Evaluating $\int_0^{\infty} \frac{\sin x}{x} dx$ with Fubini theorem. I have to calculate $\int_0^{\infty} \frac{\sin x}{x} dx$ using Fubini theorem. I tried to find some integrals with property that $\int_x^{\infty} F(t) dt = \frac{1}{x}$, but I cant find anything else but $F(t) = -\frac{1}{x^2}$ and unfortunetely I can't use this.
If I choose $F(t) = -\frac{1}{x}$, then I have:
$\displaystyle \int_0^{\infty} \frac{\sin x}{x} dx = -\int_0^{\infty} \sin x \int_{x}^{\infty} \frac{1}{t^2} dt$,
but I don't know what next
 A: The expression $\int_0^\infty \frac{\sin x}{x} dx$ is an abuse of notation, the precise meaning of this expression should be:
$$\lim_{t \to \infty} \int_0^t \frac{\sin x}{x} dx = \frac{\pi}{2} \tag{1}$$
since in fact $x^{-1} \sin x$ is not Lebesgue integrable over $(0, +\infty)$, because its positive and negative parts integrate to $\infty$. 
To show $(1)$ using the Fubini's theorem, see, for example, Example $18.4$ of Probability and Measure ($3$rd edition) by Patrick Billingsley. 
A: Integrating by parts twice, we get
$$
\begin{align}
&\int_0^L\sin(x)\,e^{-ax}\,\mathrm{d}x\tag{0}\\
&=-\int_0^L e^{-ax}\,\mathrm{d}\cos(x)\tag{1}\\
&=1-e^{-aL}\cos(L)-a\int_0^L\cos(x)\,e^{-ax}\,\mathrm{d}x\tag{2}\\
&=1-e^{-aL}\cos(L)-a\int_0^L e^{-ax}\,\mathrm{d}\sin(x)\tag{3}\\
&=1-e^{-aL}(\cos(L)+a\sin(L))-a^2\int_0^L\sin(x)\,e^{-ax}\,\mathrm{d}x\tag{4}\\
&=\frac{1-e^{-aL}(\cos(L)+a\sin(L))}{1+a^2}\tag{5}
\end{align}
$$
Explanation:
$(1)$: prepare to integrate by parts; $u=e^{-ax}$ and $v=\cos(x)$
$(2)$: integrate by parts
$(3)$: prepare to integrate by parts; $u=e^{-ax}$ and $v=\sin(x)$
$(4)$: integrate by parts
$(5)$: add $\frac{a^2}{1+a^2}$ times $(0)$ to $\frac1{1+a^2}$ times $(4)$
Now we can use Fubini
$$
\begin{align}
\int_0^L\frac{\sin(x)}x\,\mathrm{d}x
&=\int_0^L\sin(x)\int_0^\infty e^{-ax}\,\mathrm{d}a\,\mathrm{d}x\tag{6}\\
&=\int_0^\infty\int_0^L\sin(x)\,e^{-ax}\,\mathrm{d}x\,\mathrm{d}a\tag{7}\\
&=\int_0^\infty\frac{1-e^{-aL}(\cos(L)+a\sin(L))}{1+a^2}\,\mathrm{d}a\tag{8}\\[4pt]
&=\frac\pi2-\int_0^\infty\frac{L\cos(L)+a\sin(L)}{L^2+a^2}e^{-a}\,\mathrm{d}a\tag{9}
\end{align}
$$
Explanation:
$(6)$: $\int_0^\infty e^{-ax}\,\mathrm{d}a=\frac1x$
$(7)$: Fubini
$(8)$: apply $(5)$
$(9)$: arctangent integral and substitute $a\mapsto\frac aL$
Then, by Dominated Convergence, we have that
$$
\begin{align}
\lim_{L\to\infty}\int_0^\infty\left|\frac{L\cos(L)+a\sin(L)}{L^2+a^2}\right|\,e^{-a}\,\mathrm{d}a
&\le\lim_{L\to\infty}\int_0^\infty\frac{L+a}{L^2+a^2}\,e^{-a}\,\mathrm{d}a\\[4pt]
&=0\tag{10}
\end{align}
$$
Therefore, combining $(9)$ and $(10)$, we get
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{L\to\infty}\int_0^L\frac{\sin(x)}x\,\mathrm{d}x=\frac\pi2}\tag{11}
$$
A: $$\int_0^\infty dx \,\frac{\sin x}{x} = \int_0^\infty dx\, \sin x \int_0^\infty dt\, e^{-tx}$$
