Monday Math 55

How might one perform the unit-square integral
$H(s)=\int_0^1\int_0^1\frac{(1-x)(-\ln(xy))^s}{1-xy}\,dx\,dy$?
First, let us swap the order of integration:
$H(s)=\int_0^1\int_0^1\frac{(1-x)(-\ln(xy))^s}{1-xy}\,dy\,dx$.
Next, we approach the inner integral by performing the u-substitution
$y=\frac{e^{-u}}{x}$ → $dy=-\frac{e^{-u}}{x}\,du$
The limits become y=0 → u=∞, y=1 → u=-ln(x).

So, with this substitution, we find $xy=e^{-u}$, and
$\begin{eqnarray}H(s)&=&\int_0^1\int_{\infty}^{-ln(x)}\frac{(u)^s(1-x)}{1-e^{-u}}\left(-\frac{e^{-u}}{x}\right)\,du\,dx\\&=&\int_0^1\int_{-ln(x)}^{\infty}\frac{(u)^se^{-u}(1-x)}{(1-e^{-u})x}\,du\,dx\end{eqnarray}$

for 0<x<1, ln(x)<0, so -ln(x)>0; -ln(x)→∞ as x→0, and -ln(1)=0, so the integration region can be seen as between $x=e^{-u}$ and x=1, u>0. So, reversing the order of the double integral,
$\begin{eqnarray}H(s)&=&\int_0^{\infty}\int_{e^{-u}}^1\frac{(u)^se^{-u}}{1-e^{-u}}\frac{1-x}{x}\,dx\,du\\&=&\int_0^{\infty}\frac{u^se^{-u}}{1-e^{-u}}\left[\ln{x}-x\right]_{x=e^{-u}}^{1}\,du\\&=&\int_0^{\infty}\frac{u^se^{-u}}{1-e^{-u}}(u+e^{-u}-1)\,du\\&=&\int_0^{\infty}\frac{u^{s+1}e^{-u}}{1-e^{-u}}+\frac{(u)^se^{-u}(e^{-u}-1)}{1-e^{-u}}\,du\\&=&\int_0^{\infty}\frac{u^{s+1}e^{-u}}{1-e^{-u}}\,du-\int_0^{\infty}(u)^se^{-u}\,du\end{eqnarray}$

We found previously that $\int_0^{\infty}\frac{(u)^{s-1}}{e^u-1}\,du=\operatorname{\Gamma}(s)\zeta(s)$, and the gamma function is defined as $\operatorname{\Gamma}(s)=\int_0^{\infty}\(u)^{s-1}e^{-u}\,du$; this means
$\begin{eqnarray}H(s)&=&\operatorname{\Gamma}(s+2)\zeta(s+2)-\operatorname{\Gamma}(s+1)\\&=&\operatorname{\Gamma}(s+2)\zeta(s+2)-\frac{\operatorname{\Gamma}(s+2)}{s+1}\\&=&\operatorname{\Gamma}(s+2)\left[\zeta(s+2)-\frac{1}{s+1}\right]\end{eqnarray}$

This is called Hadjicostas’s formula, and holds for s any complex number with $\Re(s)\gt-2$.
Note that as the pole of ζ(s) at s=1 is of order one: the Laurent series is of the form $\zeta(s)=\frac{1}{s-1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\gamma_n(s-1)^n$,
where the $\gamma_n$ are called the Stieltjes constants, and $\gamma_0=\gamma$, the Euler-Mascheroni constant; the singularity in the term $\left[\zeta(s+2)-\frac{1}{s+1}\right]$ is thus removable, with $\lim_{s\to-1}\left[\zeta(s+2)-\frac{1}{s+1}\right]=\gamma$

Thus we have
$H(-1)=\int_0^1\int_0^1\frac{x-1}{(1-xy)\ln(xy)}\,dx\,dy=\gamma$.

Similarly, with the related integral
$I(s)=\int_0^1\int_0^1\frac{(-\ln(xy))^s}{1-xy}\,dy\,dx$,
we can use the same subtitution and change of order:
$\begin{eqnarray}I(s)&=&\int_0^{\infty}\int_{e^{-u}}^1\frac{(u)^se^{-u}}{1-e^{-u}}\frac{1}{y}\,dy\,du\\&=&\int_0^{\infty}\frac{(u)^se^{-u}}{1-e^{-u}}\left[\ln{y}\right]_{y=e^{-u}}^{1}\,du\\&=&\int_0^{\infty}\frac{(u)^se^{-u}}{1-e^{-u}}u\,du\\&=&\int_0^{\infty}\frac{(u)^{s+1}e^{-u}}{1-e^{-u}}\,du\\&=&\int_0^{\infty}\frac{(u)^{s+1}}{e^u-1}\,du\\&=&\operatorname{\Gamma}(s+2)\zeta(s+2)\end{eqnarray}$.

What if we try instead try the integral
$J(s)=\int_0^1\int_0^1\frac{(1-x)(-\ln(xy))^s}{1+xy}\,dy\,dx$?
Performing the same u-substitution:
$J(s)=\int_0^1\int_{-ln(x)}^{\infty}\frac{(u)^se^{-u}(1-x)}{(1+e^{-u})x}\,du\,dx$.

Reversing the order of the double integration, and performing the inner integral,
$\begin{eqnarray}J(s)&=&\int_0^{\infty}\int_{e^{-u}}^1\frac{(u)^se^{-u}}{1+e^{-u}}\frac{1-x}{x}\,dx\,du\\&=&\int_0^{\infty}\frac{u^se^{-u}}{1+e^{-u}}\left(u-2+(1+e^{-u})\right)\,du\\&=&\int_0^{\infty}\frac{u^{s+1}e^{-u}}{1-e^{-u}}-2\int_0^{\infty}\frac{u^se^{-u}}{1-e^{-u}}+\int_0^{\infty}(u)^se^{-u}\,du\end{eqnarray}$

We found previously that $\int_0^{\infty}\frac{(u)^{s-1}}{e^u+1}\,du=\operatorname{\Gamma}(s)\eta(s)$, so then
$\begin{eqnarray}J(s)&=&\operatorname{\Gamma}(s+2)\eta(s+2)-2\operatorname{\Gamma}(s+1)\eta(s+1)+\operatorname{\Gamma}(s+1)\\&=&\operatorname{\Gamma}(s+2)\left[\eta(s+2)+\frac{1-2\eta(s+1)}{s+1}\right]\end{eqnarray}$
(for $\Re(s)\gt-3$)
and, in analogy to I(s), we can also find that.
$\int_0^1\int_0^1\frac{(-\ln(xy))^s}{1+xy}\,dx\,dy=\operatorname{\Gamma}(s+2)\eta(s+2)$