We need to evaluate the integral

$I_n = \int_{-\infty}^{\infty} c^n e^{-\alpha c^2} \, dc$

where $n$ is a non-negative integer, and $\alpha > 0$.

Case 1: $n = 0$

The integral becomes:

$I_0 = \int_{-\infty}^{\infty} e^{-\alpha c^2} \, dc$

This is a standard Gaussian integral, which has the known result:

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

Case 2: $n = 1$

The integral is:

$I_1 = \int_{-\infty}^{\infty} c e^{-\alpha c^2} \, dc$

Since the integrand is an odd function (i.e., $c e^{-\alpha c^2}$ is odd with respect to $c$), the integral over the symmetric limits $-\infty$ to $\infty$ will be zero:

$I_1 = 0$

Case 3: $n = 2$

The integral is:

$I_2 = \int_{-\infty}^{\infty} c^2 e^{-\alpha c^2} \, dc$

We can solve this integral using a known result for Gaussian integrals. Specifically, the result is:

$I_2 = \frac{1}{2\alpha} \int_{-\infty}^{\infty} e^{-\alpha c^2} \, dc = \frac{1}{2\alpha} \sqrt{\frac{\pi}{\alpha}}$

Thus:

$I_2 = \frac{\sqrt{\pi}}{2\alpha^{3/2}}$

Case 4: $n = 3$

The integral is:

$I_3 = \int_{-\infty}^{\infty} c^3 e^{-\alpha c^2} \, dc$

Again, the integrand is odd, so the integral is zero:

$I_3 = 0$

Case 5: $n = 4$

The integral is:

$I_4 = \int_{-\infty}^{\infty} c^4 e^{-\alpha c^2} \, dc$

We can use the recurrence relation for Gaussian integrals of even powers. Specifically:

$I_4 = \frac{3}{4\alpha^2} \sqrt{\frac{\pi}{\alpha}}$

General Result

For even $n = 2k$, we have:

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

where $(2k-1)!!$ denotes the double factorial.

For odd $n$, the integral is zero:

$I_n = 0 \quad \text{for odd } n$

Summary

1. $I_0 = \sqrt{\frac{\pi}{\alpha}}$
2. $I_1 = 0$
3. $I_2 = \frac{\sqrt{\pi}}{2\alpha^{3/2}}$
4. $I_3 = 0$
5. $I_4 = \frac{3\sqrt{\pi}}{4\alpha^{5/2}}$

Would you like more details or have any questions?

Here are 5 related questions:

1. How is the Gaussian integral derived?
2. What is the significance of the double factorial in the general result?
3. How can this result be applied in probability theory?
4. How does this integral relate to the moment-generating function of a normal distribution?
5. Can you extend this to complex-valued functions?

Tip: For odd powers of $c$, integrals involving Gaussian functions typically result in zero due to symmetry considerations.