How to prove $\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) \, dx=\ln 2$ How to prove the integral
$$\begin{align}&\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\=&\int_0^\infty\left (\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right)dx\\=&\ln 2\end{align}$$
I try to use integration by parts
$$\begin{align}
&\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\
=&\int_0^\infty \int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\\
=&x\left.\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right|_0^\infty+\int_0^\infty \cos x \int_x^\infty\frac{\sin t}{t}dx+\int_0^\infty \sin x  \int_x^\infty\frac{\cos t}{t}dx
\end{align}$$
But the last two integrals are divergent.

The Laplace transform is useful for providing integral representations in terms of non-oscillating functions:
$$\text{si}(x)=\int_{x}^{+\infty}\frac{\sin t}{t}\,dt=\int_{0}^{+\infty}\frac{e^{-sx}}{1+s^2}\left(\cos(x)+s\sin(x)\right)\,ds $$
$$\text{Ci}(x)=\int_{x}^{+\infty}\frac{\cos t}{t}\,dt=\int_{0}^{+\infty}\frac{e^{-sx}}{1+s^2}\left(s\cos(x)-\sin(x)\right)\,ds $$
This allows to apply Fubini's theorem:
$$\begin{eqnarray*} \int_{0}^{+\infty}\text{si}(x)\text{Ci}(x)\,dx &=& \iiint_{(0,+\infty)^3}\frac{e^{-(a+b)x}}{(1+a^2)(1+b^2)}(\cos x+a\sin x)(b\cos x-\sin x)\,dx\,da\,db\\&=&\iint_{(0,+\infty)^2}\frac{1}{(1+a^2)(1+b^2)}\left(1-\frac{a}{a+b}+\frac{ab-3}{4+(a+b)^2}\right)\,da\,db\end{eqnarray*} $$
and the last double integral is pretty simple to be computed:
$$ \iint_{(0,+\infty)^2}\frac{da\,db}{(1+a^2)(1+b^2)}=\left(\frac{\pi}{2}\right)^2 $$
$$ \iint_{(0,+\infty)^2}\frac{a}{(1+a^2)(1+b^2)(a+b)}\,da\,db=\iint_{(0,+\infty)^2}\frac{b}{(1+a^2)(1+b^2)(a+b)}\,da\,db =\frac{\pi^2}{8}$$
and the difficult part
$$ \iint_{(0,+\infty)^2}\frac{ab-3}{(1+a^2)(1+b^2)(4+(a+b)^2)}\,da\,db=\int_{0}^{+\infty}\frac{-\pi(a^2+1)+2(a^2-1)\arctan\frac{a}{2}+2a\log(a^2+4)}{4(a^2+1)^2}\,da$$
can be broken by integration by parts.
