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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.

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2 Answers 2

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Hint:

Where does $h(t)$ achieve its maximum?

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  • $\begingroup$ when $t=5$. so $cos 2 \pi =1$ $\endgroup$
    – Manjushree
    Commented Apr 19, 2014 at 8:52
  • $\begingroup$ @user134576, $t = 5$ isn't inside the domain of integration... $\endgroup$ Commented Apr 19, 2014 at 14:16
  • $\begingroup$ when $t=1$. it has maximum $=1$. $\endgroup$
    – Manjushree
    Commented Apr 19, 2014 at 17:08
  • $\begingroup$ Ok, now expand $h(t)$ and the factor $1+t$ in series about the point $t=1$ as prescribed by the Laplace method. $\endgroup$ Commented Apr 19, 2014 at 17:10
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$\int_0^2(1+t)e^{x\cos\frac{\pi(t-1)}{2}}~dt$

$=\int_{-1}^1(t+2)e^{x\cos\frac{\pi t}{2}}~dt$

$=\int_{-1}^0(t+2)e^{x\cos\frac{\pi t}{2}}~dt+\int_0^1(t+2)e^{x\cos\frac{\pi t}{2}}~dt$

$=\int_1^0(-t+2)e^{x\cos\frac{\pi(-t)}{2}}~d(-t)+\int_0^1(t+2)e^{x\cos\frac{\pi t}{2}}~dt$

$=\int_0^1(-t+2)e^{x\cos\frac{\pi t}{2}}~dt+\int_0^1(t+2)e^{x\cos\frac{\pi t}{2}}~dt$

$=4\int_0^1e^{x\cos\frac{\pi t}{2}}~dt$

$=4\int_0^\frac{\pi}{2}e^{x\cos t}~d\left(\dfrac{2t}{\pi}\right)$

$=\dfrac{8}{\pi}\int_0^\frac{\pi}{2}e^{x\cos t}~dt$

$\approx\dfrac{8}{\pi}\sqrt{\dfrac{2\pi}{x}}e^x$ as $x\rightarrow\infty$

$=\dfrac{8\sqrt{2}e^x}{\sqrt{\pi x}}$ as $x\rightarrow\infty$

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