Definite Integration
Gamma-function type integral — recurrence
MJAT_TS3_P2
Grade 12
Question:
Let $a=\displaystyle\int_0^\infty e^{-x^2}\,dx$ and $I_n=\displaystyle\int_0^\infty x^n e^{-x^2}\,dx$, $n\in\mathbb{N}$. Then:
A) $I_n = \dfrac{n-1}{2}\,I_{n-2}$
B) $I_{2m} = \dfrac{1}{2^m}\cdot\dfrac{(2m-2)!!}{(2m)!!}\cdot a$, if $n=2m$
C) $I_n = \dfrac{1}{2}\cdot\dfrac{(n-2)!!}{n!!}\cdot a$, if $n$ is even
D) $I_n = \dfrac{(2m)!}{2^{2m+1}m!}\cdot a$, if $n=2m$ is even
Step-by-Step Solution
Key Concept: IBP: $I_n=\int_0^\infty x^{n-1}\cdot(xe^{-x^2})dx$. With $u=x^{n-1}$, $dv=xe^{-x^2}dx\Rightarrow v=-e^{-x^2}/2$. Boundary term vanishes, giving $I_n=\frac{n-1}{2}I_{n-2}$ (A ✓).
A ✓ (IBP recurrence). B ✓ (recursive application). C, D: different forms, may have errors. Answer: A, B.
Correct Answer: AB