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Considering the function $f(x,y)=\frac{1}{(1+y)(1+x^2 y)}$ for $x,y>0$. Proof that $f$ is integrable and deduce the value of $\int_{0}^{+\infty}\frac{\log(x)}{x^2 - 1}$

I don't completely understand how to do this, because my teacher has only taught us to calculate integrals in finite rectangles R=$[a_1,b_1] \times [a_2,b_2]$. $\int_0^\infty \int_0^\infty f \ \ dx dy $ is the only way I've thought about it.

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Hints:

You're on the right track. Integrate w.r.t. $y$ first: as a function of $y$, decompose $f(x,y)$ into partial fractions: $$\frac{1}{(1+y)(1+x^2 y)}=\frac{A}{(1+y)}+\frac{B}{(1+x^2y)}.$$ You should obtain $\;A=\dfrac 1{1-x^2}, \enspace B=\dfrac{x^2}{x^2-1}$, so that $$\int_0^{+\infty}f(x,y)\,\mathrm dy=\frac1{1-x^2}\log\frac{1+y}{1+x^2y} \Biggl|_0^{+\infty}=\frac1{1-x^2}\log\frac1{x^2}=\frac{2\log x}{x^2-1}.$$ Thus, the sought integral is half the double integral $\iint_{\mathbf R_+\times\mathbf R_+}f(x,y)\,\mathrm d x\,\mathrm d y. $

Now , applying Fubini's theorem, we also have $$\iint_{\mathbf R_+\times\mathbf R_+}\mkern-27muf(x,y)\,\mathrm d x\,\mathrm dy= \int_0^{+\infty}\!\biggl(\int_0^{+\infty}f(x,y)\,\mathrm dx\biggr)\mathrm dy= \int_0^{+\infty}\!\frac1{1+y}\biggl(\int_0^{+\infty}\mkern-12mu\frac1{1+x^2y}\,\mathrm dx\biggr)\mathrm dy.$$

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