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\begin{document}


{\bf Question}

Show that if $a>0,\ n>0,\ f(0)\neq 0$

$$\ds\int_0^ae^{-xt^n}f(t)\,dt \sim
\ds\frac{\Gamma(\frac{1}{n})}{nx^{\frac{1}{n}}}f(0),\
x\to+\infty$$

\vspace{.5in}

{\bf Answer}

\hspace{1in} No singularity at $t=0$.

$I(x)=\ds\int_0^a e^{ixt^n}f(t)\,dt.\ \ \undb{a>0},\
\overbrace{n>0},\ \undb{f(0)\ne 0}$

\hspace{1.5in}Integral along \ \ No vanishing

\hspace{1.6in}real line.\ \ \ \ \ \ \ \ \ \ at endpoint.

This has an endpoint minimum at $t=0$:

$h(t)=t^n \Rightarrow h'(0)=\left.nt^{n-1}\right|_{t=0}=0$

Indeed $h'(0)=h''(0)=h'''(0)=\cdots=h^{(n-1)}(0)=0$

$$h^{(n)}(0)=n!$$

Thus set $\begin{array}{rcl} u & = & h(t)-h(0)=t^n\\ du & = &
nt^{n-1}\,dt\ \rm{(exactly)}\\ & = & nu^{\frac{n-1}{n}}\,dt\
\rm{(exactly)}\end{array}$

\newpage
Therefore
$I(x)=\ds\int_0^{a_n}e^{-xu}\df{f(u^{\frac{1}{n}})}{nu^{\frac{n-1}{n}}}\,du$

Now for fixed finite $n$ the upper limit can be put to infinity
with only exponentially small error. (NB. ???? if $a>1$)

\begin{eqnarray*}
I(x) & \approx & \ds\int_0^\infty
e^{-xu}\df{f(u^{\frac{1}{n}})}{n(u^{\frac{n-1}{n}})}\,du\ \
(\rm{limit} \to +\infty)\ \ \ x\to + \infty\\ & \approx &
\ds\int_0^\infty \df{e^{-xu}}{u^{1-\frac{1}{n}}}\,du \df{f(0)}{n}
(\rm{expand\ about}\ u=0\ \ x \to +\infty\\ & \sim &
\df{\Gamma(\frac{1}{n}-1+1)}{nx^{\frac{1}{n}-1+1}}f(0)\\ & \sim &
\df{\Gamma(\frac{1}{n})f(0)}{nx^{\frac{1}{n}}}\ \ as x\to +\infty
\end{eqnarray*}

e.g., $n=2 \sim \sqrt{\pi} \df{f(0)}{2\sqrt{x}}$ in agreement with
lecture notes.

\end{document}