Show that $\int_0^{\infty} x^2e^{-x^2} dx = \frac{1}{2} \int_0^{\infty} e^{-x^2}dx$

Question

UPD: I see the question wrongly, thanks for you guys.

I use integration by parts to calculate the left side of the equation. Like $$\int x^2e^{-x^2} dx$$ $$= -x^2e^{-x} - \int 2x(-e^{-x}) dx$$ $$= -x^2e^{-x} - 2xe^{-x} - 2e^{-x} + C$$

So $\int_0^{\infty} x^2e^{-x^2} dx = \lim_{t \to \infty} ((-t^2e^{-t} - 2xe^{-t} - 2e^{-t}) - (-2)) = 2$

But I don't know how to calculate the right side of the equation.

Can anyone give me some tips? Thanks in advance!

Answer 1

You wrote $e^{-x^2}$ as $e^{-x}$ in your second line, so your working is wrong. Instead:

\begin{align*} \int_0^\infty x^2e^{-x^2} \mathrm{d}x &= \int_0^\infty -\frac{x}{2}(-2xe^{-x^2}) \mathrm{d}x \\ &= \left[-\frac{x}{2}e^{-x^2}\right]_{x = 0}^{x = \infty} - \int_0^\infty -\frac{1}{2}e^{-x^2} \mathrm{d}x \\ &= \frac{1}{2} \int_0^\infty e^{-x^2} \mathrm{d}x \end{align*} Several details were omitted, and I'll leave you to figure them out.

Answer 2

Hint : Try integration by parts by taking $u=x, dv = xe^{-x^2}dx$.

Answer 3

For a probabilistic interpretation of this identity, note that after scaling, it is equivalent to saying that a standard Gaussian random variable has variance $1$.

Answer 4

The difference between the two sides is $\int_0^\infty\left(x^2-\frac12\right)\exp-x^2dx=\left[-\frac12x\exp-x^2\right]_0^\infty=0$.