Fubini's Theorem double integral with sin and $e^{-x}$

Question

I need to calculate $\int_{E} \frac{y}{x} e^{-x} \sin x d \mu$, where $\mu$ is the product of Lebesgue measure on $\mathbb{R}$ with itself, and $E = \{(x, y): 0 \leq y \leq \sqrt{x} \}$. So, as a double integral, it looks like:

$$\int_{0}^{\infty} \int_{0}^{\sqrt{x}} \frac{y}{x} e^{-x}\sin x\, dy\, dx.$$

I'd like to be able to apply Fubini's Theorem so I can change order of integration, but in order to do that I need some helpful bound for the integrand. Is there some surprising way I can do that, say, for the $e^{-x}$ function? And once I do that, does the fact that $\frac{\sin x}{x}$ goes to $0$ as $x$ goes to infinity. Can you help me out somehow?

Whoops, fixed, I was confusing it with the limit as x goes to 0. — BMac, Mar 30 '19 at 18:51
The integrand is bounded by $ye^{-x}$, which is non-negative and integrable on $E$. So Fubini's theorem is applicable. — Sangchul Lee, Mar 30 '19 at 19:15
In this case, on reflection, I guess I'm not (though a lot of times that's the goal). — BMac, Mar 30 '19 at 19:16
Sangchul Lee, is there a more efficient way to prove $ye^{-x}$ integrable on $E$ than directly calculating the integral? — BMac, Mar 30 '19 at 19:18
@BMac Given the upper limit $\sqrt{x}$ for $y$, you need only note $\int_0^\infty\sqrt{x}e^{-x}dx=\Gamma(3/2)$ is finite. — J.G., Mar 30 '19 at 21:35

score 0 · Answer 1 · edited Feb 25 '20 at 04:15

$$\int_0^\infty \int_0^{\sqrt{x}} \frac{y}{x} e^{-x}\sin x\, dy\, dx.$$

Since $y \le \sqrt{x},\ y^2 \le x$ so $x \ge y^2.$

So

\begin{align} I &=\int_0^\infty \int_0^{\sqrt{x}} \frac{y}{x} e^{-x}\sin x\, dy\, dx\\[6pt] &=\int_0^\infty \int_{y^2}^\infty \frac{y}{x} e^{-x}\sin x\, dx\, dy \end{align}

Don't see where to go from here.

Looking at the original problem,

$\begin{array}\\ I &=\int_{0}^{\infty} \int_{0}^{\sqrt{x}} \dfrac{y}{x} e^{-x}\sin x\, dy\, dx\\ &=\int_{0}^{\infty} \dfrac{e^{-x}\sin x}{x} \int_{0}^{\sqrt{x}}y \, dy\, dx\\ &=\int_{0}^{\infty} \dfrac{e^{-x}\sin x}{x}\dfrac{x}{2} dx\\ &=\dfrac12\int_{0}^{\infty} e^{-x}\sin xdx\\ &=\dfrac12(-\dfrac12) ( e^{-x} (\sin(x) + \cos(x)))|_0^{\infty}\\ &=\dfrac14\\ \end{array} $

This doesn't answer the question OP asked: how to bound the integrand in order to apply Fubini. — Umberto P., Mar 30 '19 at 21:21

Fubini's Theorem double integral with sin and $e^{-x}$

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