TAOCP 5.1.4 Exercise 26
We evaluate the integral \[ I_t = \int_{-\infty}^{\infty} x^t \exp(-2x^2) \sqrt{n} \, dx, \] where \(t\) is a nonnegative integer and \(n > 0\) is a constant.
Section 5.1.4: Tableaux and Involutions
Exercise 26. [M21] Evaluate $\int_{-\infty}^{\infty} x^t \exp(-2x^2) \sqrt{n} , dx$ when $t$ is a nonnegative integer.
Verified: yes
Solve time: 14m10s
Solution
We evaluate the integral [ I_t = \int_{-\infty}^{\infty} x^t \exp(-2x^2) \sqrt{n} , dx, ] where (t) is a nonnegative integer and (n > 0) is a constant. The factor (\sqrt{n}) is independent of (x), so [ I_t = \sqrt{n} \int_{-\infty}^{\infty} x^t e^{-2x^2} , dx. ]
Case 1: (t) odd.
The integrand (x^t e^{-2x^2}) is an odd function, so the integral over the symmetric interval ((-\infty,\infty)) vanishes:
[
I_t = 0 \qquad (t \text{ odd}).
]
Case 2: (t = 2k) even ((k \ge 0)).
We use the standard Gaussian moment formula. For (a > 0),
[
\int_{-\infty}^{\infty} x^{2k} e^{-ax^2} , dx
= \frac{(2k)!}{4^k k! , a^k} \sqrt{\frac{\pi}{a}}.
]
Setting (a = 2) gives
[
\int_{-\infty}^{\infty} x^{2k} e^{-2x^2} , dx
= \frac{(2k)!}{4^k k! , 2^k} \sqrt{\frac{\pi}{2}}
= \frac{(2k)!}{8^k k!} \sqrt{\frac{\pi}{2}}.
]
Equivalently, using the double factorial ((2k-1)!! = \frac{(2k)!}{2^k k!}),
[
\int_{-\infty}^{\infty} x^{2k} e^{-2x^2} , dx
= \frac{(2k-1)!!}{4^k} \sqrt{\frac{\pi}{2}}.
]
Multiplying by (\sqrt{n}) yields the even‑(t) value.
Combining the two cases, the final answer is
[ \boxed{\int_{-\infty}^{\infty} x^t \exp(-2x^2) \sqrt{n} , dx = \begin{cases} 0, & t \text{ odd},\[6pt] \displaystyle \sqrt{\frac{n\pi}{2}} ; \frac{(t-1)!!}{2^t}, & t = 2k \text{ even}. \end{cases}} ]
(For (t = 0) the double factorial ((-1)!!) is taken as (1).) ∎