tough integral involving the Cosine integral I ran across an integral on a German math site that has a friend of mine and I quite stuck.
They give, without derivation, 
$$\int_0^\infty \mathrm{Ci}(\alpha x)\mathrm{Ci}(\beta x)dx=\frac{\pi}{2 \max(\alpha,\beta)}$$
The Cosine Integral is defined as $\displaystyle \mathrm{Ci}(x)=-\int_x^\infty\frac{\cos(t)}{t}dt$
Does anyone know how this is derived?. We have looked around but can not find anything.
I ran it through Maple using specific values for $\alpha$ and $\beta$.
For instance, I used $\alpha=2, \;\ \beta=3$ and it gave $\dfrac{\pi}{6}$. Which indeed relates to the formula. The max of $\alpha$ and $\beta$ in this case is $\beta=3$.
So, $\dfrac{\pi}{2\cdot 3}=\dfrac{\pi}{6}$.
Does anyone know of this integral or its derivation?.  Thanks very much.
If anyone is interested, here is a link to the site:
http://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R%28x,Ci%29
 A: We can write your integral as
$$I=\int_0^\infty \int_{\alpha x}^\infty \int_{\beta x}^\infty \frac{\cos u \cos v}{uv } du\: dv\: dx,$$
or, what is the same,
$$\int_0^\infty \int_{x}^\infty \int_{x}^\infty \frac{\cos \beta u \cos \alpha v}{uv } du\: dv\: dx.$$
Now the region of integration is $$\{(x,u,v): x<u, x<v\}$$
which, up to a set of measure zero, we can write as the disjoint union of the regions $$\{(x,u,v): x<u<v\}$$ and $$\{(x,u,v): x<v<u\}.$$
Now for example, for the second region we have
$$\int_0^\infty \int_{x}^\infty \int_{v}^\infty \square\:   du\: dv\: dx = \int_0^\infty \int_{0}^u \int_{0}^v \square\:   dx\: dv\: du $$
and for us, this gives
$$ \int_0^\infty \int_{0}^u \int_{0}^v \frac{\cos \beta u \cos \alpha v}{uv } dx\: dv\: du = \int_0^\infty \int_{0}^u \frac{\cos \beta u \cos \alpha v}{u } dv\: du = \int_0^\infty \frac{\cos \beta u \sin \alpha u}{\alpha u } du$$
Now switching the roles of $u$, $v$, and the roles of $\alpha$ and $\beta$, and adding the resulting two integrals, we get that 
$$I(\alpha, \beta)=\int_0^\infty \frac{\beta \cos \beta t \sin \alpha t + \alpha \sin \beta t \cos \alpha t}{\alpha \beta t } dt.$$
Someone may be able to take it from there. The resemblance with the sine integral suggests to me that adapting one of the methods used to evaluate $\int_0^\infty \frac{\sin t}{t} dt$ may work. 

Edit: thanks to Didier Piau, here is the final step of the solution (direct quote from his post)
Starting from the penultimate expression in Bruno's solution, namely
$$I(\alpha, \beta)=\int_0^\infty \frac{\beta \cos \beta t \sin \alpha t + \alpha \sin \beta t \cos \alpha t}{\alpha \beta t } \mathrm dt,$$
let us use the trigonometric relations
$$
2\cos \beta t \sin \alpha t =\sin(\alpha+\beta)t+\sin(\alpha-\beta)t,\quad
2\sin \beta t \cos \alpha t =\sin(\alpha+\beta)t+\sin(\beta-\alpha)t,
$$
and the fact that for every $\gamma\ne0$, the change of variables $t\to\gamma t$ yields
$$
\int_0^\infty \frac{\sin \gamma t}{t} \mathrm dt=\text{sgn}(\gamma)\int_0^\infty \frac{\sin t}{t} \mathrm dt=\text{sgn}(\gamma)\frac{\pi}2,
$$
where $\text{sgn}(\gamma)$ is $+1$ if $\gamma\gt0$, $-1$ if $\gamma\lt0$, and $0$ if $\gamma=0$. This yields
$$
I(\alpha,\beta)=\frac\pi{4\alpha}(1+\text{sgn}(\alpha-\beta))+\frac\pi{4\beta}(1+\text{sgn}(\beta-\alpha)).
$$
This expression is symmetric in $(\alpha,\beta)$, as it should be. If $\alpha\gt\beta$, the second term is zero and the first one is $\pi/(2\alpha)=\pi/(2\max(\alpha,\beta))$. Finally, if $\alpha=\beta$, both terms are $\pi/(4\alpha)=\pi/(4\beta)$ hence the sum is $\pi/(2\alpha)=\pi/(2\beta)$. This proves the desired formula.
