<s>[INST] Calculate the integral $\int_0^{\infty} x^{11} e^{-x^3}dx$ Calculate the integral $\int_0^{\infty} x^{11} e^{-x^3}dx$.
I wanted to approach this using the reduction formula. Here's what I have tried
$$\int x^2e^{-x^3}dx = -\frac{1}{3}e^{-x^3}$$
Then extending this to $n$
$$I_n = \int_0^{\infty}x^ne^{-x^3}dx $$
Using substitution I can get
$$u = x^{n-2}; du = (n-2)x^{n-3}; dv = x^2e^{-x^3}; v = -\frac{1}{3}e^{-x^3}$$
Plugging this in
$$I_n = \int_0^{\infty}x^ne^{-x^3}dx  = \left[-\frac{x^{n-2}}{3}e^{-x^3} \right]_0^{\infty}+\frac{(n-3)}{3} \int_0^{\infty}x^{n-3}e^{-x^3}dx $$
$$\implies I_n = \int_0^{\infty}x^ne^{-x^3}dx =\frac{(n-3)}{3} \int_0^{\infty}x^{n-3}e^{-x^3}dx$$
$$=\frac{n-3}{3}I_{n-3}$$
Then starting from $n=11$ I get
$$\frac{8}{3} \cdot \frac{5}{3} \cdot\frac{2}{3} \int_0^{\infty} x^2e^{-x^3}dx$$
However this does not produce the answer which is $2$. How should I go about the reduction formula for this?
 [/INST] Your recurrence formula has a little error on the first line in the 2nd term on the right should be $n-2$ not $n-3$ and the result should read:
$$
I_n=\frac{n-2}3 I_{n-3}
$$
Now we get:
$$
I_{11}=\frac93\cdot\frac63\cdot\frac33 I_2=2
$$
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