The given integral is:

$I = \int x^{2r} e^{-\frac{x^2}{2\sigma^2}} \, dx$

where $r$ and $\sigma$ are constants. This resembles a Gaussian integral with an additional factor $x^{2r}$.

To solve this, we use the general result for integrals of the form:

$\int_{-\infty}^{\infty} x^{2r} e^{-\frac{x^2}{2\sigma^2}} \, dx = (2\sigma^2)^r \cdot \sqrt{2\pi \sigma^2} \cdot \frac{(2r-1)!!}{2^r}$

where $(2r-1)!!$ is the double factorial, which is defined as:

$(2r-1)!! = (2r-1)(2r-3)(2r-5)\dots(1)$

Full Steps:

1. Substitute: Start by substituting $u = \frac{x^2}{2\sigma^2}$, then simplify the exponential term.

2. General Gaussian Integral Formula: Apply the known formula for integrals of Gaussian functions with powers of $x$.

3. Solution: $I = (2\sigma^2)^r \cdot \sqrt{2\pi \sigma^2} \cdot \frac{(2r-1)!!}{2^r}$ This provides the result in terms of $r$ and $\sigma$.

Would you like further details or clarifications on any part?

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Here are 5 related questions:

1. How is the double factorial defined for odd and even numbers?
2. Can the same technique be applied if the exponent on $x$ is an odd number instead of $2r$?
3. How does the presence of the Gaussian exponential influence the integrability of the function?
4. What happens to the integral when $r = 0$?
5. How does changing the constant $\sigma$ affect the result of the integral?

Tip: Gaussian integrals are highly symmetric and appear frequently in probability theory, especially in the context of normal distributions.