Integration of Gaussian Functions

Gaussian function is named after Carl Friedrich Gauss and is often referred to as a "bell curve" due to its characteristic shape. It's widely used in various fields such as probability theory, statistics, signal processing, and physics due to its properties.

In this article, we will discuss how we will Integrate Gaussian functions.

What are Gaussian Functions?

A Gaussian function is a function of the form:

\bold{f(x) = a e^{\left(-\frac{(x - b)^2}{2c^2}\right)}}

Where,

• a is the amplitude (the peak value of the function),
• b is the mean (the position of the center of the peak),
• c controls the width of the "bell" shape.

Integral of Gaussian Functions

To compute the integral of the Gaussian function over the entire real line, we focus on the special case where the function is centered at zero (b = 0) and normalized:

\int_{-\infty}^{\infty} e^{-\frac{x^2}{2c^2}} dx

The solution to this integral is:

\int_{-\infty}^{\infty} e^{-\frac{x^2}{2c^2}} dx = c\sqrt{2\pi}

This result is derived using a clever trick of converting the integral to polar coordinates, where the integral of the square of the Gaussian function in two dimensions reduces to a simpler form. This is a classic method used in calculus.

If you are dealing with a general Gaussian function:

f(x) = ae^{-\frac{(x-b)^2}{2c^2}}

The integral over the entire real line is:

\int_{-\infty}^{\infty} ae^{-\frac{(x-b)^2}{2c^2}} dx = a \cdot c\sqrt{2\pi}

This result holds regardless of the values of a, b, and c, as long as the integral is over the entire real line.

Derivation of the Gaussian Integral

The Gaussian integral is one of the most famous integrals in mathematics. The integral we are interested in is:

\int_{-\infty}^{\infty} e^{-x^2} \, dx

To find this integral, we use a clever trick involving polar coordinates. Here are the steps:

Step 1: Square the Integral

Consider the square of the integral III:

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

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

This can be written as a double integral over the entire (x, y) plane:

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

Step 2: Switch to Polar Coordinates

We convert the integral to polar coordinates, where x = r cos⁡θ and y = r sinθ. The Jacobian determinant for this transformation is r, so dx dy = r dr dθ. The limits change accordingly:

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

Step 3: Evaluate the Integral

First, integrate with respect to r:

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

Let u=r². Then du=2r dr, so r dr=1/2 du. The integral becomes:

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

Now integrate with respect to θ\thetaθ:

\int_{0}^{2\pi} d\theta = 2\pi

Putting it all together:

I^2 = \left( \frac{1}{2} \right) \cdot (2\pi) = \pi

Step 4: Take the Square Root

Finally, take the square root to find III:

I = \sqrt{\pi}

Applications of Integration of Gaussian Functions

Some of the common applications of Integration of Gaussian Functions:

• Probability and Statistics: The Gaussian distribution is used to model random variables with a normal distribution. It appears in the Central Limit Theorem, which states that the sum of many independent and identically distributed random variables tends toward a Gaussian distribution, regardless of the original distributions of the variables.

• Signal Processing: Gaussian functions are used to smooth signals and reduce noise.

• Physics: The Gaussian distribution describes phenomena such as diffusion processes and thermal equilibrium in statistical mechanics.

Conclusion

Gaussian functions play a critical role in various fields due to their unique properties, such as their bell-shaped curve and their occurrence in the Central Limit Theorem. Integrating Gaussian functions, especially over the entire real line, results in elegant solutions like \sqrt{\pi}, which highlight the profound interconnectedness between different areas of mathematics.

Read More,

• Integral Calculus
• Application of Integral
• Calculus

Solved Problems on Integration of Gaussian Functions

Problem 1: Evaluate the integral \int_{-\infty}^{\infty} e^{-x^2} \, dx.

Solution:

Let I=\int_{-\infty}^{\infty} e^{-x^2} \, dx. We calculate I² by considering:

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

Switching to polar coordinates, where x=rcos⁡θ and y=rsin⁡θ, we get:

x²+y²=r² dx \, dy = r \, dr \, d\theta

Thus,

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

The integral becomes:

I^2 = \int_{0}^{2\pi} d\theta \int_{0}^{\infty} re^{-r^2} \, dr

The inner integral can be solved using substitution u=r², du=2r :

I^2 = \int_{0}^{2\pi} d\theta \int_{0}^{\infty} \frac{1}{2} e^{-u} \, du

I^2 = \int_{0}^{2\pi} d\theta \cdot \frac{1}{2} \left[-e^{-u}\right]_0^\infty

I^2 = \int_{0}^{2\pi} d\theta \cdot \frac{1}{2} (1 - 0)

I^2 = \pi

Therefore,

I = \sqrt{\pi}

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

Problem 2: Evaluate the integral \int_{-\infty}^{\infty} e^{-ax^2} \, dx for a>0.

Solution:

Let I = \int_{-\infty}^{\infty} e^{-ax^2} \, dx. Use the substitution x = \frac{u}{\sqrt{a}},dx = \frac{du}{\sqrt{a}}​:

I = \int_{-\infty}^{\infty} e^{-a\left(\frac{u}{\sqrt{a}}\right)^2} \frac{du}{\sqrt{a}}

I = \frac{1}{\sqrt{a}} \int_{-\infty}^{\infty} e^{-u^2} \, du

Since \int_{-\infty}^{\infty} e^{-u^2} \, du = \sqrt{\pi}​, we have:

I = \frac{1}{\sqrt{a}} \sqrt{\pi}

Thus, \int_{-\infty}^{\infty} e^{-ax^2} \, dx = \sqrt{\frac{\pi}{a}}​​.

Practice Problems on Integration of Gaussian Functions

Problem 1: Evaluate the integral \int_{-\infty}^{\infty} x^2 e^{-x^2} \, dx.

Problem 2: Find the value of the integral \int_{-\infty}^{\infty} e^{-(x - \mu)^2/2\sigma^2} \, dx, where μ and σ are constants.

Problem 3: Compute \int_{0}^{\infty} x e^{-ax^2} \, dx for a>0.

Problem 4: Determine \int_{-\infty}^{\infty} \cos(kx)e^{-x^2} \, dx, where k is a constant.

Problem 5: Compute \int_{0}^{\infty} x e^{-ax^2} \, dx for a>1.

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Article Tags :

• Engineering Mathematics
• Mathematics
• Calculus