Laplace's Method (Integration)

Consider the integral

\begin{equation}

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

\end{equation}

Use Laplace's Method to show that

\begin{equation}

I(x) \sim \frac{4\sqrt{2}e^{x}}{\sqrt{\pi x}} \end{equation}

as $x\rightarrow\infty$.

=> I have tried using the expansion of $I(x)$ in McLaurin series but did not get the answer.

here,

\begin{equation}

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

\end{equation}

$h(0)= 0$

$h'(0)= \frac {\pi}{2}$

Also $f(t)= (1+t) \approx f(0) =1$, so that

\begin{equation}

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

\end{equation}

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

please help me.

Re: Laplace's Method (Integration)

Quote:

Originally Posted by

**grandy** Consider the integral

\begin{equation}

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

\end{equation}

Use Laplace's Method to show that

\begin{equation}

I(x) \sim \frac{4\sqrt{2}e^{x}}{\sqrt{\pi x}} \end{equation}

as $x\rightarrow\infty$.

=> I have tried using the expansion of $I(x)$ in McLaurin series but did not get the answer.

here,

\begin{equation}

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

\end{equation}

$h(0)= 0$

$h'(0)= \frac {\pi}{2}$

Also $f(t)= (1+t) \approx f(0) =1$, so that

\begin{equation}

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

\end{equation}

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

please help me.

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.