How can I compute $\int_0^\infty {\sin x \over x} dx$ from computing $\int_0^\infty e^{-xt} {{\sin x} \over x} dx$

Question

This is what I tried.

I let $\int_0^\infty e^{-xt} {\sin x \over x} dx= F(t)$

and computed $f(t) = \int_0^\infty (-x) * e^{-xt} * (sinx/x) dx$

but I couldn't get anything more.

Please help me.

Answer 1

For any $t>0$, $\frac{\sin x}{x}e^{-tx}$ is a (Lebesgue- and Riemann-) integrable function over $\mathbb{R}^+$, bounded in absolute value by the integrable function $e^{-tx}$. By the dominated convergence theorem it follows that we may apply differentiation under the integral sign:

$$ F(t) = \int_{0}^{+\infty}\frac{\sin x}{x}e^{-tx}\,dx\quad\Longrightarrow\quad F'(t) = \frac{d}{dt} F(t)=-\int_{0}^{+\infty}\sin(x)e^{-tx}\,dx = -\frac{1}{t^2+1} $$ where the last identity follows from integration by parts. Since $\lim_{t\to +\infty}F(t)=0$, for any $t>0$ we have: $$ F(t) = -\int_{t}^{+\infty}\frac{du}{u^2+1} = \arctan\frac{1}{t} $$ hence $$ \lim_{t\to 0^+}\int_{0}^{+\infty}\frac{\sin x}{x}e^{-tx}\,dx = \frac{\pi}{2}.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\sin\pars{x} \over x}\,\expo{-xt}\,\dd x & = \int_{0}^{\infty}\expo{-xt}{1 \over 2}\int_{-1}^{1}\expo{-\ic k x}\,\dd k\,\dd x = {1 \over 2}\int_{-1}^{1}\int_{0}^{\infty}\expo{-\pars{t + k\ic}x}\,\dd x\,\dd k \\[5mm] & = {1 \over 2}\int_{-1}^{1}{1 \over t + k\ic}\,\dd k = \int_{0}^{1}{t \over k^{2} + t^{2}}\,\dd k = \int_{0}^{1/t}{\dd k \over k^{2} + 1} = \bbx{\ds{\arctan\pars{1 \over t}}} \end{align}

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$$\bbox[#ffe,10px,border:1px groove navy]{\ds{% \lim_{t \to 0^{+}}\int_{0}^{\infty}{\sin\pars{x} \over x}\,\expo{-xt}\,\dd x = \lim_{t \to 0^{+}}\arctan\pars{1 \over t} = {\pi \over 2}}} $$