Evaluate $$\int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x y) \ln (x y)} d x d y$$
I couldn't get very far on this one, so I would appreciate some help =)
My attempt so far (transcribed from the comments):
By extending it to the Dirichlet Eta Function I evaluated this integral to be $\ln 2$. I arrived at the identity below after differentiating a unit square integral expression for $\eta(2)$.
$$\eta(s)\Gamma(s)=\int_0^1\int_0^1 \frac{(-\ln(xy))^{s-2}}{1+xy}dxdy$$
But I can't for the love of it solve it any differently than that. I would love to find an "elementary approach" if possible.
Best Answer
I think it is possible to avoid the Dirichlet eta function. Put\begin{equation*} I=\int_{0}^{1}\int_{0}^{1}\dfrac{1}{(1+xy)\ln(xy)}\,dxdy. \end{equation*} By symmetry \begin{equation*} I = 2\int_{0}^{1}\left(\int_{0}^{x}\dfrac{1}{(1+xy)\ln(xy)}\,dy\right)\, dx. \end{equation*} Via the substitution $y=\dfrac{z}{x}$ we get \begin{equation*} I = \int_{0}^{1}\left(\int_{0}^{x^2}\dfrac{1}{x(1+z)\ln(z)}\,dz\right)\,dx. \end{equation*} We have a double integral integrated over the domain $0<z<x^2, \, 0<x<1$. If we change the order of integration then we first have to integrate with respect to $x$ over $\sqrt{z}<x<1$ and then with respect to $z$ over $0<z<1.$ Thus \begin{gather*} I = 2\int_{0}^{1}\left(\int_{\sqrt{z}}^{1}\dfrac{1}{x(1+z)\ln(z)}\, dx\right)\, dz = 2\int_{0}^{1}\dfrac{1}{(1+z)\ln(z)}\left[\ln(x)\right]_{\sqrt{z}}^{1}\, dz =\\[2ex] \int_{0}^{1}\dfrac{-1}{1+z}\, dz = -\ln(2). \end{gather*}