1. ## Laplace's Method (Integration)

Consider the integral

I(x)=\int^{2}_{0} (1+t) \exp\left(x\cos\left(\frac{\pi(t-1)}{2}\right)\right) dt

Use Laplace's Method to show that

I(x) \sim \frac{4\sqrt{2}e^{x}}{\sqrt{\pi x}}
as $x\rightarrow\infty$.
=> I have tried using the expansion of $I(x)$ in McLaurin series but did not get the answer.
here,

h(t)=cos(\frac{\pi(t-1)}{2})

$h(0)= 0$
$h'(0)= \frac {\pi}{2}$
Also $f(t)= (1+t) \approx f(0) =1$, so that

I(x)\sim \int^{\delta}_{0} e^{x \frac{\pi}{2}t} dt

after that I tried doing integration by substitution $\tau = x \frac{\pi}{2} t$ but did not get the answer.

2. ## Re: Laplace's Method (Integration)

Originally Posted by grandy
Consider the integral

I(x)=\int^{2}_{0} (1+t) \exp\left(x\cos\left(\frac{\pi(t-1)}{2}\right)\right) dt

Use Laplace's Method to show that

I(x) \sim \frac{4\sqrt{2}e^{x}}{\sqrt{\pi x}}
as $x\rightarrow\infty$.
=> I have tried using the expansion of $I(x)$ in McLaurin series but did not get the answer.
here,

h(t)=cos(\frac{\pi(t-1)}{2})

$h(0)= 0$
$h'(0)= \frac {\pi}{2}$
Also $f(t)= (1+t) \approx f(0) =1$, so that

I(x)\sim \int^{\delta}_{0} e^{x \frac{\pi}{2}t} dt

after that I tried doing integration by substitution $\tau = x \frac{\pi}{2} t$ but did not get the answer.
Laplace's Method says

$\Large I=\displaystyle{\int_a^b}e^{-x g(t)}h(t) ~dt \approx e^{-x g(\hat{t})}\sqrt{\dfrac{2 \pi}{x g''(\hat{t})}}$

here

$g(t)=\cos\left(\dfrac{\pi (t-1)}{2}\right)$

$h(t)=1+t$

$\hat{t}=1$

see if you can finish from here.