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

• 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..

• To be clear: We're viewing $y$ as a constant, right? Commented Jan 10, 2021 at 0:56
• $y\gt 0$ OK. $y=0$ then $\int_0^\infty \frac{sinx}{x}dx=\frac{\pi}{2}$ Using absolute value doesn't work, integral in infinite... Commented Jan 10, 2021 at 0:58
• 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? Commented Jan 10, 2021 at 1:05
• As a hint, $\sin(x)/x$ is not Lebesgue integrable. See here for instance Commented Jan 10, 2021 at 1:06
• For $y > 0$, try comparing to $e^{-yx}/x$ rather than $\sin(x)/x$ to see that you're integrable. Commented Jan 10, 2021 at 1:06

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.