Half-Gaussian Integral
Evaluate the integral: $$\int_{0}^{\infty} e^{-x^2/2} \, dx$$
You may use the fact that the full Gaussian integral is $\int_{-\infty}^{\infty} e^{-x^2/2} \, dx = \sqrt{2\pi}$.
Hints
- The integrand $e^{-x^2/2}$ is an even function: $f(-x) = f(x)$. How does this relate the integral over $[0, \infty)$ to the integral over $(-\infty, \infty)$?
- The full Gaussian integral $\int_{-\infty}^{\infty} e^{-x^2/2} \, dx = \sqrt{2\pi}$ is the normalizing constant of the standard normal distribution. Use symmetry to get the half-integral.
- Alternatively, use the substitution $t = x/\sqrt{2}$ to reduce to the standard form $\int_0^\infty e^{-t^2} dt = \sqrt{\pi}/2$, then multiply by $\sqrt{2}$.
Worked Solution
How to Think About It: The integrand $e^{-x^2/2}$ is the kernel of the standard normal distribution. The full integral over the real line equals $\sqrt{2\pi}$ -- this is the normalizing constant of the standard normal density. Since the integrand is even (symmetric about zero), the integral from 0 to infinity is exactly half the full integral. That is all there is to this problem. Alternatively, you can derive it from the classic Gaussian integral $\int_0^\infty e^{-t^2} dt = \sqrt{\pi}/2$ via a simple substitution.
Quick Estimate: $\sqrt{\pi/2} = \sqrt{1.571} \approx 1.253$. A quick sanity check: the integrand starts at $e^0 = 1$ and decays to 0. Most of its mass is in $[0, 2]$, and the average value over that range is around 0.5-0.6. So the integral is roughly $0.55 \times 2 \approx 1.1$, which is consistent with 1.25.
Approach: Use symmetry of the even integrand.
Formal Solution:
Method 1 (Symmetry):
The integrand $f(x) = e^{-x^2/2}$ satisfies $f(-x) = f(x)$, so it is even. Therefore: $$\int_{-\infty}^{\infty} e^{-x^2/2} \, dx = 2 \int_{0}^{\infty} e^{-x^2/2} \, dx$$
Dividing both sides by 2: $$\int_{0}^{\infty} e^{-x^2/2} \, dx = \frac{1}{2} \sqrt{2\pi} = \sqrt{\frac{\pi}{2}}$$
Method 2 (Substitution from the standard Gaussian integral):
It is known that $\int_0^\infty e^{-t^2} \, dt = \frac{\sqrt{\pi}}{2}$.
Substitute $t = x/\sqrt{2}$, so $dt = dx/\sqrt{2}$ and $x = \sqrt{2}\, t$: $$\int_0^\infty e^{-x^2/2} \, dx = \int_0^\infty e^{-t^2} \cdot \sqrt{2} \, dt = \sqrt{2} \cdot \frac{\sqrt{\pi}}{2} = \frac{\sqrt{2\pi}}{2} = \sqrt{\frac{\pi}{2}}$$
Answer: $$\int_{0}^{\infty} e^{-x^2/2} \, dx = \sqrt{\frac{\pi}{2}} \approx 1.2533$$
Intuition
This integral appears constantly in probability and statistics because $e^{-x^2/2}$ is the kernel of the standard normal distribution. The reason $\sqrt{2\pi}$ shows up as the normalizing constant of the normal density is precisely that $\int_{-\infty}^{\infty} e^{-x^2/2} dx = \sqrt{2\pi}$. Whenever you work with normal distributions, you are implicitly using this result.
The broader lesson is that many apparently hard integrals involving Gaussians can be solved by reduction to known forms via substitution and symmetry arguments. The polar coordinates trick that derives $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$ in the first place (squaring the integral and converting to polar) is one of the most elegant moves in all of calculus, and it is worth understanding -- not just memorizing the result.