Prove that limit of integral is 0 (without dominated convergence theorem)

Question

Let $r,x,\sigma \in \mathbb{R}^+$.

Let $$f(x) = x\left(\frac{x^2}{2\sigma^2}-1\right)\exp\left(-\frac{x^2}{2\sigma^2} \right)$$

It can be shown that $$ \int_0^{+\infty} f(x)~dx = 0$$

I would like to prove that

$$ \lim_{r\rightarrow +\infty} \int_0^{+\infty} f(x)\sqrt{r^2+x^2}~dx = 0, $$

How to do that ? (I guess I cannot use the dominated convergence theorem here, to interchange limit and integral, since $\lim_{r\rightarrow +\infty} f(x)\sqrt{r^2+x^2}=\pm\infty$, right ?)

Answer 1

Under your assumptions, $$\int^\infty_0f(x)\sqrt{r^2+x^2}\,dx=\int^\infty_0f(x)\left(\sqrt{r^2+x^2}-r\right)\,dx.$$ Now $\sqrt{r^2+x^2}-r\rightarrow0$ as $r\rightarrow\infty$, and it's dominated by $x$. So if you can prove $\int^\infty_0|x\,f(x)|\,dx<\infty$, you're done (and that should be easy).