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The problem is to solve the following integrals. The parameters are all positives.

\begin{align} & \int \sqrt{ a^2 - x^2} \, J_1\left(r \sqrt{a^2-x^2}\right) \cos (x z) dx \\ & \int x J_0\left(r \sqrt{a^2-x^2}\right) \sin (x z) dx \end{align}

I try using trigonometric variable change and Euler formula. I think it makes them more complex to solve Here. I show how the integrals are transformed.

\begin{align} & x=a sin\theta , \,\,\,\,\, dx=a\cos\theta d \theta\\ \\ & e^{ i a \sin \theta z}= \cos (a \sin \theta z) + i\sin (a \sin \theta z) \\ \\ & a^2 \int \cos^2\theta J_1\left(r a \cos\theta \right) \cos (a \sin \theta z) d\theta \\ & a \int \sin \theta J_0\left(r a \cos \theta\right) \sin (a \sin \theta z) d\theta \end{align} The goal is to obtain close solutions for both integrals.

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  • $\begingroup$ Is it sufficient for you to compute the corresponding definite integrals on $(0,a)$? $\endgroup$ – Gary Sep 16 at 5:57
  • $\begingroup$ Yes, It would be very helpful to know that they can be integrated into that interval (0, a) $\endgroup$ – irondonio Sep 16 at 6:11
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You can evaluate them on $(0,a)$ in terms of spherical Bessel functions (which are expressible in terms of trigonometric functions) as follows: $$ \int_0^a {\sqrt {a^2 - x^2 } J_1 \left( {r\sqrt {a^2 - x^2 } } \right)\cos (xz)dx} = a^2 r\frac{{\mathsf{j}_1 (a\sqrt {r^2 + z^2 } )}}{{\sqrt {r^2 + z^2 } }} \\ = r\frac{{\sin (a\sqrt {r^2 + z^2 } )}}{{r^2 + z^2}} - ar\frac{{\cos (a\sqrt {r^2 + z^2 } )}}{{\sqrt {r^2 + z^2 } }} $$ and $$ \int_0^a {xJ_0 \left( {r\sqrt {a^2 - x^2 } } \right)\sin (xz)dx} = a^2 z\frac{{\mathsf{j}_1 (a\sqrt {r^2 + z^2 } )}}{{\sqrt {r^2 + z^2 } }} \\ = z\frac{{\sin (a\sqrt {r^2 + z^2 } )}}{{r^2 + z^2 }} - az\frac{{\cos (a\sqrt {r^2 + z^2 } )}}{{\sqrt {r^2 + z^2 } }}. $$ See (1.13.50) and (2.13.50) in A. Erdélyi, W. Magnus, F. Oberhettinger, F. G. Tricomi, Tables of Integral Transforms. Vol. I., McGraw-Hill Book Company, Inc., New York-Toronto-London, 1954.

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  • $\begingroup$ This book is very useful, it helps a lot, thanks $\endgroup$ – irondonio yesterday

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