Consider the integral
$\qquad \qquad \qquad \qquad \qquad \qquad$ $I\left(x\right) = ∫_0^\frac{π}{2} e^{-x \tan ^2\left(θ\right)} dθ $ as x $→\infty$
I'm looking for the first three terms of the asymptotic expansion of this. So far I have taken this to be of the form
$$∫ f\left(t\right)e^{-x\phi\left(t\right)}dt = \frac{1}{x} ∫ \frac{f\left(t\right)}{\phi'\left(t\right)} \frac{d}{dt}\left(e^{-x\phi\left(t\right)}\right) dt$$
$\qquad \qquad \qquad$ with f($θ$) = 1 $\quad$ $\phi\left(θ\right)$ = -tan$^2\left(θ\right)$ $\quad$ $\phi'\left(θ\right) = -\tan(θ)\sec^2(θ)$
So, $\phi\left(θ\right)$ has its maximum at $θ = 0$, and using Laplace's method, I set up my integral and did integration by parts as
$$-\frac{1}{2} ∫_0^\epsilon \frac{1}{x\tan \left(θ\right)\sec ^2\left(θ\right)} \frac{d}{dθ}\left(e^{-x\tan ^2\left(θ\right)}\right) dθ = -\frac{e^{-x\tan ^2\left(θ\right)}}{2x\tan \left(θ\right)\sec ^2\left(θ\right)}|^\epsilon_0 + \frac{1}{2}∫_0^\epsilon \frac{2\tan ^2\left(θ\right) + \sec ^2\left(θ\right)}{2x\sec ^2\left(θ\right)\tan ^2\left(θ\right)} e^{-x\tan ^2\left(θ\right)} dθ$$
But now, when I evaluate it at this maximum point, the denominator is zero. This seems to imply integration by parts fails for this function but in that case I don't know what else to do. Also, while it's not a mathematically sound point, it seems to get awfully ugly awfully fast which usually leads away from that particular solution. If it helps, I believe it has the general, one-term solution of
$$I\left(x\right) \sim \frac{1}{2} \frac{\sqrt{2π}f\left(c\right)e^{x\phi\left(c\right)}}{\sqrt{-x\phi''\left(c\right)}} = \frac{1}{2} \frac{\sqrt{2π}*1*e^{-x\tan \left(0\right)}}{\sqrt{-x\left(4\sec ^2\left(0\right)-6\sec ^4\left(0\right)\right)}} = \frac{1}{2}\frac{\sqrt{2π}}{\sqrt{-x\left(-2\right)}} = \frac{1}{2}\sqrt{\frac{π}{x}}$$
How do I get three terms from this?
Best Answer
Are you obliged to use Laplace's method?
The substitution $\tan^2\theta=y$ transforms the integral into $$I(x)=\frac{1}{2}\int_0^\infty\frac{e^{-xy}\,dy}{(1+y)\sqrt{y}}$$ which fits Watson's lemma, giving immediately $$I(x)\asymp\frac{1}{2}\sum_{n=0}^{(\infty)}(-1)^n\frac{\Gamma(n+1/2)}{x^{n+1/2}}\qquad(x\to\infty)$$