The given problem is asking to determine the general formula for the integral:

$\int_{-\infty}^{\infty} x^n e^{-\alpha x^2} \, dx$

where $n = 0, 1, 2, 3, 4, \dots$, and $\alpha$ is a constant.

Step-by-Step Solution:

1. Case for $n = 0$:

First, consider the case where $n = 0$:

$I_0 = \int_{-\infty}^{\infty} e^{-\alpha x^2} \, dx$

This is a standard Gaussian integral, and the result is:

$I_0 = \sqrt{\frac{\pi}{\alpha}}$

2. Case for $n = 1$:

Next, consider the case when $n = 1$:

$I_1 = \int_{-\infty}^{\infty} x e^{-\alpha x^2} \, dx$

Since $x e^{-\alpha x^2}$ is an odd function, and the limits of integration are symmetric about zero, this integral evaluates to zero:

$I_1 = 0$

3. Case for general even $n = 2m$:

For even powers of $x$, the integral becomes:

$I_{2m} = \int_{-\infty}^{\infty} x^{2m} e^{-\alpha x^2} \, dx$

This can be solved using the standard formula for Gaussian integrals involving even powers of $x$:

$I_{2m} = \frac{(2m-1)!!}{(2\alpha)^m} \sqrt{\frac{\pi}{\alpha}}$

where $(2m-1)!!$ is the double factorial of $2m-1$.

4. Case for general odd $n = 2m + 1$:

For odd powers of $x$, the integral evaluates to zero due to symmetry:

$I_{2m+1} = 0$

Final General Formula:

• For even $n = 2m$: $I_{2m} = \frac{(2m-1)!!}{(2\alpha)^m} \sqrt{\frac{\pi}{\alpha}}$

• For odd $n = 2m + 1$: $I_{2m+1} = 0$

This provides the general result for the integral of $x^n e^{-\alpha x^2}$ over $(-\infty, \infty)$.

Would you like further clarification or details on any step? Here are some related questions for further exploration:

1. How do you calculate the double factorial $(2m-1)!!$?
2. Can you derive the Gaussian integral from scratch?
3. How would the integral change if the limits were finite instead of infinite?
4. What is the significance of odd vs. even powers of $x$ in integrals?
5. How does the value of $\alpha$ affect the result of these integrals?

Tip: Symmetry in integrals often simplifies the calculation, especially when dealing with odd or even functions over symmetric limits.