标准正态分布PDF积分等于1的证明：积分存在性的疑问

Great question—this is such a common gap in introductory probability and stats courses. We often rush to compute (I^2 = 1) to show (I = 1), but we skip over the critical first step: verifying that (I = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-x2/2} dx) actually converges (i.e., exists as a finite value). Let's walk through this clearly:

Step 1: Use Symmetry to Simplify the Problem #

The standard normal PDF (f(x) = \frac{1}{\sqrt{2\pi}} e^{-x2/2}) is an even function—meaning (f(-x) = f(x)) for all real (x). This lets us rewrite the integral as:
[
I = 2\int_{0}^{\infty} f(x) dx
]
Now we only need to prove that (\int_{0}^{\infty} f(x) dx) converges; the symmetry will take care of the negative half of the integral.

Step 2: Split the Integral into Manageable Parts #

Break the integral from 0 to ∞ into two segments:
[
\int_{0}^{\infty} f(x) dx = \int_{0}^{1} f(x) dx + \int_{1}^{\infty} f(x) dx
]

Part 1: The Finite Interval ([0,1])

On the closed interval ([0,1]), (f(x)) is a continuous function (it's a combination of exponential and polynomial terms, which are both continuous everywhere). By the fundamental theorem of calculus, definite integrals of continuous functions over finite intervals are always finite. So (\int_{0}^{1} f(x) dx) is just a regular, finite value—no convergence issues here.

Part 2: The Infinite Interval ([1, \infty))

Here's where we need to use the comparison test for improper integrals. The key observation is that the exponential term (e^{-x2/2}) decays extremely quickly as (x) grows—faster than any polynomial, and even faster than simpler exponential functions like (e^{-x/2}).

For (x \geq 1), notice that (x^2/2 \geq x/2) (since (x \geq 1) implies (x^2 = x \cdot x \geq x \cdot 1 = x); divide both sides by 2). Since the exponential function is strictly decreasing, this means:
[
e^{-x2/2} \leq e^{-x/2}
]
Multiply both sides by the positive constant (\frac{1}{\sqrt{2\pi}}):
[
0 < f(x) \leq \frac{1}{\sqrt{2\pi}} e^{-x/2}
]

Now compute the improper integral of the upper bound:
[
\int_{1}^{\infty} e^{-x/2} dx = \lim_{b \to \infty} \int_{1}^{b} e^{-x/2} dx = \lim_{b \to \infty} \left[ -2e^{-x/2} \right]{1}^{b} = \lim{b \to \infty} \left( -2e^{-b/2} + 2e^{-1/2} \right) = 2e^{-1/2}
]
This result is a finite, positive number—so (\int_{1}^{\infty} e^{-x/2} dx) converges. By the comparison test, since (0 < f(x) \leq \frac{1}{\sqrt{2\pi}} e^{-x/2}) for all (x \geq 1), (\int_{1}^{\infty} f(x) dx) must also converge to a finite value.

Step 3: Combine the Results #

Since both (\int_{0}^{1} f(x) dx) (finite) and (\int_{1}^{\infty} f(x) dx) (convergent, finite) are finite, their sum (\int_{0}^{\infty} f(x) dx) is finite. Multiply by 2 (from the symmetry step) and we get that (I) exists as a finite real number.

As an extra check, you could also use the limit comparison test with (\frac{1}{x^2}) (since (e^{-x2/2}) decays faster than (x^{-2}) as (x \to \infty)), but the exponential comparison is more straightforward for most learners.

内容的提问来源于stack exchange，提问作者IntegrateThis