Prove that $\int _0^{\infty }\frac{xe^{-x^2}}{x+2}dx\:$ converges without finding the value or using dirichlet test

Question

How can i prove that $\int _0^{\infty }\frac{xe^{-x^2}}{x+2}dx\:$ coverges without finding the value or using dirichlet test ?

Thanks.

Answer 1

$$ \frac{x}{x+2}<1, \qquad\forall x\geq0 $$ so that $$ \frac{x}{x+2}e^{-x^2}<e^{-x^2}, \qquad\forall x\geq0 $$ and finally $$ \int_0^\infty \frac{x}{x+2}e^{-x^2}dx<\int_0^\infty e^{-x^2}dx $$ The right-hand side is an integral that is known to be convergent.

Answer 2

Since $\dfrac{x}{x+2}<1$ when $x\geq 0$ we have $$f(x)=\dfrac{xe^{-x^2}}{x+2}<e^{-x^2}$$ for such $x$. Now $e^{-x^2}\leq e^{-x}$ if $x\geq 1$ hence $$0\leq f(x)\leq e^{-x}$$ if $x\geq 1$. The integral $\int_1^{\infty} e^{-x}dx$ is convergent, hence $\int_1^{\infty} f(x)dx$ too. Since $f$ is continuous on $[0,1]$ the integral over $[0,\infty[$ converges too.