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

  • $\begingroup$ Actually, it's expressible as a Bessel function... $\endgroup$ – J. M. is a poor mathematician Aug 8 '12 at 14:47
  • $\begingroup$ aja, thanks what bessel function if possible :) thanks again $\endgroup$ – Jose Garcia Aug 8 '12 at 14:48


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

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:

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

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

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