The problem asks to determine the general formula for the integral:

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

where $n = 0, 1, 2, 3, 4, \dots$.

This is a Gaussian-type integral and can be evaluated using techniques related to the properties of Gaussian integrals and moments of the normal distribution.

Key Considerations:

• When $n$ is odd, the integral will evaluate to zero due to the symmetry of the integrand. This is because $x^n e^{-\alpha x^2}$ is an odd function, and the integral over a symmetric interval from $-\infty$ to $\infty$ of an odd function is zero.
• When $n$ is even, the integral will be non-zero and related to the Gamma function or factorials, based on powers of the Gaussian distribution.

General Approach:

1. Case when $n = 0$:

The integral becomes: $I_0 = \int_{-\infty}^{\infty} e^{-\alpha x^2} \, dx$ This is a standard Gaussian integral, and its value is known: $I_0 = \sqrt{\frac{\pi}{\alpha}}$

2. Case when $n = 2m$ (even):

For even powers $n = 2m$, the integral is: $I_{2m} = \int_{-\infty}^{\infty} x^{2m} e^{-\alpha x^2} \, dx$ This can be related to the Gamma function or factorials. Specifically, for even $n$, the result can be expressed as: $I_{2m} = \frac{(2m-1)!!}{(2\alpha)^{m}} \sqrt{\frac{\pi}{\alpha}}$ where $(2m-1)!!$ is the double factorial of $(2m-1)$, which is the product of all odd numbers up to $2m-1$.

General Formula:

The general formula for the integral is: