I am unsure how to proceed with the following question due to the fact I am unsure how to obtain a Taylor series for the following function.
Question:
Using Watson’s Lemma, obtain the first three terms in the asymptotic expansions as x → ∞ of $$\int_0^{\pi/4}e^{-xt}(1+\cos t)^{\frac{1}{2}}dt$$
Currently I am taking $\phi(t) = (1+\cos t)^\frac{1}{2}$ with $\lambda = 0$ so $g(t) = (1+\cos t)^\frac{1}{2}$
I currently take the expansion to be $\sqrt{(1+\sum_{n=0}^\infty \frac{(-1)^n t^{2n}}{(2n)!}}$
Here I am unsure how to translate this expansion into the approximation $f(x) = \sum_{n=0}^N a_n\frac{\Gamma(\lambda+n+1)}{x^{\lambda+n+1}}$
EDIT:: Going from what user254665 said. Does this seem correct?
letting $(1+cost)^{1/2} = \sqrt{2}cos\frac{t}{2} = \sqrt(2) \sum_{k=0}^\infty\frac{(\frac{-1}{4})^kt^{2k}}{(2k)!}$
$f(x) \sim \sum_{n=0}^\infty\frac{\Gamma(2n+1)}{x^{2n+1}}\frac{(\frac{-1}{4})^n}{(2n)!}$
$\Gamma(2n+1)$ cancels with $(2n)!$
$\sum_{n=0}^\infty\frac{(\frac{-1}{4})^n}{x^{2n+1}}$
$n=0, \frac{1}{x}$
$n=1, \frac{-1}{4x^3}$
$n=2, \frac{1}{16x^5}$