I'm working through the exercises of Bender and Orszag's famous book, but I got stuck in 6.25 (a), in which it is asked to prove that

$$J_\nu (x) \sim (x/2)^\nu / \Gamma(\nu+1) \; \text{as} \; \nu \rightarrow \infty,$$

by using the following integral representation

$$J_\nu(x)=\frac{(x/2)^\nu}{\sqrt{\pi}\Gamma(\nu+1/2)} \int^\pi_0 \cos(x\cos\theta) \sin^{2\nu}\theta \, d\theta,$$

which is valid for $\nu > -1/2.$ ($J_\nu(x)$ is the $\nu$th-order Bessel function of the first kind.)

As the exercise belongs to section 6.4, which deals with Laplace's method and Watson's lemma, I thought I first had to perform a change of variables in order to get an integral of the form

$$I(x)=\int^b_a f(t)e^{x\phi(t)} \, dt.$$

So, I took $t=\cos\theta$ and obtained

$$\frac{(x/2)^\nu}{\sqrt{\pi}\Gamma(\nu+1/2)} \int^{1}_{-1} (1-t^2)^{p-\frac{1}{2}} e^{ixt} \, dt.$$

However, I cannot apply either Laplace's method or Watson's lemma, because the function $\phi$ I got is complex: $\phi(t)=it$.

What am I missing?

  • $\begingroup$ A suggestion based on a large amount of ignorance: Can you consider the real and imaginary parts separately? $\endgroup$ May 24 '14 at 4:25
  • $\begingroup$ Thanks for your comment, but I don't think it makes anything easier as I would end up with $I(x)=\int^b_a f(t)\cos(xt) dt$ and again I would not be able to apply either Laplace's method or Watson's lemma. $\endgroup$ May 24 '14 at 4:44
  • $\begingroup$ For reference, the large $\nu$ behavior is $x^\nu/\left(\sqrt{\nu}2^{\nu}\Gamma(\nu+1/2)\right)$. $\endgroup$ May 24 '14 at 14:31
  • $\begingroup$ @AntonioVargas, are you sure? where did you get that from? I'm just asking because your expression contradicts the point of the exercise. $\endgroup$ May 24 '14 at 14:42
  • $\begingroup$ Yes, I am sure. I calculated it using the Laplace method and then verified it numerically. As the answer below indicates, there must have been a typo in the statement of the exercise. If you'd like I can write up a short answer with some details of the calculation. $\endgroup$ May 24 '14 at 14:49

To begin, rewrite the integral as

$$ \int_0^\pi \cos(x\cos \theta) \exp\Bigl[2\nu \log \sin \theta\Bigr]\,d\theta. $$

The quantity $\log \sin \theta$ has a maximum at $\theta = \pi/2$, and near there

$$ \log\sin\theta = -\frac{1}{2} \left(\theta - \frac{\pi}{2}\right)^2 + \cdots. $$


$$ \cos(x\cos\theta) = 1 + \cdots $$

there, so by the Laplace method we have

$$ \int_0^\pi \cos(x\cos \theta) \exp\Bigl[2\nu \log \sin \theta\Bigr]\,d\theta \sim \int_{-\infty}^{\infty} 1\cdot \exp\left[-2\nu \cdot \frac{1}{2} \left(\theta - \frac{\pi}{2}\right)^2\right]\,d\theta $$

for large $\nu$. Now simplify.

  • $\begingroup$ the secret is always in the first step! thanks a lot, Antonio. $\endgroup$ May 24 '14 at 15:15


As Antonio Vargas stated in the comments above, when $v$ is large, $\sin^{2v} \theta $ is small everywhere on the interval $[0, \pi]$ except $\theta = \frac{\pi}{2}$ (where $\sin (\theta) = 1$).

So a slightly modified argument, noting that $\cos (x \cos \theta)=1$ when $\theta = \frac{\pi}{2}$, shows that it is also an asymptotic expansion of $J_{v}(x)$ as $v \to \infty$.

$ $

Both Wikipedia and Wolfram MathWorld state that is an asymptotic expansion as $x \to 0$.


http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html (57)

$ $

Notice that for small $x$, $\cos (x \cos \theta)$ is essentially $1$.

Therefore, $$J_{v}(x) \sim \frac{(\frac{x}{2})^{v}}{\sqrt{\pi}\ \Gamma (v + \frac{1}{2})} \int_{0}^{\pi} \sin^{2v} \theta \ d \theta$$

$$ = 2 \ \frac{(\frac{x}{2})^{v}}{\sqrt{\pi}\ \Gamma (v + \frac{1}{2})} \int_{0}^{\pi /2} \sin^{2v} \theta \ d \theta$$

$$ = 2 \ \frac{(\frac{x}{2})^{v}}{\sqrt{\pi}\ \Gamma (v + \frac{1}{2})} \int_{0}^{\pi /2} \sin^{2(v+1/2)-1} (\theta) \cos^{2(1/2)-1} (\theta) \ d \theta$$

$$ = \frac{(\frac{x}{2})^{v}}{\sqrt{\pi}\ \Gamma (v + \frac{1}{2})} B \left(v+\frac{1}{2},\frac{1}{2} \right)$$

$$ =\frac{(\frac{x}{2})^{v}}{\sqrt{\pi}\ \Gamma (v + \frac{1}{2})} \frac{\Gamma(v+\frac{1}{2}) \Gamma(\frac{1}{2})}{\Gamma(v+1)}$$

$$ = \frac{(\frac{x}{2})^{v}}{\Gamma(v+1)}$$

  • $\begingroup$ your answer makes sense. I've also found in a couple of other books that this is the asymptotic expansion for $x \rightarrow 0$. however, there are two things that make me uneasy about your solution: i) the fact that $\nu \rightarrow \infty$ isn't used anywhere; ii) I've found the exact same exercise in another book and they give two additional hints: $\int^{\infty}_{-\infty} e^{-\nu t^2} dt = \sqrt{\pi / \nu}$ and $\Gamma(\nu+1/2) / \Gamma(\nu) \sim \sqrt{\nu} \;\text{as}\; \nu \rightarrow \infty.$ I'll give it another try and if I don't arrive anywhere, I'll accept your answer. $\endgroup$ May 24 '14 at 13:52
  • $\begingroup$ @Pringoooals The expansion is valid even if $v$ is small. Wikipedia just says it has to be positive. $\endgroup$ May 24 '14 at 13:59
  • $\begingroup$ you're are 100% right. what bothers me right now is the fact that I'm still not using either Laplace's method or Watson's lemma, or even the method of steepest descent, which is what this exercise should be all about. anyway, your take on this is so beautifully simple that doing it any other way now seems counterproductive. thanks for enlightening me! $\endgroup$ May 24 '14 at 14:05
  • $\begingroup$ It also happens to be the expansion as $\nu \to \infty$ with $x$ fixed. In fact your proof, with some modified reasoning, is sufficient to prove it for that case as well ;) $\endgroup$ May 24 '14 at 15:02

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