I was wondering if one were to integrate $$\int_0^{b_2}\int_0^{b_1} \frac{1}{y-x} \mathrm{d}x\ \mathrm{d}y $$ where $b_2\geq b_1$ and both are strictly positive, would it diverge or not? I feel a bit uncomfortable with this because we know $\frac{1}{x}$ is divergent when we integrate across 0 but that does not seem to be the case here. Perhaps we first integrate with respect to $x$ to get $$\int_0^{b_2} ln(y)-ln|y-b_1|\ \mathrm{d}y = \int_0^{b_2}ln(y)\ \mathrm{d}y - \int_0^{b_2}\ln|y-b_1|\ \mathrm{d}y.$$ The second term can be split into two $$\int_0^{b_1} \ln(b_1-y) \mathrm{d}y + \int_{b_1}^{b_2}\ln(y-b_1)\mathrm{d}y.$$ So now we have $$\int_0^{b_2}ln(y)\ \mathrm{d}y - \int_0^{b_1} \ln(b_1-y)\ \mathrm{d}y - \int_{b_1}^{b_2}\ln(y-b_1)\ \mathrm{d}y$$ and thus $$ y\ln(y)-y\Big|_0^{b_2} - (y-b_1)(\ln(b_1-y)-1)\Big|_0^{b_1} - (y-b_1)(\ln(y-b_1)-1)\Big|_{b_1}^{b_2} $$ But we know that $\lim_{x\to0}x\ln(x)=0$, so we can evaluate this to $$b_2(\ln(b_2)-1) - (b_1)(\ln(b_1)-1) - (b_2-b_1)(\ln(b_2-b_1)-1).$$ Some terms cancel and we can get rid of the $-1$'s to get: $$b_2\ln(b_2)-b_1\ln(b_1)-(b_2-b_1)\ln(b_2-b_1).$$ But I'm not quite sure if this is all rigorous and when I chugged the double integral into Wolfram Alpha, it tells me that it diverges, but if it in fact does, can somebody explain to me where I went wrong on here? Would changing the integral into something like $$\int_c^{b_2}\int_c^{b_1} \frac{1}{y-x} \mathrm{d}x\ \mathrm{d}y $$ make it converge because we avoid the critical point of $0$?
This question came up when I was trying to find $\mathbb{E}[1/(Y-X)]$ where $Y\sim \textrm{Unif}\ [0,b_2]$ and $X\sim \textrm{Unif}\ [0,b_1]$ uniform distributions. If there's a better method to find that instead please let me know. Thanks!
Your answer has flaws in the first part. If you want to say \begin{equation} \int_0^{b_2}\int_0^{b_1} \frac{1}{y-x} \mathrm{d}x\ \mathrm{d}y = \int_0^{b_2} ln(y)-ln|y-b_1|\ \mathrm{d}y \end{equation} $y$ should lie outside of $[0, b_1]$ since $1/(y-x)$ goes to $\infty$ or $-\infty$ depending on the direction. Since there always exists some $y\in[0,b_2]$ such that $y\in[0,b_1]$, this integral cannot be defined. Because of this, the integral you initially propose cannot be defined.