How can I compute $\int_0^\infty {\sin x \over x} dx$ from computing $\int_0^\infty e^{-xt} {{\sin x} \over x} dx$ 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.
 A: 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
$$ \lim_{t\to 0^+}\int_{0}^{+\infty}\frac{\sin x}{x}e^{-tx}\,dx = \frac{\pi}{2}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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\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}

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