Evaluating $\int_0^1dx\int_x^{\frac{1}{x}}\frac{y\,dy}{(1+xy)^2(1+y^2)}$

Question

Evaluate the integral (change of order of integration may be useful):$$\int_0^1dx\int_x^{\frac{1}{x}}\frac{y\,dy}{(1+xy)^2(1+y^2)}$$

I searched the region, I got this graph

But, can't identify the region exactly. In fact, the lower bound of y and upper bound does not match to a region. I can be wrong. Please give any hints or you can add a solution.

Answer 1

The region of interest is bounded between the y-axis, the line $y=x$, and the line $y=\frac{1}{x}$, where $0<x<1$. Maybe this image will help you visualize it better:

To change the order of integration, first notice that we need to split the region into two along the line $y=1$ because of the shape of the region. Consider the first region $0 \leq y \leq 1$ and $0 \leq x \leq y$ and the second region $1 \leq y < \infty$ and $0 \leq x \leq \frac{1}{y}$. These two regions together are equal to the original region.

Rewrite your integral as follows: $$\int_0^1 \int_x^{\frac{1}{x}} \frac{y}{{(1+xy)}^2{(1+y^2)}} \; dy \; dx=\int_0^1 \int_0^{y} \frac{y}{{(1+xy)}^2{(1+y^2)}} \; dx \; dy + \int_1^{\infty} \int_0^{\frac{1}{y}} \frac{y}{{(1+xy)}^2{(1+y^2)}} \; dx \; dy$$ $$=\int_0^1 -\frac{1}{{(1+y^2)}^2} + \frac{1}{1+y^2}\; dy+ \frac{1}{2} \int_1^{\infty} \frac{1}{1+y^2} \; dy$$ $$=-\frac{\pi}{8}-\frac{1}{4}+\frac{\pi}{4}+\frac{\pi}{8}=\boxed{\frac{\pi}{4}-\frac{1}{4}}$$