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Created May 12, 2026 16:50
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some tasks in integration carried out with local LLM
Raw
doGaussianIntegral.Rmd
> library(ellmer)
1/4 packages newly attached/loaded, see sessionInfo() for details.
> ch = chat_ollama(model="qwen3-coder:30b")
> ch$chat("Discuss how to evaluate the integral of the Gaussian density function over the whole real line.")
# Evaluating the Gaussian Integral over the Entire Real Line
The integral of the Gaussian density function over the whole real line is a fundamental result in mathematics and
statistics:
$$\int_{-\infty}^{\infty} e^{-x^2/2} dx = \sqrt{2\pi}$$
## The Classic Proof: Squaring and Polar Coordinates
The most elegant approach involves squaring the integral and converting to polar coordinates.
### Step 1: Set up the squared integral
Let $I = \int_{-\infty}^{\infty} e^{-x^2/2} dx$
Then $I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2/2} dx\right)\left(\int_{-\infty}^{\infty} e^{-y^2/2} dy\right)$
### Step 2: Combine into a double integral
$$I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)/2} dx\,dy$$
### Step 3: Convert to polar coordinates
Using $x = r\cos\theta$, $y = r\sin\theta$, $dx\,dy = r\,dr\,d\theta$:
$$I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2/2} \cdot r\,dr\,d\theta$$
### Step 4: Evaluate the radial integral
$$\int_{0}^{\infty} r e^{-r^2/2} dr = \left[-e^{-r^2/2}\right]_{0}^{\infty} = 0 - (-1) = 1$$
### Step 5: Complete the calculation
$$I^2 = \int_{0}^{2\pi} 1\,d\theta = 2\pi$$
Therefore: $I = \sqrt{2\pi}$
## Alternative Approaches
### 1. Using the Gamma Function
The integral can be related to the Gamma function:
$$\int_{-\infty}^{\infty} e^{-x^2/2} dx = \sqrt{2} \int_{-\infty}^{\infty} e^{-u^2} du = \sqrt{2} \cdot
\frac{\Gamma(1/2)}{\sqrt{2}} = \sqrt{2\pi}$$
### 2. Using Fourier Transforms
The Fourier transform of the Gaussian function leads to the same result through duality properties.
## Generalized Case
For the standard Gaussian:
$$\int_{-\infty}^{\infty} e^{-ax^2} dx = \sqrt{\frac{\pi}{a}}$$
where $a > 0$.
For the general Gaussian density with mean $\mu$ and variance $\sigma^2$:
$$\int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx = 1$$
This result is crucial because it ensures that the Gaussian is a valid probability density function, with total
probability equal to 1.
## Key Insights
1. **The factor $\sqrt{2\pi}$** emerges naturally from the relationship between the Gaussian function and the area
under a semicircle
2. **Rotation invariance** of the exponential term $e^{-(x^2+y^2)/2}$ is key to the polar coordinate approach
3. **The result is independent of the variable** - this reflects the fundamental symmetry of the Gaussian function
4. **This integral is the cornerstone** of many statistical applications, including the normal distribution's
normalization and various probability calculations
The elegance of this result demonstrates the deep connections between different areas of mathematics, particularly
analysis, geometry, and probability theory.
>
[1]+ Stopped R
PC002566:~ vincentcarey$ pwd
/Users/vincentcarey
PC002566:~ vincentcarey$ vi bag.tex
PC002566:~ vincentcarey$ fg
R
> ch$chat("Can you explain how Gamma(1/2) is evaluated?")
# Evaluating Γ(1/2)
The Gamma function Γ(z) is defined as:
$$\Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} dt$$
For Γ(1/2), we have:
$$\Gamma(1/2) = \int_{0}^{\infty} t^{-1/2} e^{-t} dt = \int_{0}^{\infty} \frac{e^{-t}}{\sqrt{t}} dt$$
## Method 1: Connection to the Gaussian Integral
We can relate this to our previous Gaussian integral. Let's make a substitution:
Let $u = \sqrt{t}$, so $t = u^2$ and $dt = 2u\,du$
$$\Gamma(1/2) = \int_{0}^{\infty} \frac{e^{-u^2}}{u} \cdot 2u\,du = 2\int_{0}^{\infty} e^{-u^2} du$$
We know that $\int_{0}^{\infty} e^{-u^2} du = \frac{\sqrt{\pi}}{2}$ (this is the Gaussian integral from 0 to ∞)
Therefore:
$$\Gamma(1/2) = 2 \cdot \frac{\sqrt{\pi}}{2} = \sqrt{\pi}$$
## Method 2: Using the Gaussian Integral Directly
We can also derive this more directly. Consider:
$$\Gamma(1/2) = \int_{0}^{\infty} t^{-1/2} e^{-t} dt$$
Let's make the substitution $t = x^2$, so $dt = 2x\,dx$ and $t^{-1/2} = x^{-1}$:
$$\Gamma(1/2) = \int_{0}^{\infty} x^{-1} e^{-x^2} \cdot 2x\,dx = 2\int_{0}^{\infty} e^{-x^2} dx$$
As mentioned above, $\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt{\pi}}{2}$
So: $\Gamma(1/2) = 2 \cdot \frac{\sqrt{\pi}}{2} = \sqrt{\pi}$
## Method 3: Using the Beta Function
We can also use the relationship between Gamma and Beta functions:
$$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
Setting $x = y = 1/2$:
$$B(1/2,1/2) = \frac{\Gamma(1/2)\Gamma(1/2)}{\Gamma(1)} = \frac{[\Gamma(1/2)]^2}{1}$$
We know that $B(1/2,1/2) = \int_{0}^{1} t^{-1/2}(1-t)^{-1/2} dt = \int_{0}^{1} \frac{1}{\sqrt{t(1-t)}} dt$
This integral equals π (can be evaluated using trigonometric substitution), so:
$$[\Gamma(1/2)]^2 = \pi$$
$$\Gamma(1/2) = \sqrt{\pi}$$
## The Fundamental Nature of Γ(1/2) = √π
This result is quite remarkable because:
1. It connects the Gamma function (which generalizes factorials) to π
2. It shows that $\Gamma(1/2) = \frac{(1/2-1)!}{1} = \frac{(-1/2)!}{1}$, which is consistent with the idea that
$\sqrt{\pi}$ represents the "half-factorial" of 1
3. It provides the normalization constant that makes the Gaussian integral yield $\sqrt{2\pi}$
This connection between π and the Gamma function is one of the beautiful relationships that appear throughout
mathematics, linking analysis, geometry, and probability theory.
> ch$chat("Would you be able to develop some R code to illustrate visually the process of converting the one-dimensional integration task to evaluating an integral over the plane using polar coordinates?")
```r
# Visualization of the Gaussian integral evaluation using polar coordinates
library(ggplot2)
library(dplyr)
library(plotly)
# Create a grid of points in Cartesian coordinates
x <- seq(-3, 3, length.out = 50)
y <- seq(-3, 3, length.out = 50)
grid_cartesian <- expand.grid(x = x, y = y)
# Calculate the Gaussian function values in Cartesian coordinates
grid_cartesian$z <- exp(-0.5 * (grid_cartesian$x^2 + grid_cartesian$y^2))
# Create the visualization
p1 <- ggplot(grid_cartesian, aes(x = x, y = y, z = z)) +
geom_raster(aes(fill = z)) +
scale_fill_gradient(low = "white", high = "darkblue") +
geom_contour(bins = 10, color = "black", alpha = 0.3) +
labs(title = "Gaussian Function in Cartesian Coordinates",
x = "x", y = "y", fill = "e^(-x²-y²)/2") +
theme_minimal() +
