Prove $f\in L^{1}(\lambda_2)$

Question

Let $(\mathbb{R}^{2},B(\mathbb{R}^{2}),\lambda_2)\; $ , where $\lambda_2$ is the Lebesgue measure for $\mathbb{R}^{2}$

We define :

$$f(x,y) = ye^{-y^{2}(1+x^{2})}1_{(x,y)\,\in\,\mathbb{R}_+\times \,\mathbb{R}_+}$$

Show that $f\in L^{1}(\lambda_2)$

** Edit : Without calulating the value of the integral **

I know that I have to bound f from above but with what ?

Thanks for your help.

Answer 1

Essentially, \begin{align*} \|f\|_{L^{1}(\lambda_{2})}=\int_{0}^{\infty}\int_{0}^{\infty}ye^{-y^{2}(1+x^{2})}dydx, \end{align*} a change of variable $u=y^{2}(1+x^{2})$ will do the computation in the inner integral.

Edit:

$e^{u}\geq\dfrac{1}{2}u^{2}$ for $u\geq 0$, so \begin{align*} e^{-y^{2}(1+x^{2})}\leq\dfrac{2}{y^{4}(1+x^{2})^{2}}, \end{align*} and \begin{align*} &\int_{0}^{\infty}\int_{0}^{\infty}ye^{-y^{2}(1+x^{2})}dydx\\ &\leq\int_{0}^{\infty}\int_{0}^{1}ye^{-y^{2}(1+x^{2})}dydx+\int_{0}^{\infty}\int_{1}^{\infty}\dfrac{2}{y^{3}(1+x^{2})^{2}}dydx, \end{align*} and these integrals converge.