<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}
\begin{description}
\item[(i)]
Show that if

$$I_n=\ds\int_0^{\infty}x^ne^{-x} \,dx,\ \ n \geq
-\ds\frac{1}{2},$$

then

$$I_n=nI_{n-1}$$

{\bf{Hint:}} you may assume that $\lim_{x \to +\infty}
\left(x^pe^{-x}\right)=0$ for any value of p.

\item[(ii)]
Evaluate $I_0$ and hence calculate $I_8$.

\item[(iii)]
Given that when $n=-\ds\frac{1}{2},\ I_{-\frac{1}{2}}=\sqrt{\pi}$,
calculate $I_\frac{5}{2}$.

\end{description}

\medskip

{\bf Answer}
\begin{description}
\item[(i)]
$I_n=\ds\int_0^{\infty} x^ne^{-x} \,dx\ \ \ n \geq 0$

$\begin{array} {ll} u=x^n &amp; \ds\frac{dv}{dx}=e^{-x}\\ du=nx^{n-1}
&amp; v=-e^{-x} \end{array}$

Integrate by parts:

\begin{eqnarray*} I_n &amp; = &amp;
\left[-x^ne^{-x}\right]_0^{\infty}+n\ds\int_0^{\infty}x^{n-1}e^{-x}
\,dx\\ &amp; = &amp; \lim_{x \to
\infty}\left[-x^ne^{-x}\right]+0+n\ds\int_0^{\infty}x^{n-1}e^{-x}
\,dx\\ &amp; = &amp; n\ds\int_0^{\infty}x^{n-1}e^{-x} \,dx \end{eqnarray*}

by hint

$\Rightarrow \un{I_n=nI_{n-i}}$

\item[(ii)]
$I_0=\ds\int_0^{\infty}x^0e^{-x}\,dx
=\ds\int_0^{\infty}e^{-x}\,dx=\left[-e^{-x}\right]_0^{\infty}=+1$

Therefore \begin{eqnarray*} I_8=8I_7=8\times 7 I_6= \cdots &amp; = &amp;
8\times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
\times 1\\ &amp; = &amp; 8!=40320\end{eqnarray*}

\item[(iii)]
If $I_{-\frac{1}{2}}=\sqrt{\pi}$,

$I_{\frac{5}{2}}=\ds\frac{5}{2}I_{\frac{3}{2}}=
\ds\frac{5}{2}\times\ds\frac{3}{2}I_{\frac{1}{2}}=\ds\frac{5}{2}\times\ds\frac{3}{2}
\times \ds\frac{1}{2}I_{-\frac{1}{2}}=\ds\frac{15}{8}\sqrt{\pi}$

\end{description}

\end{document}
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