The standard normal distribution has the probability density function (PDF):

\[ f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \]

To show this is a valid PDF, we need to verify:

\[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \]

Step 1: Square the Integral

Let:

\[ I = \int_{-\infty}^{\infty} e^{-x^2/2} \, dx \]

Then:

\[ I^2 = \left( \int_{-\infty}^{\infty} e^{-x^2/2} \, dx \right)^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)/2} \, dx \, dy \]

Switch to polar coordinates:

\[ x = r \cos\theta,\quad y = r \sin\theta,\quad dx\,dy = r\,dr\,d\theta \]

So:

\[ I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2/2} \cdot r \, dr \, d\theta \]

Step 2: Evaluate the Inner Integral

\[ \int_{0}^{\infty} r e^{-r^2/2} \, dr \]

Substitute \( u = r^2/2 \Rightarrow du = r \, dr \):

\[ = \int_{0}^{\infty} e^{-u} \, du = 1 \]

Step 3: Evaluate the Full Integral

\[ I^2 = \int_{0}^{2\pi} 1 \, d\theta = 2\pi \Rightarrow I = \sqrt{2\pi} \]

Thus:

\[ \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dx = \frac{1}{\sqrt{2\pi}} \cdot \sqrt{2\pi} = \boxed{1} \]

Final Result

\[ \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dx = \boxed{1} \]

So the standard normal PDF is properly normalized.

Reference