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It is well-know that the Bessel function has the asymptotic expansion $J_n(\omega) \sim \left( \frac 2 {\pi \omega} \right)^{1/2} \left( \cos \left(\omega -\frac 1 2 n \pi - \frac 1 4 \pi\right) - \sin \left(\omega -\frac 1 2 n \pi - \frac 1 4 \pi\right) \frac {4n^2 - 1} {8 \omega} ...\right)$. We also have the integral representation $$J_n(\omega) = \int_0^1 \cos (n \pi x - \omega \sin \pi x) dx = \Re \int_0^1 e^{n \pi i x} e^{ - \omega \sin \pi x } dx.$$

Now we look at Hörmander's Theorem 7.7.5(The Analysis of Linear Partial Differential Operators I). The phase function is $f(x) = - \sin \pi x $, and has stationary point $x_0 = 1/2$. The weight function is $u(x) = e^{n \pi i x} $. Hörmander's theorem says \begin{equation} \begin{split} \int_0^1 u(x) e^{i \omega f(x)} d x & \sim e^{i \omega f(1/2)} (\det (\omega f''(1/2)/ 2 \pi i))^{-1/2} (u (1/2) + L_1 u \omega^{-1} )\\ & = \left( \frac 2 {\pi x} \right)^{1/2} e^{- i \omega + \pi i /4} (e^{n \pi i /2} + L_1 u \omega^{-1} ). \end{split} \end{equation} The first term is right. Let us look at the second term.

We have $f(1/2) = -1$, $f''(1/2) = \pi^2$, so $g_{1/2} (x) = - \sin \pi x + 1 - (\pi^2/2) (x-1/2)^2$. We note that $g_{1/2} (x)$ vanishes not only of third order but of fourth order at $1/2$, that is, $g(1/2) = g'(1/2) = g''(1/2) = g'''(1/2) = 0$, and $g''''(1/2) = - \pi^4$. $L_1 u $ is defined as $$i^{-1} \sum_{\nu - \mu = 1} \sum_{2\nu \geq 3 \mu} 2^{-\nu} \langle \pi^{-2} D, D \rangle^{\nu} (g_{1/2}^{\mu} u) (1/2) /\mu! \nu! .$$ We see that $$L_1 u = i^{-1} \left( 2^{-1} \pi^{-2} (n \pi i)^2 e^{n \pi i /2} + 2^{-3} \pi^{-4} (- \pi^4 ) e^{n \pi i /2} + 0\right) = i e^{n \pi i /2} \left( \frac 1 2 n ^2 + \frac 1 8 \right).$$ Therefore the second term is $ \sin \left(\omega -\frac 1 2 n \pi - \frac 1 4 \pi\right) \frac {4n^2 + 1} {8 \omega}$ according to Hömander instead of $ - \sin \left(\omega -\frac 1 2 n \pi - \frac 1 4 \pi\right) \frac {4n^2 - 1} {8 \omega}$. My guess is that there is a factor $(-1)^{\nu}$ missing in Hörmander's theorem, then also in Hörmander's Lemma 7.7.3., but this lemma looks right.

Could anyone tell me what goes wrong here? I would really appreciate it!

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I haven't looked too carefully at your description above, but looking at your final paragraph makes me wonder: are you using the right $D$? Hormander always used $D = -i \partial$ so that $D^2 = -\triangle$. This would easily account for the missing factor of $(-1)^\nu$ that you are looking for. – Willie Wong Jul 1 '13 at 11:17
up vote 8 down vote accepted

Looking closely again, I am pretty sure my comment gives the answer. Note that Hörmander uses the notation $D = - i \partial$. So in particular

$$ Du = -i (n\pi i) u $$

when $u = e^{n\pi i x}$. This gives an extra minus sign in the first term ($\nu = 1$ and $\mu = 0$) of the definition for $L_1 u$, while leaving the second term ($\nu = 2$ and $\mu = 1$) unchanged, providing the extra $(-1)^\nu$ you are looking for.

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Thanks very much. I need this result for my research, but I am not an expect in analysis, so I took the conventional D for granted. – Zhi Qi Jul 1 '13 at 12:48
@ZhiQi As a side remark: in literature concerning harmonic/fourier analysis (including pseudo/para differential calculus), the normalisation $D = -i \partial$ is very common, as it has the advantage $\widehat{Df} = \xi \widehat{f}$ which is somewhat more convenient when you do algebraic manipulations of the symbol. – Willie Wong Jul 1 '13 at 13:00
Thanks. I really have learned. I learned PDE by Evans' book with the conventional $D f = \nabla f$, so I did not realize the difference. I am a number and representation theorist. Bessel functions comes in naturally in my field, and I am trying to understand it via stationary phase method. That is why I encountered this "inconsistency". – Zhi Qi Jul 1 '13 at 13:14

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