I am trying to prove that $$ F(y)=\int_\pi^\infty\frac{e^{-xy}\sin x}{x}\,dx $$ is differentiable on $(0,\infty)$. My first though was to try and show that $F$ is Lipschitz which led me to the following inequality: $$ \lvert F(y_1)-F(y_2)\rvert\le\int_\pi^\infty \left\lvert\frac{\sin x}{x} \right\rvert\cdot\lvert e^{-xy_1}-e^{-xy_2}\rvert\,dx. $$ Then I wanted to apply Holder's inequality to get $$ \lvert F(y_1)-F(y_2)\rvert\le\int_\pi^\infty \left\lvert\frac{\sin x}{x}\right\rvert\cdot\lvert e^{-xy_1}-e^{-xy_2}\rvert\,dx\le\left\|\frac{\sin x}{x}\right\|_1\cdot\|e^{-xy_1}-e^{-xy_2}\|_\infty, $$ but it occurs to me that I don't know that $\left\|\frac{\sin{x}}{x}\right\|_1$ exists. I know that $$ \int_{\pi}^\infty \frac{\sin{x}}{x}\,dx\le\int_0^\infty\frac{\sin x}{x}\,dx=\frac{\pi}{2}, $$ but that doesn't help me with $$ \int_\pi^\infty \left\lvert\frac{\sin x}{x}\right\rvert\,dx. $$ Any thoughts?
UPDATE:
With some input from my friends, I have decided I can switch the norms on my functions and get $$ \begin{align*} \lvert F(y_1)-F(y_2)\rvert&\le\int_{\pi}^\infty \left\lvert\frac{\sin x}{x}\right\rvert\cdot\lvert e^{-xy_1}-e^{-xy_2}\rvert\,dx\\ &\le\left\|\frac{\sin x}{x}\right\|_\infty\cdot\|e^{-xy_1}-e^{-xy_2}\|_1\\ &=\frac{1}{\pi}\cdot\lvert e^{-y_1\pi}-e^{-y_2\pi}\rvert\\ &\le\frac{1}{\pi}\cdot\lvert y_1\pi-y_2\pi\rvert\\ &=\lvert y_1-y_2\rvert. \end{align*} $$ Thus $F(y)$ is Lipschitz and therefore differentiable a.e. on $(0,\infty)$.
$$\begin{eqnarray*}\int_{\pi n}^{\pi(n+1)}\left|\frac{\sin(x)}{x}\right|\,dx \geq \int_{\pi \left(n+\frac{1}{4}\right)}^{\pi \left(n+\frac{3}{4}\right)}\left|\frac{\sin(x)}{x}\right|\,dx&\geq& \frac{1}{\sqrt{2}}\int_{\pi \left(n+\frac{1}{4}\right)}^{\pi \left(n+\frac{3}{4}\right)}\frac{dx}{x}\\&\geq&\frac{1}{\sqrt{2}}\int_{\pi \left(n+\frac{1}{4}\right)}^{\pi \left(n+\frac{3}{4}\right)}\frac{dx}{\pi \left(n+\frac{3}{4}\right)}\\&=&\frac{\sqrt{2}}{4n+3}\end{eqnarray*}$$ but since $\sum_{n\geq 1}\frac{\sqrt{2}}{4n+3}$ is divergent, $\frac{\sin x}{x}\color{red}{\not\in} L^1(\pi,+\infty).$
As pointed by Dr.MV in the comments, integration by parts leads to the stronger inequality: $$ \int_{n\pi}^{(n+1)\pi}\left|\frac{\sin x}{x}\right|\,dx \geq \frac{2}{\pi(n+1)}.$$
About differentiability, it is enough to get rid of the sine through: $$\left|F(y_1)-F(y_2)\right|\leq\int_{\pi}^{+\infty}\frac{\left|e^{-xy_1}-e^{-x y_2}\right|}{x}\,dx.$$