Let be a positive integer. We define .
Using integration by parts determine the constant such that .
What is the value of p?
$\displaystyle I_{n+3}=\int_0^{\infty}e^{-3u^3}u^{n+3}du=-\frac{1}{9}\int_0^{\infty}\left(e^{-3u^3}\right)'u^{n+1}du=$
$\displaystyle =\left.-\frac{1}{9}\cdot\frac{u^{n+1}}{e^{3u^3}}\right|_0^ {\infty}+\frac{n+1}{9}I_n=\frac{n+1}{9}I_n$
Then $\displaystyle (n+1)I_n=9I_{n+3}\Rightarrow p=9$