theme(plot.title = element_text(hjust = 0.5))
print(p1)
# Now create the polar coordinate visualization
# Transform to polar coordinates
grid_polar <- grid_cartesian
grid_polar$r <- sqrt(grid_polar$x^2 + grid_polar$y^2)
grid_polar$theta <- atan2(grid_polar$y, grid_polar$x)
grid_polar$z_polar <- exp(-0.5 * grid_polar$r^2)
p2 <- ggplot(grid_polar, aes(x = r, y = theta, z = z_polar)) +
geom_raster(aes(fill = z_polar)) +
scale_fill_gradient(low = "white", high = "darkblue") +
labs(title = "Same Gaussian Function in Polar Coordinates",
x = "r (radius)", y = "θ (angle)", fill = "e^(-r²/2)") +
theme_minimal() +
theme(plot.title = element_text(hjust = 0.5))
print(p2)
# Create a more detailed demonstration using the integration approach
# Demonstrate the double integral setup
n <- 100
x_vals <- seq(-3, 3, length.out = n)
y_vals <- seq(-3, 3, length.out = n)
grid_vals <- expand.grid(x = x_vals, y = y_vals)
# Calculate the function values for different z-slices
grid_vals$z_values <- exp(-0.5 * (grid_vals$x^2 + grid_vals$y^2))
# Show a 3D surface plot
if(require(plotly)) {
# Create 3D plot
p3 <- plot_ly(grid_vals, x = ~x, y = ~y, z = ~z_values,
type = "surface", colors = "Blues") %>%
layout(title = "3D Surface of Gaussian Function",
scene = list(xaxis = list(title = "x"),
yaxis = list(title = "y"),
zaxis = list(title = "z")))
print(p3)
# Show the integrand vs radius
radius_vals <- seq(0, 3, length.out = 100)
integrand_vals <- exp(-0.5 * radius_vals^2)
p4 <- ggplot(data.frame(r = radius_vals, z = integrand_vals),
aes(x = r, y = z)) +
geom_line(color = "blue", size = 1) +
geom_ribbon(aes(ymin = 0, ymax = z), alpha = 0.3, fill = "lightblue") +
labs(title = "Integrand in Polar Coordinates: e^(-r²/2)",
x = "r (radius)", y = "e^(-r²/2)") +
theme_minimal() +
theme(plot.title = element_text(hjust = 0.5))
print(p4)
}
# Numerical verification of the integral
# Method 1: Direct numerical integration of the 1D case
int_1d <- integrate(function(x) exp(-x^2/2), lower = -Inf, upper = Inf)
print(paste("1D Gaussian integral:", int_1d$value))
print(paste("Expected value (sqrt(2*pi)):", sqrt(2*pi)))
print(paste("Difference:", abs(int_1d$value - sqrt(2*pi))))
# Method 2: Numerical integration of radial component
# Integral from 0 to infinity of (2*pi*r)*e^(-r²/2) dr
# which equals 2*pi * integral from 0 to infinity of r*e^(-r²/2) dr
# The integral from 0 to infinity of r*e^(-r²/2) dr = 1
# So the whole integral = 2*pi*1 = 2*pi
int_radial <- integrate(function(r) r * exp(-r^2/2), lower = 0, upper = Inf)
print(paste("Radial integral:", int_radial$value))
print(paste("2*pi:", 2*pi))
# The full 2D integral verification - we'll compute a finite approximation
# Using rectangular approximation
rectangular_approx <- function(x_vals, y_vals) {
dx <- x_vals[2] - x_vals[1]
dy <- y_vals[2] - y_vals[1]
total <- 0
for(i in 1:(length(x_vals)-1)) {
for(j in 1:(length(y_vals)-1)) {
x1 <- x_vals[i]
x2 <- x_vals[i+1]
y1 <- y_vals[j]
y2 <- y_vals[j+1]
# Use midpoint rule
mid_x <- (x1 + x2)/2
mid_y <- (y1 + y2)/2
total <- total + exp(-0.5 * (mid_x^2 + mid_y^2)) * dx * dy
}
}
return(total)
}
# Test with a smaller grid for demonstration
small_x <- seq(-2, 2, length.out = 20)
small_y <- seq(-2, 2, length.out = 20)
rectangular_result <- rectangular_approx(small_x, small_y)
