If $f(x,y)=e^{-xy}\sin x\sin y$, $\int_0^\infty\int_0^\infty f(x,y)dxdy$ exists but $\iint_{\mathbb{R}_{\geq 0}^2} f(x,y) dxdy$ does not.

Question

Let $f(x,y)=e^{-xy}\sin x\sin y$ for $(x,y)\in\mathbb{R_{\geq 0}^2}$ and $f(x,y)=0$ wherever else.

I want to prove that

$$\int_{\mathbb{R}}\int_{\mathbb{R}} f(x,y)dxdy$$ exists but $$\iint_{\mathbb{R}^2} f(x,y)dxdy$$ does not, the first as Lebesgue integrals over $\mathbb{R}$, and the second as a Lebesgue integral over $\mathbb{R}^2$. Furthermore I want to prove that $$\int_{\mathbb{R}}\int_{\mathbb{R}} |f(x,y)|dxdy$$ does not exist (In a way that does not involve the inexistence of the second integral as hypothesis).

For the first integral, I considered this:

---

First, if $y=0$, the function is inmediately Lebesgue integrable with respect to $x$. If $y\neq 0$, let's consider that $$|f(x,y)|=|e^{-xy}\sin x\sin y|=e^{-xy}|\sin x||\sin y|\leq |\sin y|e^{-xy}$$ and in this case $f(x,y)$ is dominated by a Lebesgue integrable function $|\sin y| e^{-xy}$ which has an integral $$\int_0^\infty |\sin y|e^{-xy}dx=|\sin y| \int_0^\infty e^{-xy}dx=\frac{|\sin y|}{y}$$ so the inner integral $F(y)=\int_0^\infty f(x,y) dx$ exists and

$$|F(y)|=\left|\int_{0}^{\infty} f(x,y)dx\right|\leq \frac{|\sin y|}{y}.$$

I want to prove that $F(y)$ is Lebesgue-Integrable with respect to $y$, but even if it's dominated by $\frac{|\sin y|}{y}$, it is not enough, since $\frac{|\sin y|}{y}$ is not Lebesgue-Integrable.

How can I do this?

Answer 1

For $y>0$, $\int_0^\infty e^{-xy}\sin x\,dx$ is the imaginary part of $\int_0^\infty e^{-(y-i)x}\,dx=1/(y-i)$. Thus $\int_0^\infty e^{-xy}\sin x\,dx=1/(1+y^2)$. The iterated integral of $f$ is thus $\int_0^\infty(\sin y)/(1+y^2)\,dy$ which is finite.

Here's an incomplete attempt to prove that $f$ is not an $L^1$ function.

We can write the (possibly divergent) double integral of $|f|$ as $$\begin{align} \int_0^\infty\int_0^\infty|\sin x\sin y|e^{-xy}\,dx\,dy &=\sum_{m,n=0}^\infty\int_0^\pi\int_0^\pi \sin x\sin y\, e^{-(x+m\pi)(y+n\pi)}\,dx\,dy\\ &=\int_0^\pi\int_0^\pi\sin x\sin y\sum_{m,n=0}^\infty e^{-(x+m\pi)(y+n\pi)}\,dx\,dy\\ &=\int_0^\pi\int_0^\pi\sin x\sin y\,G(x,y)\,dx\,dy\end{align}$$ say, where $$G(x,y)=\sum_{m,n=0}^\infty e^{-(x+m\pi)(y+n\pi)}=\sum_{m=0}^\infty \frac{e^{-xy}e^{-m\pi y}}{1-e^{-\pi(x+m\pi)}}.$$ All we need to do now is to show that $G(x,y)$ tends to infinity "quickly" enough as $(x,y)\to0$ to ensure the integral diverges. I'm not seeing how at the moment!

Answer 2

I think that your function actually belongs to $L^1(\mathbb R^2)$.

Let $y>0$. Since $$\int_a^be^{-xy}\sin x\,dx=-\frac{e^{-xy}}{1+y^2}(\cos x+y\sin x)\Big|_a^b,$$ we have that, for every $k\in\mathbb N$, $$\int_{2k\pi}^{(2k+1)\pi}e^{-xy}\sin x\,dx=\frac{e^{-2k\pi y}}{1+y^2}\left(e^{-\pi y}+1\right),$$ and also $$\int_{(2k+1)\pi}^{(2k+2)\pi}e^{-xy}\sin x\,dx=-\frac{e^{-2k\pi y}}{1+y^2}\left(e^{-2\pi y}+e^{-\pi y}\right).$$ Therefore, for $y>0$, $$\begin{align*}\int_0^{\infty}e^{-xy}|\sin x|\,dx&=\sum_{k=0}^{\infty}\left(\int_{2k\pi}^{(2k+1)\pi}e^{-xy}|\sin x|\,dx+\int_{(2k+1)\pi}^{(2k+2)\pi}e^{-xy}|\sin x|\,dx\right)\\ &=\sum_{k=0}^{\infty}\left(\int_{2k\pi}^{(2k+1)\pi}e^{-xy}\sin x\,dx-\int_{(2k+1)\pi}^{(2k+2)\pi}e^{-xy}\sin x\,dx\right)\\ &=\frac{1}{1+y^2}\left(e^{-2\pi y}+2e^{-\pi y}+1\right)\sum_{k=0}^{\infty}e^{-2k\pi y}\\ &=\frac{(1+e^{-\pi y})^2}{y^2+1}\frac{1}{1-e^{-2\pi y}}\\ &=\frac{1}{y^2+1}\frac{1+e^{-\pi y}}{1-e^{-\pi y}},\end{align*}$$ which shows that $$\int_0^{\infty}\int_0^{\infty}e^{-xy}|\sin x\sin y|\,dxdy=\int_0^{\infty}\frac{|\sin y|}{1+y^2}\frac{1+e^{-\pi y}}{1-e^{-\pi y}}\,dy.$$ The last integrand is bounded as $y\to 0$, and it is bounded above by $C/(y^2+1)$ as $y\to\infty$, so it converges.