1
$\begingroup$

By chance, I see this formula

$\int_0^1 T_{2n+1}(x)\sin(ax) { dx \over \sqrt{1-x^2}}=(-1)^n\frac{\pi}{2}J_{2n+1}(a)$ but what is the closed form if we have

$\int_0^1 T_{2n}(x)\sin(ax) { dx \over \sqrt{1-x^2}}=?$

Here $T_n$ is Chebyshev polynomials of first kind and $J_n$ is Bessel function of first kind

$\endgroup$
  • $\begingroup$ I hope that this result will prove slightly enlightening. $\endgroup$ – Lucian Apr 6 '15 at 0:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.