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I am trying to understand how to approximate integrals with Bessel functions. In particular I have something like:

$$I_{\ell} = \int_{0}^{\infty} j_{\ell}(pr) dr = \frac{\sqrt{\pi} \Gamma[(1+\ell)/2] }{2p \Gamma(1+\ell/2) } \approx \frac{1}{p} \sqrt{ \frac{\pi}{2 \ell} } .$$

We can reduce $I_{\ell}$ into a suitable form via definitions:

$$I_{\ell} = \frac{1}{p} \sqrt{ \frac{\pi \nu}{2} } \int_{0}^{\infty} z^{-1/2} J_{\nu}(\nu z) dz$$

where $\nu = \ell +1/2$.

Then there is this representation http://dlmf.nist.gov/10.9.E17, where we can write :

$$J_{\nu }(z) = \frac{1}{2\pi i} \int_{\infty - i\pi}^{\infty + i \pi} \exp ( \nu (z \sinh t- t)),$$ which puts $I_{\ell}$ of the form:

$$I_{\ell} = \frac{1}{p} \sqrt{ \frac{\pi \nu}{2} } \int_{0}^{\infty} dz z^{-1/2} \int_{\infty - i\pi}^{\infty + i \pi} dt\exp ( \nu (z \sinh t- t)) .$$

I guess we could use $\Phi(z,t) = z \sinh t -t$.

Is this integral amenable to saddle point techniques? https://en.wikipedia.org/wiki/Method_of_steepest_descent

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    $\begingroup$ So you would like to approximate $I_\ell$ without relying on its closed form in terms of gamma functions? $\endgroup$ Commented Nov 22 at 11:22
  • $\begingroup$ yes exactly, i want to get an approximate answer using some approximation technique. I’m trying to understand it for this case since we already know what the answer is, we can check it $\endgroup$ Commented Nov 22 at 12:29

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