Is $f(x) = \frac{\sin(x)}{x}e^{-xy}$ integrable for $x\in ]0,\infty[$ and $y \geq 0$?

Question

Is $\left]0,\infty\right[ \to \mathbb{R},\,\,\, \operatorname{f}\left(x\right) = \frac{\sin\left(x\right)}{x}\,\,{\rm e}^{-xy}\,\,\,$ Lebesgue-integrable for $\,\,\, y \geq 0\ ?$.
I tried $\,\,\,\left\vert\operatorname{f}\left(x\right)\right\vert \leq \left\vert\,{\frac{\sin\left(x\right)}{x}}\,\right\vert,\,\,\,$ but I don't know whether $\,\,\,\left\vert\,{\frac{\sin\left(x\right)}{x}}\,\right\vert\,\,\,$ is integrable.

Maybe I am taking a wrong direction..

$y\gt 0$ OK. $y=0$ then $\int_0^\infty \frac{sinx}{x}dx=\frac{\pi}{2}$ Using absolute value doesn't work, integral in infinite... — herb steinberg, Jan 10 '21 at 00:58
yes, y is a constant. bigger or equal 0. I don't understand you - so you are saying, that for y>0 it definitely is integrable? — Mathpiano, Jan 10 '21 at 01:05
As a hint, $\sin(x)/x$ is not Lebesgue integrable. See here for instance — HallaSurvivor, Jan 10 '21 at 01:06
For $y > 0$, try comparing to $e^{-yx}/x$ rather than $\sin(x)/x$ to see that you're integrable. — HallaSurvivor, Jan 10 '21 at 01:06
This seems also being very difficult to integrate - but if its integrable, then I'm happy enough :) — Mathpiano, Jan 10 '21 at 01:12
If you know $\left|\frac{\sin(x)}x\right| \leq 1$, it may instead be easier to directly compare with $|f(x)| \leq e^{-yx}$ instead (the only true difference is for $x\approx 0$, where $e^{-yx}/x$ has problems) — Brian Moehring, Jan 10 '21 at 03:40

score 1 · Answer 1 · answered Jan 10 '21 at 02:05

Just for curiosity. Observe that \begin{align} \int^\infty_0 \frac{\sin (\alpha x)}{x} e^{-yx}\ dx =&\ \int^\infty_0 \frac{e^{i\alpha x}-e^{-i\alpha x}}{2i x}e^{-xy}\ dx\\ =&\ \int^\infty_0 \frac{e^{i\alpha x}}{2i x}e^{-xy}\ dx- \int^\infty_0 \frac{e^{-ix}}{2i x}e^{-xy}\ dx\\ =&\ \int^\infty_0 \frac{e^{i\alpha x}}{2i x}e^{-xy}\ dx+\int^0_{-\infty} \frac{e^{ix}}{2ix}e^{xy}\ dx\\ =&\ \frac{1}{2}\int^\infty_{-\infty} e^{i\alpha x}\frac{e^{-y|x|}}{ix}\ dx. \end{align} Notice that \begin{align} \frac{d}{d\alpha}\int^\infty_0 \frac{\sin (\alpha x)}{x} e^{-yx}\ dx = \frac{1}{2}\int^\infty_{-\infty} e^{i\alpha x}e^{-y|x|}\ dx = \frac{y}{y^2+\alpha^2}. \end{align} Finally, we see that \begin{align} \int^\infty_0 \frac{\sin (\alpha x)}{x} e^{-yx}\ dx = \tan^{-1}\left(\frac{\alpha}{y} \right). \end{align} Set $\alpha = 1$ gives us the desired result.

Is $f(x) = \frac{\sin(x)}{x}e^{-xy}$ integrable for $x\in ]0,\infty[$ and $y \geq 0$?

1 Answers1