Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:,Ci%29

share|cite|improve this question 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
up vote 4 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.