Using the method of stationary phase, I was able to obtain the first term of the asymptotic expansion of the following integral, as $x \rightarrow \infty$:
$$\int\limits_0^{\pi / 2} {e^{ix\cos t}}dt \sim \sqrt{\frac{\pi}{2x}} e^{i({x-\pi/4})}.$$
However, I'm interested in finding a few more terms of the asymptoptic expansion, so I guess I have to apply the method of steepest descent.
This was my reasoning:
We have an integral of the form $$\int \limits_0^{\pi/2} e^{x\phi (t)}dt $$
with
$$\tag{1} \phi(t) = i\cos t.$$
We want to find paths for which $\operatorname{Im} \phi(t)$ is constant and determine which directions along these curves cause $\operatorname{Re} \phi(t)$ to decrease.
Let’s start by taking $t=x +iy$ on (1), so we can find the real and imaginary parts of $\phi(t)$; we get:
$$\phi(x + iy) = i\cos(x + iy) = \sin x\sinh y + i\cos x\cosh y$$
so $\operatorname{Re} \phi(t) = \sin x\sinh y$ and $\operatorname{Im} \phi(t) = \cos x\cosh y.$
Thus, the paths that make $\operatorname{Im} \phi(t)$ constant constitute the family of curves
$$\tag{2} \cos x\cosh y = k, \, k \in \mathbb{R}.$$
There are two curves which belong to family (2) that are useful in the context of this problem, as they will allow us to follow a path from $t=0$ to $t=\pi/2$ in the steepest descent sense, namely
$$\gamma_1(x,y) = \cos x\cosh y – 1$$
which will takes us from $(x,y)=(0,0)$ to $(x,y)=\left( \frac{\pi}{2}-\epsilon,-R \right)$, with $R$ very large and $\epsilon$ a small positive quantity (this path will be named $C_1$), and
$$\gamma_3(x,y) = \cos x\cosh y $$
which will takes us from $(x,y)=\left( \frac{\pi}{2},-R \right)$ to $(x,y)=\left( \frac{\pi}{2},0 \right)$ (this path will be named $C_3$).
To join paths $C_1$ and $C_3$, we need to resort to another path $C_2$, parallel to the x-axis, which will takes us from $(x,y)=\left( \frac{\pi}{2}-\epsilon,-R \right)$ to $(x,y)= \left( \frac{\pi}{2},-R \right)$.
These are the contour lines of $\operatorname{Re} \phi(t) = \sin x\sinh y$ (darker means smaller)
So, to apply the method of steepest descent we should take the following contour
We have then
$$\int\limits_0^{\pi / 2} {e^{ix\cos t}}dt = \int\limits_{{C_1}} {{e^{ix\cos t}}dt} + \int\limits_{{C_2}} {{e^{ix\cos t}}dt} + \int\limits_{{C_3}} {{e^{ix\cos t}}dt}.$$
Now, making $R \rightarrow \infty$, the lenght of $C_2$ becomes arbitrarily small as does the integrand, which decays exponentially in the direction of steepest descent, so the contribution of this integral is negligible.
To compute the contour integral along path $C_1$, we use a substitution of the form $\phi(t) = -u + ik$, so that the exponential in the integrand becomes $\exp (-xu + ix)$, as $k=1$ on $C_1$. We then have
$$t = \arccos (1 + iu) \to \frac{dt}{du} = – \frac{i}{\sqrt{1- (1 + iu)^2}} \Rightarrow dt =
\frac{1 – i}{2 \sqrt{u} \sqrt{1 + iu/2}}du$$
and, from $u(t) = i(1- \cos t) = i – \sin x\sinh y – i\underbrace {\cos x\cosh y}_{ = 1} = – \sin x\sinh y$, we get
$$u(0 + i0) = – \sin 0 \sinh 0 = 0$$
and
$$u\left(\frac{\pi}{2}-i\infty \right) = – \sin \left(\frac{\pi}{2} \right) \sinh(-\infty) = + \infty.$$
We finally arrive at
$$\tag{3} \int\limits_{C_1} {{e^{ix\cos t}}dt} = e^{ix} \int\limits_0^{+\infty} {\frac{1 – i}{2\sqrt{u} \sqrt{1 + iu/2}} e^{-xu} du}.$$
Regarding the contour integral along path $C_3$, we use a substitution of the form $\phi(t) = -u + ik$, so that the exponential in the integrand becomes $\exp (-xu)$, as $k=0$ on $C_3$. We then have
$$t = \arccos (iu) \to \frac{dt}{du} = – \frac{i}{\sqrt{1 + u^2}} \Rightarrow dt =
– \frac{i}{\sqrt{1 + u^2}}du$$
and, from $u(t) = – i \cos t) = – \sin x\sinh y – i\underbrace {\cos x\cosh y}_{ = 0} = – \sin x\sinh y$, we get
$$u\left(\frac{\pi}{2}-i\infty \right) = – \sin \left(\frac{\pi}{2}\right) \sinh (-\infty) = +\infty$$
and
$$u\left(\frac{\pi}{2} + i0 \right) = – \sin \left(\frac{\pi}{2} \right) \sinh(0) = 0.$$
We finally arrive at
$$\tag{4} \int\limits_{C_3} {{e^{ix\cos t}}dt} = \int\limits_0^{+\infty} {\frac{i}{\sqrt{1 + u^2}} e^{-xu} du}.$$
We are now able to find the asymptotic series for both (3) and (4), using Watson’s lemma.
The first result we need is the following MacLaurin series
$$\tag{5} \frac{1}{\sqrt{1 – 4x}} \sim \sum\limits_{k=0}^\infty {\frac{(2k)!x^k}{(n!)^2}}. $$
Now, making the substitution $x=-iu/8$ in (5) and multiplying by $(1-i) / u^{1/2} = \sqrt{2} e^{-i \pi/4} / u^{1/2}$, we get
$$\tag{6} \frac{1 – i}{2 \sqrt{u} \sqrt{1 + iu/2}} \sim \frac{e^{-i \pi/4}}{\sqrt{2}} \sum\limits_{k=0}^\infty {\frac{(2k)! (-i)^k u^{k-1/2}}{8^k (k!)^2}}.$$
Similarly, making the substitution $x=-u^2/4$ in (5) and multiplying by $i$, we get
$$\tag{7} \frac{i}{\sqrt{1 + u^2}} \sim \sum\limits_{k=0}^\infty {\frac{i(2k)! (-1)^k u^{2k}}{4^k (k!)^2}}.$$
Finally, applying Watson’s lemma we arrive at
$$\begin{align} \int\limits_0^{\pi / 2} {e^{ix\cos t}}dt &\sim \int\limits_{{C_1}} {{e^{ix\cos t}}dt} + \int\limits_{{C_3}} {{e^{ix\cos t}}dt} \\
&\sim {\frac{e^{i(x – \pi/4)}}{\sqrt{2}} \sum\limits_{k=0}^\infty \frac{(2k)! (-i)^k \Gamma(k+1/2)}{8^k (k!)^2 x^{k+1/2}}} + \sum\limits_{k=0}^\infty {\frac{i (-1)^k (2k)! \Gamma(2k+1)}{4^k (k!)^2 x^{2k+1}}}.\end{align}$$
So, for example, the first two terms in the asymptotic series would be
$$\int\limits_0^{\pi / 2} {e^{ix\cos t}}dt \sim \sqrt{\frac{\pi}{2x}} e^{i({x-\pi/4})} + \frac{i}{x}.$$
Is my reasoning correct?
Is there any way to check the final result?
Thanks in advance!
Best Answer
This can be done with Maple 18.01 by
$$ (1/2)\pi \mathop{\rm BesselJ} (0, x)+(1/2i)\pi \mathop{\rm StruveH} (0, x)$$
and then:
$$\left( 1/2\,\sqrt {2}\sqrt {\pi }\sin \left( x+\pi /4 \right) -i/2 \sqrt {\pi }\sqrt {2}\cos \left( x+\pi /4 \right) \right) \sqrt {{x}^ {-1}}+{\frac {i}{x}}+ \left( -1/16\,\sqrt {\pi }\sqrt {2}\cos \left( x +\pi /4 \right) -i/16\sqrt {\pi }\sqrt {2}\sin \left( x+\pi /4 \right) \right) \left( {x}^{-1} \right) ^{3/2}+O \left( \left( {x} ^{-1} \right) ^{5/2} \right)$$
as $x \to \infty$.