print(paste("Rectangular approximation:", rectangular_result))
# Show the relationship with the full 2D integral
print(paste("Should equal (sqrt(2pi))² =", (sqrt(2*pi))^2))
print(paste("Which equals", 2*pi))
print(paste("Rectangle result compared to 2π:", rectangular_result, "vs", 2*pi))
# Visual demonstration of area under a curve in polar coordinates
# Show how integrating in polar coordinates works
theta_vals <- seq(0, 2*pi, length.out = 100)
r_vals <- seq(0, 3, length.out = 100)
radius_func <- function(r) r * exp(-0.5 * r^2)
area_function <- sapply(r_vals, radius_func)
# Create a plot showing how the area element changes
theta_plot <- seq(0, 2*pi, length.out = 20)
r_plot <- 1:10
area_element <- r_plot * exp(-0.5 * r_plot^2) * 0.1 # approximating the area
p5 <- ggplot(data.frame(r = r_plot, area = area_element),
aes(x = r, y = area)) +
geom_bar(stat = "identity", fill = "blue", alpha = 0.7) +
labs(title = "Area Elements in Polar Coordinates",
x = "r (radius)", y = "Area contribution") +
theme_minimal() +
theme(plot.title = element_text(hjust = 0.5))
print(p5)
# Mathematical verification that integral gives sqrt(2*pi)
# Let's do the symbolic verification with R's integrate function
sqrt_2pi <- sqrt(2 * pi)
print(paste("sqrt(2*pi) =", sqrt_2pi))
# Compare with numerical integration
analytical_result <- int_1d$value
print(paste("Numerical result =", analytical_result))
print(paste("Difference =", abs(analytical_result - sqrt_2pi)))
# Demonstrate the key relationship
# If I = ∫_{-∞}^{∞} e^{-x²/2} dx, then I² = ∫_{-∞}^{∞} ∫_{-∞}^{∞} e^{-(x²+y²)/2} dx dy
# This equals ∫_{0}^{2π} ∫_{0}^{∞} r * e^{-r²/2} dr dθ = 2π * 1 = 2π
# So I = √(2π)
```
Let me provide a simpler, more focused demonstration:
```r
# Simple demonstration of the Gaussian integral and polar coordinate conversion
# Create a simple visualization showing the transformation
library(ggplot2)
# Data for the 2D Gaussian
x_seq <- seq(-3, 3, length.out = 30)
y_seq <- seq(-3, 3, length.out = 30)
grid <- expand.grid(x = x_seq, y = y_seq)
grid$z <- exp(-0.5 * (grid$x^2 + grid$y^2))
# The relationship in polar coordinates
# For each r, we integrate over θ from 0 to 2π
# Since the integrand depends only on r, this becomes:
# ∫₀^{2π} ∫₀^{∞} r * e^{-r²/2} dr dθ = 2π * ∫₀^{∞} r * e^{-r²/2} dr
# Let's compute the radial integral
r_vals <- seq(0, 4, length.out = 100)
radial_integral <- sapply(r_vals, function(r) r * exp(-0.5 * r^2))
# Plot showing how we're computing the integral
ggplot(data.frame(r = r_vals, integrand = radial_integral),
aes(x = r, y = integrand)) +
geom_line(color = "blue", size = 1) +
geom_ribbon(aes(ymin = 0, ymax = integrand), fill = "lightblue", alpha = 0.5) +
geom_vline(xintercept = 0, linetype = "dashed", color = "gray") +
labs(title = "Integrand in Polar Coordinates: r * e^{-r²/2}",
x = "r (radius)", y = "Integrand value") +
theme_minimal() +
theme(plot.title = element_text(hjust = 0.5)) +
annotate("text", x = 1, y = 0.3, label = "∫₀^{∞} r * e^{-r²/2} dr = 1",
size = 4)
# Now let's show the final step
print("Verification:")
print("∫₀^{∞} r * e^{-r²/2} dr = 1 (this is the key integration)")
print("So the full 2D integral becomes:")
print("∫₀^{2π} ∫₀^{∞} r * e^{-r²/2} dr dθ = 2π * 1 = 2π")
print("Therefore: I² = 2π, so I = √(2π)")
```
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