Study the integrability of $\frac{e^{-x}(1-\cos xy)}{y^2}$ on $[0,\infty)\times [0,\infty)$.

Question

Study the integrability of $\frac{e^{-x}(1-\cos xy)}{y^2}$ on $[0,\infty)\times [0,\infty)$.

I am having some trouble proving or disproving the integrability of the function above. Appealing to the Dominated Convergence Theorem seemed at first a good approach by letting $$F(y) = \int_0^{\infty} \frac{e^{-x}(1-\cos xy)}{y^2}dx$$ and then trying to find a dominating function $g(x) \in \mathcal{L}([0,\infty))$ for the absolute value of the integrand. I got this $$\left \vert \frac{e^{-x}(1-\cos xy)}{y^2}\right \vert= \left \vert \frac{2e^{-x}\sin^2(\frac{xy}{2})}{y^2} \right \vert \leq \left \vert \frac{2e^{-x}\frac{x^2y^2}{4}}{y^2} \right \vert = \left \vert \frac{x^2 e^{-x}}{2}\right \vert = \frac{x^2e^{-x}}{2} = g(x)$$ and $g(x) \in \mathcal{L}([0,\infty))$. Therefore, $F(y)$ is in $\mathcal{L}([0,\infty))$ and somehow, I should be able to conclude that $\int_0^{\infty} F(y) dy$ exists, which I can't (or I am missing some super obvious thing). Any help would be appreciated. Thank you!

Answer 1

Dominated convergence is an unnecessary complication, your method works just as well directly. The only problem is that $\frac12x^2e^{-x}$ is only integrable on $[0,\infty)\times[0,T]$, not $[0,\infty)\times[0,\infty)$. There's a simple workaround, though; the estimate $\sin t\le t$ is useful for $t\le1$, but $\sin t\le1$ is obviously an improvement for $t>1$. Hence we have $$\left|\frac{e^x(1-\cos y)}{y^2}\right|\le\begin{cases} \frac12x^2e^{-x}&\text{if }y\le1,\\ 2\frac{e^{-x}}{y^2}&\text{if }y>1. \end{cases}$$ Integrating, we find

$$\iint\left|\frac{e^x(1-\cos y)}{y^2}\right|\,\mathrm dx\mathrm dy\le\int_0^\infty\int_0^1\tfrac12x^2e^{-x}\,\mathrm dy\mathrm dx+\int_0^\infty\int_1^\infty2\tfrac{e^{-x}}{y^2}\,\mathrm dy\mathrm dx\\ =\int_0^\infty\tfrac12x^2e^{-x}\,\mathrm dx+\left(\int_0^\infty e^{-x}\,\mathrm dx\right)\left(\int_1^\infty\frac2{y^2}\,\mathrm dy\right)<\infty$$ and we're done.

Answer 2

We are dealing with a non-negative integrand function, hence: $$\int_{0}^{+\infty}\int_{0}^{+\infty}\frac{1-\cos(xy)}{y^2} e^{-x}\,dx\,dy \stackrel{Fubini}{=} \int_{0}^{+\infty}e^{-x}\int_{0}^{+\infty}\frac{1-\cos(xy)}{y^2}\,dy\,dx $$ and for every $x>0$ we have

$$ \int_{0}^{+\infty}\frac{1-\cos(xy)}{y^2}\,dy \stackrel{y\mapsto z/x}{=} x\int_{0}^{+\infty}\frac{1-\cos z}{z^2}\,dz =\frac{\pi}{2}x$$ from which it follows that $$\int_{0}^{+\infty}\int_{0}^{+\infty}\frac{1-\cos(xy)}{y^2} e^{-x}\,dx\,dy=\frac{\pi}{2}\int_{0}^{+\infty}xe^{-x}\,dx = \color{red}{\frac{\pi}{2}}.$$