A: Assuming $\alpha,\beta>0$,
$$
\begin{align}
&\int_0^\infty\int_{\alpha x}^\infty\int_{\beta x}^\infty\frac{\cos(t)}{t}\frac{\cos(s)}{s}\;\mathrm{d}s\;\mathrm{d}t\;\mathrm{d}x\tag{1}\\
&=\int_0^\infty\int_x^\infty\int_x^\infty\frac{\cos(\alpha t)}{t}\frac{\cos(\beta s)}{s}\;\mathrm{d}s\;\mathrm{d}t\;\mathrm{d}x\tag{2}\\
&=\int_0^\infty\int_0^t\int_{x}^\infty\frac{\cos(\alpha t)}{t}\frac{\cos(\beta s)}{s}\;\mathrm{d}s\;\mathrm{d}x\;\mathrm{d}t\tag{3}\\
&=\int_0^\infty\int_0^\infty\int_0^{\min(s,t)}\frac{\cos(\alpha t)}{t}\frac{\cos(\beta s)}{s}\;\mathrm{d}x\;\mathrm{d}s\;\mathrm{d}t\tag{4}\\
&=\int_0^\infty\int_0^\infty\min(s,t)\frac{\cos(\alpha t)}{t}\frac{\cos(\beta s)}{s}\;\mathrm{d}s\;\mathrm{d}t\tag{5}\\
&=\int_0^\infty\int_0^ts\frac{\cos(\alpha t)}{t}\frac{\cos(\beta s)}{s}\;\mathrm{d}s\;\mathrm{d}t+\int_0^\infty\int_t^\infty t\frac{\cos(\alpha t)}{t}\frac{\cos(\beta s)}{s}\;\mathrm{d}s\;\mathrm{d}t\tag{6}\\
&=\int_0^\infty\int_0^ts\frac{\cos(\alpha t)}{t}\frac{\cos(\beta s)}{s}\;\mathrm{d}s\;\mathrm{d}t+\int_0^\infty\int_0^s t\frac{\cos(\alpha t)}{t}\frac{\cos(\beta s)}{s}\;\mathrm{d}t\;\mathrm{d}s\tag{7}\\
&=\int_0^\infty\int_0^ts\frac{\cos(\alpha t)\cos(\beta s)+\cos(\beta t)\cos(\alpha s)}{ts}\;\mathrm{d}s\;\mathrm{d}t\tag{8}\\
&=\int_0^\infty\frac{\cos(\alpha t)\sin(\beta t)/\beta+\cos(\beta t)\sin(\alpha t)/\alpha}{t}\;\mathrm{d}t\tag{9}\\
&=\int_0^\infty\frac{(\sin((\beta{+}\alpha)t)+\sin((\beta{-}\alpha)t))/\beta+(\sin((\alpha{+}\beta)t)+\sin((\alpha{-}\beta)t))/\alpha}{2t}\;\mathrm{d}t\tag{10}\\
&=\frac{\pi}{2}\left(\frac{1+\operatorname{signum}(\beta{-}\alpha)}{2\beta}+\frac{1+\operatorname{signum}(\alpha{-}\beta)}{2\alpha}\right)\tag{11}\\
&=\frac{\pi}{2}\frac{1}{\max(\alpha,\beta)}\tag{12}
\end{align}
$$
$(2)$ is a change of variables.
$(3)$ and $(4)$ are changes of order of integration.
$(5)$ is integration in $x$.
$(6)$ splits the domain where $s<t$ and $s>t$.
$(7)$ is a change of order of integration in the second integral.
$(8)$ is a change of variables in the second integral.
$(9)$ is integration in $s$.
$(10)$ is the trig identity: $2\sin(x)\cos(y)=\sin(x+y)+\sin(x-y)$.
$(11)$ is $\int_0^\infty\frac{\sin(\alpha t)}{t}\mathrm{d}t=\frac{\pi}{2}\operatorname{signum}(\alpha)$.
$(12)$ is just rewriting.
