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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. $\endgroup$
    – amWhy
    Feb 23, 2018 at 14:44
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    $\begingroup$ Why do you believe this integral has a closed form? $\endgroup$
    – Clayton
    Feb 23, 2018 at 14:45
  • $\begingroup$ What is bessel function of first kind and zero order? $\endgroup$
    – ablmf
    Feb 23, 2018 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 ?" $\endgroup$ Feb 23, 2018 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 $\endgroup$ Feb 23, 2018 at 14:51

1 Answer 1

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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 $\endgroup$ Feb 24, 2018 at 15:00
  • $\begingroup$ @BrunoFelipe. You are welcome. $\endgroup$ Feb 24, 2018 at 15:41
  • $\begingroup$ To [tired, Mariusz Iwaniuk] Sir, I am dealing with the similar integration $\int_0^1 \sin{bx} J_0(\sqrt{1-x^2})\, dx$..No result of this integral mentioned in Gradshteyn and Ryzhik's book 7th edition, section 6.677. Is the result will be zero.Would you kindly tell me what will be the closed form result of this integration or the methods how I can proceed to solve it. $\endgroup$ Apr 25, 2020 at 5:59
  • $\begingroup$ @R.Bhattacharya. Write a new question. $\endgroup$ Apr 25, 2020 at 10:05
  • $\begingroup$ Okay Sir...I shall post the link (of the new question) here for your perusal... $\endgroup$ Apr 25, 2020 at 10:18

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