Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 believe this integral $$\int_0^a \frac{\cos(ux)}{\sqrt{a^2-x^2}}\mathrm dx$$

can not be computed exactly. However is there a method or transformation to express this integral in terms of the cosine integral or similar? I am referring to the integrals here.

$a$ is real number; with the change of variable this integral becomes

$$ \int_0^a\cos(u\sin t) \ \mathrm dt $$ with $$ x=a\sin t, $$ So, the new integral is $$ \int_0^{\pi /2}\cos(ua\sin t) \ \mathrm dt $$

share|cite|improve this question
Actually, it's expressible as a Bessel function... – J. M. Aug 8 '12 at 14:47
aja, thanks what bessel function if possible :) thanks again – Jose Garcia Aug 8 '12 at 14:48
up vote 5 down vote accepted


$$\int_0^a \frac{\cos(ux)}{\sqrt{a^2-x^2}}\mathrm dx$$

you were able to transform it into

$$\int_0^{\pi/2}\cos(au\sin\,t)\mathrm dt$$

which is expressible in terms of the Anger function $\mathscr{J}_\nu(z)$, which is equivalent to the more familiar Bessel function of the first kind $J_\nu(z)$ for integer orders:

$$\int_0^{\pi/2}\cos(au\sin\,t)\mathrm dt=\frac12\int_0^\pi\cos(au\sin\,t)\mathrm dt=\frac{\pi}{2}J_0(au)$$

share|cite|improve this answer
shouldn't it be $ \frac{\pi }{2} J_{0}(au/2) $ due to the change of variable $ t \rightarrow t/2 $ – Jose Garcia Aug 8 '12 at 18:58
anyway thank you all for your answers :D – Jose Garcia Aug 8 '12 at 18:58
@Jose, note the limits. :) – J. M. Aug 9 '12 at 1:03

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.