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Gaussian Integral — Definition, Formula & Examples

The Gaussian integral is the definite integral of e−x2e^{-x^2} over the entire real line, and it evaluates to π\sqrt{\pi}. It serves as the foundation for the normal distribution in statistics and appears throughout physics and engineering.

The Gaussian integral is the improper integral ∫−∞∞e−x2 dx=π\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}. More generally, for a parameter a>0a > 0, the integral ∫−∞∞e−ax2 dx=π/a\int_{-\infty}^{\infty} e^{-ax^2}\,dx = \sqrt{\pi/a}. The integrand e−x2e^{-x^2} has no elementary antiderivative, so the integral cannot be computed using the Fundamental Theorem of Calculus directly; instead, it is evaluated by squaring the integral and converting to polar coordinates.

Key Formula

∫−∞∞e−x2 dx=π\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}
Where:
  • xx = Real-valued integration variable
  • ee = Euler's number, approximately 2.71828

How It Works

Since e−x2e^{-x^2} has no closed-form antiderivative, you cannot simply find an antiderivative and plug in bounds. Instead, the standard technique squares the integral: let I=∫−∞∞e−x2 dxI = \int_{-\infty}^{\infty} e^{-x^2}\,dx, then compute I2I^2 as a double integral over the entire xyxy-plane. Converting to polar coordinates (r,θ)(r, \theta) transforms the double integral into one that is straightforward to evaluate, yielding I2=πI^2 = \pi and therefore I=πI = \sqrt{\pi}. This trick — sometimes called the "Gaussian integral trick" — is one of the most elegant techniques in calculus. The result extends to integrals of the form ∫−∞∞e−ax2+bx dx\int_{-\infty}^{\infty} e^{-ax^2+bx}\,dx by completing the square in the exponent.

Worked Example

Problem: Prove that the Gaussian integral equals √π by squaring it and converting to polar coordinates.
Step 1: Square the integral: Define I=∫−∞∞e−x2 dxI = \int_{-\infty}^{\infty} e^{-x^2}\,dx. Then I2I^2 is a double integral over the plane.
I2=∫−∞∞∫−∞∞e−(x2+y2) dx dyI^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\,dx\,dy
Step 2: Convert to polar coordinates: Substitute x2+y2=r2x^2 + y^2 = r^2 and dx dy=r dr dθdx\,dy = r\,dr\,d\theta, with rr from 00 to ∞\infty and θ\theta from 00 to 2π2\pi.
I2=∫02π∫0∞e−r2 r dr dθI^2 = \int_0^{2\pi}\int_0^{\infty} e^{-r^2}\,r\,dr\,d\theta
Step 3: Evaluate the radial integral: Use the substitution u=r2u = r^2, so du=2r drdu = 2r\,dr. The inner integral becomes a standard exponential integral.
∫0∞r e−r2 dr=12∫0∞e−u du=12\int_0^{\infty} r\,e^{-r^2}\,dr = \frac{1}{2}\int_0^{\infty} e^{-u}\,du = \frac{1}{2}
Step 4: Combine and take the square root: Multiply by the angular integral ∫02πdθ=2π\int_0^{2\pi} d\theta = 2\pi and solve for II.
I2=2π⋅12=π  ⟹  I=πI^2 = 2\pi \cdot \frac{1}{2} = \pi \implies I = \sqrt{\pi}
Answer: ∫−∞∞e−x2 dx=π≈1.7725\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi} \approx 1.7725

Another Example

Problem: Evaluate ∫−∞∞e−3x2 dx\int_{-\infty}^{\infty} e^{-3x^2}\,dx.
Step 1: Apply the generalized formula: Use ∫−∞∞e−ax2 dx=π/a\int_{-\infty}^{\infty} e^{-ax^2}\,dx = \sqrt{\pi/a} with a=3a = 3.
∫−∞∞e−3x2 dx=π3\int_{-\infty}^{\infty} e^{-3x^2}\,dx = \sqrt{\frac{\pi}{3}}
Step 2: Simplify: Rationalize and approximate if needed.
π3=3π3≈1.0233\sqrt{\frac{\pi}{3}} = \frac{\sqrt{3\pi}}{3} \approx 1.0233
Answer: π/3≈1.0233\sqrt{\pi/3} \approx 1.0233

Visualization

Why It Matters

The Gaussian integral is the reason the normal (bell curve) distribution in statistics integrates to 1 — without it, probability theory as taught in every introductory statistics course would lack its most essential tool. In quantum mechanics, Gaussian wave packets and path integrals rely on this result. Mastering the polar-coordinate proof is a milestone in multivariable calculus courses (Calculus III) and a gateway to techniques used in mathematical physics and signal processing.

Common Mistakes

Mistake: Trying to find an elementary antiderivative of e−x2e^{-x^2} and getting stuck.
Correction: No elementary antiderivative exists. You must use the polar-coordinate squaring technique or recognize the standard result π\sqrt{\pi} (or π/a\sqrt{\pi/a} for e−ax2e^{-ax^2}).
Mistake: Forgetting the extra factor of rr when converting dx dydx\,dy to polar coordinates.
Correction: The area element in polar coordinates is r dr dθr\,dr\,d\theta, not dr dθdr\,d\theta. This factor of rr is exactly what makes the radial integral solvable via uu-substitution.

Related Terms