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What's the value of the integral given by : $$ I(a) = \int_0^\infty \dfrac{1}{x} e^{-a x} J_{3/2}(x)J_{3/2}(x) dx, $$ where $a$ is a positive real parameter.

I don't know if this could help, but following is the graph I obtain numerically. enter image description here

Also, taking a look at Gradshteyn and Ryzhik (edition 8), no interesting formula seems to be proposed. Except for formula 6.626 (page 711), where it's mentioned that it applies only for $a > 1$. However, the numerical computation shows that the integral does converge for all $a$ as shown in the graph above.

Mokrane
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    Have you tried using $$ J_{3/2} (x) = - \sqrt {\frac{{2x}}{\pi }} \frac{d}{{dx}}\frac{{\sin x}}{x},? $$ – Gary Mar 01 '22 at 10:47
  • Yes, but I couldn't obtain any analytical expression out of it. This is what I obtained : $$ \dfrac{2}{\pi}\int_0^\infty \dfrac{1}{x^2}e^{-a x}\left(\dfrac{\sin x}{x} - \cos x \right)^2 d x .$$ And by developing further and isolating the trigonometric functions, the three obtained integrals do not converge if calculated separately. – Mokrane Mar 01 '22 at 10:48
  • (identical to the remark by @Gary) It looks you are using (consciously ?) a classical expression for $J_{3/2}$ : see https://math.stackexchange.com/q/3527919/305862 – Jean Marie Mar 01 '22 at 10:58
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    You can try using this: $$\int_0^\infty e^{-a x} (\frac{\sin x}{x} - \cos x)^2 dx = \frac{a^2 + 2}{a^3 + 4 a} - \frac14 a \log(\frac{4}{a^2} + 1)$$ And then integrating twice w.r.t. a. Although it might not make the problem easier. – Yuriy S Mar 01 '22 at 10:58
  • @YuriyS Can I have a reference for that ? – Mokrane Mar 01 '22 at 11:04
  • Are you looking at it in the perspective of a Laplace transform ? – Jean Marie Mar 01 '22 at 11:08
  • It is easy to see that the integral converges for all $a>0$. The integrand is of order $x^2$ near $x=0$. – Gary Mar 01 '22 at 11:16
  • @JeanMarie Not necessarily, but why not.. In this case, the question would be to find $$\mathfrak{L}\left( \dfrac{1}{x^2}J_{3/2}(x)J_{3/2}(x)\right)$$. – Mokrane Mar 01 '22 at 11:18
  • @Gary Yes, definitely! – Mokrane Mar 01 '22 at 11:20
  • @YuriyS Oh, it's even more complicated this way ! – Mokrane Mar 01 '22 at 11:21
  • With the help of Mathematica, I get $I(a)= -\frac{3\left(a^2+2\right)(a-2i)\ln(a-2i)-2a\left(a^2+6\right)\ln(a)+a\left(a^2+6\right)\ln(a+2i)+2a(-2+\pi (3+ia)a)-2(a+i)(a-2i)^2\ln(-a+2i)+4i\ln(a+2i)+8\pi}{12\pi}$ – KStar Mar 01 '22 at 11:54
  • @KStarGamer Oh, this is interesting! I'll try to simplify it to get rid of the complex number $i$ and check if it matches the numerical result. Thank you very much! – Mokrane Mar 01 '22 at 12:44
  • @Mokrane We actually have the stronger generalisation: $\int_{0}^{\infty} x^{s-1} e^{-a x} J_{3/2} (x)^2 , dx = \frac{\left((a-2i)^{1-s}\left(a^2-2ia(s-1)-s^2+s+2\right)+(a+2i)^{1-s}\left(a^2+2ia(s-1)-s^2+s+2\right)-2\left(a^2+s^2-5s+6\right)a^{1-s}\right)\csc(\pi s)}{2\Gamma(4-s)}$ – KStar Mar 01 '22 at 13:44
  • interesting integral. A closed form exists and can be extracted the following way: 1.) recognize that this integral is a prime candidat to use Mellin-Parseval theorem 2.) use https://dlmf.nist.gov/10.22#E57 to obtain one of the needed Mellin transform, the second is trivial 3) close the contour in the appropriate half space of the complex plane 4) mechanically pick up residues and calculate some easy sumes related to $\log(x)$

    This technique generalizes nicely under the replacments $J_{3/2}(x)\rightarrow J_{n/2}(x), , , 1/x \rightarrow x^s $

     – asgeige Mar 02 '22 at 22:21
  • a proof for DMLF 10.22.57 should actually not too difficult: use https://dlmf.nist.gov/10.22#E43 which is relativly simple to proof togehter with https://math.stackexchange.com/questions/2340981/square-of-the-bessel-function-with-for-integer-n and a simple Beta-integral should get u a long way – asgeige Mar 02 '22 at 22:37

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Here is a direction which will give you a series solution:

Using formula 8.442 P. 960 of my edition of Gradshteyn and Ryzhik (Table of integrals, series and products, Academic Press, 1980) :

$$J_{3/2}(x)^2=\sum_{k=0}^{\infty}(-1)^k\left(\frac{x}{2}\right)^{2k+3}\frac{\Gamma(2k+4)}{k! \Gamma(k+4)\Gamma(k+\frac52)^2}$$

(take care to the last square). From there, integrate term by term to obtain a solution under the form of a series. Has this series a closed form ? Hopefully as an hypergeometric expression...

Jean Marie
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