the origin (epsilon will tend towards 0), then moving straight, parallel and slightly above to the real axis (z=x+i\delta) up to R, the radius of the second circle round the origin (R will tend to infty). So, we go round till we almsot reach the positive real axis (z=x-i\delta), then again towards the first circle.(sorry, don't know how to make a sketch with tex)
The defined curve is closed and rectifiable (Jordan curve)
We are going to apply the residue thm. on the function
f(z)=\frac{z^3\log z}{e^z-1}
The above set contour could be separated in four different paths:
\gamma_1(x):=x+i\delta, x\in[\varepsilon,R]
\gamma_2(x):=Re^{ix}, x\in[0, 2\pi]
\gamma_3(x):=x-i\delta, x\in[R,\varepsilon]
\gamma_4(x):=\varepsilon e^{-ix}, x\in[0, 2\pi]
\gamma=\gamma_1+\gamma_2+\gamma_3+\gamma_4
by the residue thm. for the zeros of the denominator of f (i.e. c_k) we have:
2\pi i\displaystyle{\sum_k Res_{c_k} f(z)}=\displaystyle{\oint_{\gamma}f(z)dz}=\displaystyle{\int_{\gamma_1}f(z)dz}+\displaystyle{\int_{\gamma_2}f(z)dz}+\displaystyle{\int_{\gamma_3}f(z)dz}+\displaystyle{\int_{\gamma_4}f(z)dz}
We will proove that the integrals of gamma_2 and gamma_4 (I will use Gamma for both) vanish for R->oo and epsilon->0:
Proof:
By the estimation lemma we get:
\displaystyle{|\oint_{\Gamma}f(z)dz|}=\displaystyle{|\oint_{\Gamma}\frac{z^3\log z}{e^z-1}dz|}\leq 2\pi R||\frac{z^3\log z}{e^z-1}||_{\Gamma}
where, ||-|| denotes the supremum norm on Gamma
Now, consider the following inequalities:
||e^z-1||_{\Gamma}\geq ||e^z||_{\Gamma}-1= ||e^{Re^{\pm ix}}||-1\geq e^R-1
||\log z||_{\Gamma}=||\log Re^{\pm ix}||=||\log R\pm 2\pi i||\leq \log R+2\pi (the - sign considers the curve gamma_3 but the result remains unchanged)
||z^3||_{\Gamma}=||R^3e^{\pm 3ix}||=R^3
so, altogether we get:
\displaystyle{|\oint_{\Gamma}\frac{z^3\log z}{e^z-1}dz|}\leq 2\pi R||\frac{z^3\log z}{e^z-1}||_{\Gamma}\leq 2\pi\frac{R^4(\log R+2\pi)}{e^R-1}=2\pi\frac{{\varepsilon}^4(\log\varepsilon+2\pi)}{e^{\varepsilon}-1}
Now, using l'Hospital's rule the proof of the statement is finished:
2\pi\frac{R^4(\log R+2\pi)}{e^R-1}\longrightarrow^{R\rightarrow\infty}0
2\pi\frac{{\varepsilon}^4(\log\varepsilon+2\pi)}{e^{\varepsilon}-1}\longrightarrow^{\varepsilon\rightarrow 0} 0
Now, consider the integrals for gamma_1 and gamma_3:
\displaystyle{\int_{\gamma_1}f(z)dz}+\displaystyle{\int_{\gamma_3}f(z)dz}=\displaystyle{\int_{\varepsilon}^R\frac{(x+i\delta)^3\log (x+i\delta)}{e^{x+i\delta}-1}}+\displaystyle{\int^{\varepsilon}_R\frac{(x-i\delta)^3\log (x-i\delta)}{e^{x-i\delta}-1}}
now, letting delta towards 0, and considering the plain cut we obtain
\displaystyle{\int_{\varepsilon}^R\frac{x^3\log x}{e^{x}-1}}-\displaystyle{\int_{\varepsilon}^R\frac{x^3\log (x+2\pi i)}{e^{x}-1}}=-2\pi i\displaystyle{\int_{\varepsilon}^R\frac{x^3}{e^{x}-1}}
which, after letting epsilon towards 0 and R towards infty, becomes the sought integral.
-2\pi i\displaystyle{\int_0^{\infty}\frac{x^3}{e^{x}-1}}=2\pi i\displaystyle{\sum_k Res_{c_k} f(z)}
As the denominator has infinite many zeroes, but discrete and countable, the residue sum becomes a series, also to be computed.
All we have to do now is to calculate the residue of f(z) for the zeroes of the denominator, i.e. c_k=2\pi ik,k\in N
From the l'Hospital's rule we conclude that the singularities are poles and have all order 3 (for k=0 the singularity is removable), so
Res_{2\pi ik} f(z)=\displaystyle{\lim_{z\rightarrow 2\pi ik}\frac{d^2}{dz^2}\frac{(z-2\pi ik)^3z^3\log z}{e^z-1}}Now, I am not sure about the residue, but to my calculations it should be
Res_{2\pi ik} f(z)=-8ik^3\pi^3\log(2\pi ik)
which does not prove the initial integral identity. That's why I suspect there is a mistake in my calculations. If you find it, I would be glad to know.I'm also still looking for the path suitable for the first integral. Marin
