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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 $(0,\infty)$.

Michael Hardy
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Laars Helenius
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    sin(x)/x is not $L^1(\pi,+\infty)$ – Tryss Mar 05 '15 at 03:42
  • In fact that $L_1$-norm is infinite. – Pedro Mar 05 '15 at 03:42
  • That is what I thought. Perhaps I should switch the norms? Certainly $\sin{x}/x$ is essentially bounded on $[\pi,\infty)$. – Laars Helenius Mar 05 '15 at 03:44
  • yes, sin(x)/x is bounded on $\mathbb{R}$ – Tryss Mar 05 '15 at 03:59
  • You are almost there! – science Mar 05 '15 at 04:07
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    I think dominated convergence theorem would be simpler and give the stronger result that it is everywhere differentiable on $(0,\infty),.$ – JLA Mar 05 '15 at 04:50
  • The followup question was to calculate the derivative. I ended up using dominated convergence to get the answer there. I suppose just calculating the derivate cuts to the chase though, right? – Laars Helenius Mar 05 '15 at 04:53
  • Yes, but you need to show that it converges to what you think it should converge to, and that's probably easiest to do using dominated convergence theorem. I'm not sure you can do this your way directly. – JLA Mar 05 '15 at 05:00
  • Well, all I wanted to show was that it was a differentiable function. I wasn't necessarily calculating the actual derivative. – Laars Helenius Mar 05 '15 at 05:08

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$$\begin{eqnarray*}\int_{\pi n}^{\pi(n+1)}\left|\frac{\sin(x)}{x}\right|\,dx \geq \int_{\pi \left(n+\frac{1}{4}\right)}^{\pi \left(n+\frac{3}{4}\right)}\left|\frac{\sin(x)}{x}\right|\,dx&\geq& \frac{1}{\sqrt{2}}\int_{\pi \left(n+\frac{1}{4}\right)}^{\pi \left(n+\frac{3}{4}\right)}\frac{dx}{x}\\&\geq&\frac{1}{\sqrt{2}}\int_{\pi \left(n+\frac{1}{4}\right)}^{\pi \left(n+\frac{3}{4}\right)}\frac{dx}{\pi \left(n+\frac{3}{4}\right)}\\&=&\frac{\sqrt{2}}{4n+3}\end{eqnarray*}$$ 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: $$\left|F(y_1)-F(y_2)\right|\leq\int_{\pi}^{+\infty}\frac{\left|e^{-xy_1}-e^{-x y_2}\right|}{x}\,dx.$$

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Jack D'Aurizio
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