Can I express the integral $\int_0^1[\cos (xt)/(1-t^2)]dt$ in terms of Bessel Polynomial? I tried by putting $t=\sin \theta$ and used the integral representation of Bessel's polynomial $J_n(x)=(1/\pi)\int_0^\pi \cos(n\theta-x\sin \theta)d\theta$. I expect the answer may be $J_0(x)(\pi/2)$. But I did not get it even now. Help is solicited.

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    $\begingroup$ Why the original integral should be convergent? The integrand function has a non-integrable singularity in $t=1$. I bet you have forgotten a square root. $\endgroup$ – Jack D'Aurizio Feb 13 '15 at 17:07

Due to issues relating to convergence, the only values of x for which the integral converges are

of the form $x=(2k+1)~\dfrac\pi2$ , and the result is $I_{2k+1}=\dfrac\pi4\sqrt{|2k+1|}\cdot J^{(1,0)}\bigg(-\dfrac12~,~|2k+1|\dfrac\pi2\bigg)$,

for all $k\in\mathbb Z$.

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    $\begingroup$ If, however, you meant $\sqrt{1-t^2}$, as @Jack suggested, then the result is $I(x)=\dfrac\pi2~J_0(x)~$ for all $x\in\mathbb R$. $\endgroup$ – Lucian Feb 13 '15 at 17:50
  • $\begingroup$ ... by just setting $t=\sin \theta$. $\endgroup$ – Jack D'Aurizio Feb 13 '15 at 18:18
  • $\begingroup$ Yes. It is a an error on my part. It is $\sqrt{1-t^2}$. How can I change the limits from $(0,\pi/2)$ to $(0,\pi)$ $\endgroup$ – Purushothaman Feb 14 '15 at 1:04
  • $\begingroup$ Hint: $~\sin\bigg(\dfrac\pi2+u\bigg)~=~?$ $\endgroup$ – Lucian Feb 14 '15 at 5:03

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