Hint/Proof Verification: $\int_{\mathbb R^{2}}|x|e^{-\sqrt{x^{2}+y^{2}}}d\lambda^{2}(x,y)$

Question

I am asked to substantiate/calculate:

$\int_{\mathbb R^{2}}|x|e^{-\sqrt{x^{2}+y^{2}}}d\lambda^{2}(x,y)$

My ideas:

Since $f(x,y):=|x|e^{-\sqrt{x^{2}+y^{2}}}$ is continuous and therefore measurable. Furthermore, $f(x,y)\geq 0$. Thus from Tonelli's theorem we get:

$\int_{\mathbb R^{2}}|x|e^{-\sqrt{x^{2}+y^{2}}}d\lambda^{2}(x,y)=\int_{\mathbb R} \int_{\mathbb R}|x|e^{-\sqrt{x^{2}+y^{2}}}d\lambda(x)d\lambda(y)$

Then from $\int_{\mathbb R} \int_{\mathbb R}|x|e^{-\sqrt{x^{2}+y^{2}}}d\lambda(x)d\lambda(y) $ the $|x|$ is rather annoying, and I want to eliminate it via symmetry (about $0$ for $x$)

thus we get $2\int_{\mathbb R} \int_{0}^{\infty}xe^{-\sqrt{x^{2}+y^{2}}}d\lambda(x)d\lambda(y)=2\int_{\mathbb R}[-e^{-\sqrt{y^2+x^2}}\sqrt{y^2+x^2}-e^{-\sqrt{y^2+x^2}}\vert ^{\infty}_{0}]d\lambda(y) $

Can I say that $\lim_{x \to \infty}-e^{-\sqrt{y^2+x^2}}\sqrt{y^2+x^2}=0$ since we have exponential growth, so the convergence to $0$ of $-e^{-\sqrt{y^2+x^2}}$ is quicker than the divergence of $\sqrt{y^2+x^2}$ to $\infty$

If this is the case, then we get $2 \int_{\mathbb R}e^{-y}(y+1)d\lambda(y)$ and in this case the lebesgue integral is equivalent to the riemann integral due to the continuity of $e^{-y}(y+1)$ on $\mathbb R$. Furthermore, then simply using riemann integration:

$2 \int_{\mathbb R}e^{-y}(y+1)d\lambda(y)= 2 \int_{\mathbb R}e^{-y}(y+1)dy=\infty$

Where did I go wrong? Or is the integral simply divergent?

Answer 1

The integrand is an even function of both $x$ and $y$. Therefore, we have

$$\begin{align} \int_{\mathbb{R}^2}|x|e^{-\sqrt{x^2+y^2}}\,d\lambda^2&=4\int_0^\infty \int_0^\infty xe^{-\sqrt{x^2+y^2}}\,dx\,dy\\\\ &=4\int_0^\infty \left.\left(-e^{-\sqrt{x^2+y^2}}\left(1+\sqrt{x^2+y^2}\right)\right)\right|_{x=0}^\infty\,dy\\\\ &=4\int_0^\infty (1+y)e^{y}\,dy\\\\ &=8 \end{align}$$

Alternatively, transformation to polar coordinates yields

$$\begin{align} \int_{\mathbb{R}^2}|x|e^{-\sqrt{x^2+y^2}}\,d\lambda^2&=\int_0^{2\pi}\int_0^\infty r|\cos(\theta)|e^{-r}\,r\,dr\,d\theta\\\\ &=4\int_0^\infty r^2 e^{-r}\,dr\\\\ &=8 \end{align}$$

as expected!

Answer 2

The problem is that you use $\sqrt{y^2}=y$ instead of $|y|$. Then your last integral diverges because $e^{-(-\infty)}=\infty$, instead of $e^{-|-\infty|}=0$.