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I need a help with integral below,

$$\int_{0}^{1}x\sin (bx) J_0\left(a\sqrt{1-x^2}\right)\,\mathrm dx$$

where $a \geq 0$ and $b$ are constants and $J_0(x)$ is the zeroth-order of Bessel function of the first kind.

This integral has a closed form solution ?

I found some integrals similar to the integral above, but I don't have any idea on how to apply it.

I found this integral expression below on Gradshteyn and Ryzhik's book 7th edition, section 6.677, number 6:

$$\int_{0}^{a} \cos (cx)J_0\left(b\sqrt{a^2-x^2}\right)\,\mathrm dx = \frac{\sin (a\sqrt{b^2+c^2})}{\sqrt{b^2+c^2}} \quad [b\geq 0]$$

I try use the definition of zeroth-order of Bessel function of the first kind to solve this integral:

$$J_0(z) = \sum_{k=0}^\infty (-1)^k\frac{(\frac{1}{4}z^2)^k}{(k!)^2}$$

then i found:

$$J_0(a) \sum_{k=0}^\infty \int_{0}^{1}x\sin (bx)\left(1-x^2\right)^k\,\mathrm dx $$

On Wolfram Alpha i found that:

$$\int_{0}^{1}x\sin (bx)\left(1-x^2\right)^k\,\mathrm dx = \frac{1}{4}\sqrt{\pi}a\Gamma (k+1)\tilde{F}_1\Big(;k+\frac{5}{2};-\frac{a^2}{4}\Big)$$

where $\tilde{F}_1$ is the Regularized Hypergeometric Function, where:

$$\tilde{F}_1\Big(;k+\frac{5}{2};-\frac{a^2}{4}\Big) = \frac{J_{k+\frac{3}{2}}(a) }{(\frac{a}{2})^{k+\frac{3}{2}}}$$

therefore:

$$\int_{0}^{1}x\sin (bx) J_0\left(a\sqrt{1-x^2}\right)\,\mathrm dx = J_0(a)\frac{1}{4}\sqrt{\pi}a \sum_{k=0}^\infty \Gamma (k+1) J_{k+\frac{3}{2}}(a)\Big(\frac{a}{2}\Big)^{-(k+\frac{3}{2})} $$

The solution found above is right ?

Thanks in advance.

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    $\begingroup$ "I want..." ? What's stopping you from doing so? Please ask a question. $\endgroup$ – Namaste Feb 23 '18 at 14:44
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    $\begingroup$ Why do you believe this integral has a closed form? $\endgroup$ – Clayton Feb 23 '18 at 14:45
  • $\begingroup$ What is bessel function of first kind and zero order? $\endgroup$ – ablmf Feb 23 '18 at 14:48
  • $\begingroup$ "What's stopping you from doing so? Please ask a question." - Sorry, this is my first post, the question is "This integral can be solved in closed form ?" $\endgroup$ – Bruno Felipe Feb 23 '18 at 14:48
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    $\begingroup$ "Why do you believe this integral has a closed form?" I don´t have indications, i prefer to be in closed form, but an approximation it is fine for me too $\endgroup$ – Bruno Felipe Feb 23 '18 at 14:51
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Suggestion by user tired:

Integral expression below on Gradshteyn and Ryzhik's book 7th edition, section 6.677, number 6:

$$\int_0^1 \cos (b x) J_0\left(a \sqrt{1-x^2}\right) \, dx=\frac{\sin \left(\sqrt{a^2+b^2}\right)}{\sqrt{a^2+b^2}}$$

Differentiate for variable b:

$$\int_0^1 \frac{\partial \left(\cos (b x) J_0\left(a \sqrt{1-x^2}\right)\right)}{\partial b} \, dx=\frac{\partial }{\partial b}\frac{\sin \left(\sqrt{a^2+b^2}\right)}{\sqrt{a^2+b^2}}$$ $$\color{blue}{\int_0^1 x \sin (b x) J_0\left(a \sqrt{1-x^2}\right) \, dx=-\frac{b \cos \left(\sqrt{a^2+b^2}\right)}{a^2+b^2}+\frac{b \sin \left(\sqrt{a^2+b^2}\right)}{\left(a^2+b^2\right)^{3/2}}}$$

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  • $\begingroup$ Thank you!, i am very grateful $\endgroup$ – Bruno Felipe Feb 24 '18 at 15:00
  • $\begingroup$ @BrunoFelipe. You are welcome. $\endgroup$ – Mariusz Iwaniuk Feb 24 '18 at 15:41

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