Show that $\int_{-\infty}^{\infty}\frac{x}{1+x^2y^2} dy$ is not continuous

We want to show that $\int_{-\infty}^{\infty}f(x,y) dy$ is not continuous on $[-a,a]$. By criterion:

$x\in I$,

$I = [-a,a]$,

$f(x,y)= \frac{x}{1+x^2y^2}$

There doesn't exist a function $h(y)$ such that for $x\in I$ the $|f(x,y)|\leq h(y)$ holds.

So besides looking at the endpoints $x=-a$ and $x= a$, what I would do here is to look at partial derivatives by $y$ and by $x$. However in the solutions there are only partial by $x$, why is that so (arent we looking to find the maximum of the whole function?) ?

Solution 1:

If $x > 0$, the substitution $u=xy$ gives $$\int_{-\infty}^{\infty}\frac{x}{1+x^2y^2} dy = \int_{-\infty}^{+\infty} \frac{1}{1+u^2} du = \pi$$

If $x < 0$, the same substitution gives $$\int_{-\infty}^{\infty}\frac{x}{1+x^2y^2} dy = -\int_{-\infty}^{+\infty} \frac{1}{1+u^2} du = -\pi$$

so the function $\displaystyle{x \mapsto \int_{-\infty}^{\infty}\frac{x}{1+x^2y^2} dy}$ is clearly not continuous at the point $x=0$.