The Bessel-Integral function of order $\nu$ is defined by the formula $$\text{Ji}_{\nu }(z)=\int_{\infty }^z \frac{J_{\nu }(t)}{t} \, dt , | \arg (z)| <\pi$$

How do we derive the asymptotic approximation $$\text{Ji}_{\nu }(x)\approx \sqrt{\frac{2}{\pi x}} \frac{\sin \left(x-\frac{\nu \pi }{2}-\frac{\pi }{4}\right)}{x} (x>0)$$

  • $\begingroup$ Spontaneously: replace $J_\nu(t)$ with almost any of the common integral representations, interchange the order of integration, and look for stationary phase. $\endgroup$ Nov 26, 2021 at 20:28
  • $\begingroup$ Slightly less spontaneously: based on the shape of the asymptotics you want, an integral representation like (Gradshteyn-Ryzhik 8.411.11) $J_\nu(t) = \frac{2}{\pi} \int_0^\infty \sin(t \cosh \xi - \frac{\nu \pi}{2} ) \cosh(\nu \xi) \, d \xi$ ought to do the trick. If you then locate a stationary phase around $x \approx t \cosh \xi$, you should get what you're looking for. $\endgroup$ Nov 27, 2021 at 0:09

1 Answer 1


Two options. Either, as suggested in the comments, replace $J_\nu(t)$ with an integral representation like (Gradshteyn-Ryzhik 8.411.11) $$ J_\nu(t) = \frac{2}{\pi} \int_0^\infty \sin\Bigl(t \cosh \xi - \frac{\nu \pi}{2} \Bigr) \cosh(\nu \xi) \, d \xi $$ and play stationary phase/saddle point method/Laplace method games.

Or, if you're lazy, take the word of someone (like the DLMF) who has already done this to $J_\nu(t)$ itself, getting (for $t$ large compared to $\nu$) $$ J_\nu(t) \approx \sqrt{\frac{2}{\pi t}} \cos\Bigl( t - \frac{\nu \pi}{2} - \frac{\pi}{4}\Bigr ). $$ Per the DLMF asymptotic, the error in this approximation saves a power of $t$ compared to this leading term (i.e., the error is $O(t^{-3/2})$) whence integrating it against $1/t$ contributes $O(x^{-3/2})$.

Insert this into $$ \mathrm{Ji}_\nu(x) \approx \int_{\infty}^x \frac{J_\nu(t)}{t} \, d t = \int_{\infty}^x \sqrt{\frac{2}{\pi t}} \frac{1}{t} \cos\Bigl( t - \frac{\nu \pi}{2} - \frac{\pi}{4}\Bigr ) \, d t $$ and integrate by parts to get $$ \mathrm{Ji}_\nu(x) \approx \sqrt{\frac{2}{\pi x}} \frac{\sin(x - \frac{\nu \pi}{2} - \frac{\pi}{4} )}{x} + \int_\infty^x \sqrt{\frac{2}{\pi t^5}} \sin\Bigl( t - \frac{\nu \pi}{2} - \frac{\pi}{4} \Bigr) \, d t. $$ Note that the integral at the end is $O(x^{-3/2})$, and so is the error from the asymptotic expansion of the Bessel function by the aforementioned power savings.

  • $\begingroup$ Perhaps it is useful to add that the last integral is $\mathcal{O}(x^{ - 3/2} )$. Note also that the error coming from the replacement of the Bessel function by its leading order asymptotics also has the same order of magnitude. $\endgroup$
    – Gary
    Nov 27, 2021 at 2:17
  • $\begingroup$ @Gary Both of these are fair and correct points. I (somewhat cheekily) assumed OP was happy with a pretty heuristic argument, what with $\approx$ and all. $\endgroup$ Nov 27, 2021 at 2:20
  • $\begingroup$ In any case, I've added notes to the above effect. $\endgroup$ Nov 27, 2021 at 2:29
  • $\begingroup$ Yes, certainly $\approx$ cannot be replaced by $\sim$ since the $\cos$ can vanish sometimes giving that $J_\nu$ is asymptotic to $0$ which makes no sense. $\endgroup$
    – Gary
    Nov 27, 2021 at 2:31

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