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..

Answer 1

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.