# 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

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vibrationdata.com/math/cosine_integral.pdf is relevant. –  Peter Phipps Nov 19 '11 at 16:25
I am almost sure that you can solve this problem by noticing that the Fourier transform of the function that is -i/2 on -1 to 0 and i/2 on 0 to 1 is $cos(x)/x$ but I don't quite see how to do it. –  user4143 Nov 19 '11 at 21:20
It was specified on the referenced site that $\alpha,\beta>0$. That should probably be carried through here. The statement is not true for general $\alpha$ and $\beta$. –  robjohn Nov 20 '11 at 10:42

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.

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After I had finished this, I saw that Bruno and Didier had done much the same thing, so I decided not to post. However, this argument, while no better than those before, was easier for me to follow, so I thought I would post anyway. If nothing else, it will bring this question back to the top of the list so that Bruno's and Didier's can get more votes :-) –  robjohn Nov 20 '11 at 15:29
Wow, robjohn, Fantastic. –  Cody Nov 20 '11 at 17:47

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.

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Bruno, please insert my post at the end of yours. Unless one of us went wrong, this should produce a complete solution. –  Did Nov 20 '11 at 1:42
Will do! I just completed my solution as well, using the same identity. You get the credit though :) –  Bruno Joyal Nov 20 '11 at 2:15
Perfect. You know you can copy-paste the source of my answer by clicking on "edit". –  Did Nov 20 '11 at 2:17
Merci beaucoup cher Didier! –  Bruno Joyal Nov 20 '11 at 2:18
Thank you Didier and Bruno. As always, very nice. I would like to give you both a greenie. I hate to slight anyone for their solutions. –  Cody Nov 20 '11 at 11:13
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