Derivation of the Gaussian Integral

Introduction

The Gaussian integral (Euler-Poisson integral) is fundamental:

\[ \int_{-\infty}^{\infty} e^{-c x^2} \, dx = \sqrt{\frac{\pi}{c}}, \quad c>0. \]

We derive this by squaring the integral and converting to polar coordinates.

Step 1: Define the integral for \(c=1\)

Let

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

We aim to show \(I = \sqrt{\pi}\).

Step 2: Square the integral

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

Since the variables are independent:

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

Step 3: Convert to polar coordinates

Let \(x = r\cos\theta\), \(y = r\sin\theta\) with \(r \geq 0\), \(0 \leq \theta \leq 2\pi\).

Then:

\(x^2 + y^2 = r^2\)
\(dx \, dy = r \, dr \, d\theta\)

The integral becomes:

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

Step 4: Evaluate the \(r\)-integral

Let \(u = r^2\), so \(du = 2r \, dr\) \(\Rightarrow\) \(r \, dr = du/2\).

When \(r=0\), \(u=0\); when \(r \to \infty\), \(u \to \infty\).

\[ \int_{0}^{\infty} e^{-r^2} \, r \, dr = \int_{0}^{\infty} e^{-u} \cdot \frac{du}{2} = \frac{1}{2} \left[ -e^{-u} \right]_{0}^{\infty} = \frac{1}{2}. \]

Step 5: Evaluate the \(\theta\)-integral

\[ \int_{0}^{2\pi} d\theta = 2\pi. \]

Step 6: Combine results

\[ I^2 = (2\pi) \cdot \frac{1}{2} = \pi. \]

Since \(I > 0\), we take the positive square root:

\[ I = \sqrt{\pi}. \]

Thus:

\[ \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}. \]

Step 7: Generalize to arbitrary \(c>0\)

Use the substitution \(t = \sqrt{c} \, x\), so \(dt = \sqrt{c} \, dx\) \(\Rightarrow\) \(dx = dt / \sqrt{c}\).

When \(x \to \pm\infty\), \(t \to \pm\infty\).

\[ \int_{-\infty}^{\infty} e^{-c x^2} \, dx = \int_{-\infty}^{\infty} e^{-t^2} \cdot \frac{dt}{\sqrt{c}} = \frac{1}{\sqrt{c}} \int_{-\infty}^{\infty} e^{-t^2} \, dt. \]

Using the result from Step 6:

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

Final Result

\[ \boxed{\int_{-\infty}^{\infty} e^{-c x^2} \, dx = \sqrt{\frac{\pi}{c}}, \quad c>0}. \]

Numerical Verification

gaussian_integral <- function(c) {
  f <- function(x) exp(-c * x^2)
  integrate(f, -Inf, Inf)$value
}

gaussian_analytic <- function(c) sqrt(pi / c)

c_values <- c(0.5, 1, 2, 3, 5)
results <- data.frame(
  c = c_values,
  Numerical = sapply(c_values, gaussian_integral),
  Analytical = sapply(c_values, gaussian_analytic)
)

results$Difference <- results$Numerical - results$Analytical
print(results)

##     c Numerical Analytical    Difference
## 1 0.5 2.5066283  2.5066283 -5.159664e-10
## 2 1.0 1.7724539  1.7724539  1.221689e-12
## 3 2.0 1.2533141  1.2533141  1.142835e-08
## 4 3.0 1.0233267  1.0233267 -1.443290e-14
## 5 5.0 0.7926655  0.7926655  1.536413e-10

Visualization

library(ggplot2)

x_vals <- seq(-3, 3, length.out = 200)
c_values <- c(0.5, 1, 2, 3, 5)
plot_data <- expand.grid(x = x_vals, c = c_values)
plot_data$y <- exp(-plot_data$c * plot_data$x^2)

ggplot(plot_data, aes(x = x, y = y, color = factor(c))) +
  geom_line(size = 1) +
  labs(title = expression("Gaussian Functions " ~ e^{-c*x^2}),
       x = "x", y = "f(x)", color = "c value")  +
  theme_minimal()

Conclusion

We have derived the Gaussian integral using the square-and-polar-coordinates method. This result is essential in probability, quantum mechanics, heat equations, and Fourier analysis.