- #1
wel
Gold Member
- 36
- 0
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 [itex]x\rightarrow\infty[/itex].
=> I have tried using the expansion of [itex]I(x)[/itex] in McLaurin series but did not get the answer.
here,
\begin{equation}
h(t)=cos(\frac{\pi(t-1)}{2})
\end{equation}
[itex]h(0)= 0[/itex]
[itex]h'(0)= \frac {\pi}{2}[/itex]
Also [itex]f(t)= (1+t) \approx f(0) =1[/itex], 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 [itex]\tau = x \frac{\pi}{2} t[/itex] but did not get the answer.
please help me.
\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 [itex]x\rightarrow\infty[/itex].
=> I have tried using the expansion of [itex]I(x)[/itex] in McLaurin series but did not get the answer.
here,
\begin{equation}
h(t)=cos(\frac{\pi(t-1)}{2})
\end{equation}
[itex]h(0)= 0[/itex]
[itex]h'(0)= \frac {\pi}{2}[/itex]
Also [itex]f(t)= (1+t) \approx f(0) =1[/itex], 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 [itex]\tau = x \frac{\pi}{2} t[/itex] but did not get the answer.
please help me.