# 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.

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