\documentclass[a4paper,12pt]{article}
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\newcommand{\pl}{\partial}
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\newcommand{\df}{\ds\frac}
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\begin{document}


{\bf Question}

The modified Bessel function of the first kind may be defined by

$$I_n(x)=\ds\frac{1}{\pi}\ds\int_0^\pi e^{x\cos t}\cos(nt)\, dt.$$

Show that

$$I_n(x) \sim \ds\frac{e^x}{\sqrt{2\pi x}},\ x\to+\infty.$$


\vspace{.5in}

{\bf Answer}

$I_n(x)=\df{1}{\pi}\ds\int_0^\pi e^{x\cos t}\cos(nt)\,dt$

On the interval $[0,\pi]$, the max. value of cos $t$ (min. value
of $-\cos t$) is at the endpoint $t=0$.

Therefore set

$u=h(t)-h(0)=-\cos t+1\ \,du=\sin t\,dt$

Thus

$I_n(x)=\df{1}{\pi}\ds\int_0^2e^{-xu+x}\df{\cos[(nt(u))]}{\sin[t(u)]}\,du$

$u \approx h''(0)\df{(t-0)^2}{2}+\cdots \Rightarrow u \approx
\df{t^2}{2^2}$

$\begin{array}{rcl} I_n(x) & \approx & \df{e^x}{\pi}\ds\int_0^2
e^{-xu}\df{\cos(n\sqrt{2u})}{\sin(\sqrt{2u})}\,du\\ & \sim &
\df{e^x}{\pi}\ds\int_0^\infty
e^{-xu}\df{\cos(n\sqrt{2u})}{\sin(\sqrt{2u})}\,du\ \ x\to +
\infty\\ & & \df{\cos(n\sqrt{2u})}{\sin(\sqrt{2u})}\\ & &  \sim
\df{1-\frac{n^2
2u}{2}+\cdots}{\sqrt{2u}-\frac{2\sqrt{2}u^\frac{3}{2}}{6}+\cdots}\,
\rm{as}\ u \to 0\ \rm{and\ hence} \sim \df{1}{\sqrt{2u}}\\ & \sim
& \df{e^x}{\pi}\ds\int_0^\infty
\df{e^{-xu}}{\sqrt{2}\sqrt{u}}\,du\\ & \sim &
\df{e^x}{\pi}\df{\Gamma(+\frac{1}{2})}{\sqrt{2}\sqrt{x}}\\ & \sim
& \df{e^x}{\sqrt{2\pi x}}\ \ x\to + \infty \end{array}$


\end{document}