The problem asks for the general formula of the integral:

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

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

This is a standard Gaussian integral problem, where the integrand involves a product of a polynomial $x^n$ and the Gaussian function $e^{-\alpha x^2}$. The behavior of the integral depends on whether $n$ is even or odd.

Solution:

1. Odd Powers $n$: For any odd $n$, the integral evaluates to zero. This is due to the fact that $x^n$ is an odd function and $e^{-\alpha x^2}$ is an even function, so their product is odd. Integrating an odd function over a symmetric interval (from $-\infty$ to $\infty$) gives zero: $I_n = 0 \quad \text{for odd } n.$

2. Even Powers $n$: For even powers of $n$, the integral can be solved using known results for Gaussian integrals. Specifically, when $n = 2k$ (where $k$ is a non-negative integer), the integral has the following general form:

   $I_{2k} = \int_{-\infty}^{\infty} x^{2k} e^{-\alpha x^2} \, dx = \frac{(2k-1)!!}{(2\alpha)^{k}} \sqrt{\frac{\pi}{\alpha}},$ where $(2k-1)!!$ denotes the double factorial, defined as: $(2k-1)!! = (2k-1)(2k-3)(2k-5)\cdots 3 \cdot 1.$

   For example:

   • For $n = 0$ (i.e., $k = 0$): $I_0 = \int_{-\infty}^{\infty} e^{-\alpha x^2} \, dx = \sqrt{\frac{\pi}{\alpha}}.$
   • For $n = 2$ (i.e., $k = 1$): $I_2 = \int_{-\infty}^{\infty} x^2 e^{-\alpha x^2} \, dx = \frac{1}{2\alpha} \sqrt{\frac{\pi}{\alpha}}.$
   • For $n = 4$ (i.e., $k = 2$): $I_4 = \frac{3}{(2\alpha)^2} \sqrt{\frac{\pi}{\alpha}} = \frac{3}{4\alpha^2} \sqrt{\frac{\pi}{\alpha}}.$

Thus, the general formula for the integral is: