How to find the leading behavior of this integral Consider the integral:
$$\int_{0}^{\pi}\frac{dx}{\pi}\frac{\cos(xt)}{\sqrt{1+a\sin^{2}(x/2)}},\,\,\,\,\,\,\mathrm{as\,\,}t\rightarrow\infty$$
with $a>0.$
I know that the answer is $\exp(-t/\xi)$, where $\xi=1/\ln(2)=1.4427$ for $a=8$
I'm interested in the leading behavior of this integral as $t\rightarrow\infty$ . I tried to applied the Laplace's Method but did not get the answer.
Please help me.
 A: For $t \to \infty$ over the reals, the integral is $\sin(\pi t)/(\pi t \sqrt {a + 1}) + o(t^{-2})$.
But I assume that you're looking for the asymptotic of the Fourier series coefficients of $f(x) = (1 + a \sin^2(x/2))^{-1/2}$, that is, $t \to \infty$ over the integers.
In that case, if we take a contour going in the direction $i$ from the points $\pm\pi$ and avoiding the branch cut of $f$ lying on the imaginary axis, the main contribution to the integral of $f(x) e^{i t x}$ comes not from the endpoints but from the part of the contour going along the branch cut, because the integrals over $[-\pi, -\pi + i \infty)$ and $(\pi + i \infty, \pi]$ cancel out.
The branch point at $x_0 = 2 i \operatorname{arcsinh}(1 / \sqrt a)$ is of the $x^{-1/2}$ type, and the integral of $f(x) e^{i t x}$ from $-\pi$ to $\pi$ is twice the integral from $x_0$ to $i \infty$, for which Laplace's method gives
$$f(x) = (a+1)^{-1/4} (i(x-x_0))^{-1/2} + O(|x-x_0|^{1/2})
 \quad \text{when } x \to x_0,\\
\frac 1 {2\pi} \int_{-\pi}^\pi f(x) \cos t x \,dx =
\frac 1 \pi \int_{x_0}^{i \infty} f(x) e^{i t x} \,dx \sim \\
\frac 1 \pi (a+1)^{-1/4}
 \int_{x_0}^{i \infty} (i(x - x_0))^{-1/2} \,e^{i t x} dx = \\
\frac i \pi (a+1)^{-1/4} e^{i x_0 t}
 \int_0^\infty (-x)^{-1/2} \,e^{-t x} dx = \\
(a+1)^{-1/4} (\pi t)^{-1/2}
 \exp \left( -2 t \operatorname{arcsinh} \frac 1 {\sqrt a} \right).$$
A: Let $f(t)$ be represented by the integral 
$$f(t)=\frac1\pi \int_0^\pi \frac{\cos(xt)}{\sqrt{1+\sin^2(x/2)}}\,dx \tag 1$$
Integrating by parts the integral in $(1)$ with $u=\frac{1}{\sqrt{1+\sin^2(x/2)}}$ and $v=\frac{\sin(xt)}{t}$ reveals
$$\begin{align}
f(t)&= \frac{\sin(\pi t)}{\pi \sqrt{2}\,t}+\frac1{4\pi t} \int_0^\pi \frac{\sin(xt)\sin(x)}{\left(1+\sin^2(x/2)\right)^{3/2}}\,dx \tag 2\\\\
&=\frac{\sin(\pi t)}{\pi \sqrt{2}\,t}+o\left(\frac1{t^2}\right) \tag3
\end{align}$$
Note that integrating by parts the integral in $(2)$ with $u=\frac{\sin(x)}{\left(1+\sin^2(x/2)\right)^{3/2}}$ and $v=-\frac{\cos(xt)}{t}$ would produce a term that is of order $o\left(\frac{1}{t^2}\right)$, thereby producing $(3)$.  
