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$$\int_0^z t^{\rho} J_{\mu}(at) J_{\nu}(bt) dt = \frac{(\frac 1 2 a z)^{\mu} (\frac 1 2 bz)^{\nu} z^{\rho + 1}}{\Gamma(\mu + 1)\Gamma(\nu + 1)} \times \sum_{k = 0}^{\infty} \frac{(-1)^k (\frac 1 2 a z)^{2k} {}_2 F_1(-k, -\mu - k; \nu + 1; b^2/a^2)}{k!(\mu + \nu + \rho + 2k + 1)(\mu + 1)_k}, \Re(\mu + \nu + \rho) > -1$$

I would like to solve infinite series, in simplified form.

Pali majumder
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  • Do you have this in a better format by chance? – Vladhagen May 16 '14 at 04:10
  • How did you arrive at this series, and what makes you assume a simplified form exists? – Nathaniel Bubis May 16 '14 at 04:16
  • Very often finite integrals of special functions do not come out to anything nice. This isn't going to reduce to anything more simple than what you have, I'm afraid. – Cameron Williams May 16 '14 at 04:30
  • I don't know that I've ever seen a nice form of a partial integral (meaning not from $0$ to $\infty$, or of that flavor) of Bessel functions, and I don't expect this one to be any different. – davidlowryduda May 16 '14 at 04:32
  • I think wolfram alpha can solve this. I plugged in the integral and got a single hypergeometric function, not a series of them. – David H May 16 '14 at 04:32
  • Scratch that. I forgot the scale factors $a$ and $b$ when I entered the integral into wolfram alpha. After adding them, WRA is no longer able to do the integral. Oops. – David H May 16 '14 at 04:38

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