# Exploring Variations of the Gaussian Integral with Techniques

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## Chapter 1: Introduction to the Gaussian Integral

In a previous article, I detailed the conventional approach using polar coordinates to solve the Gaussian integral. This time, I aim to explore several variations of the Gaussian integral, as they frequently arise in disciplines like statistics and quantum mechanics.

### Section 1.1: The Basic Integral of exp(-ax²)

Let's begin with one of the most fundamental variations of the Gaussian integral, specifically the integral of exp(-ax²), where a > 0. Although the technique used to solve this integral mirrors that of the standard Gaussian integral, it's essential to revisit it as we will leverage this result for the subsequent integrals.

To tackle this integral, we will employ the polar coordinates method, utilizing the u-substitution u = -ar² instead of u = -r². This approach leads us directly to the desired outcome.

### Section 1.2: The Integral of x²ⁿexp(-ax²)

With the first integral in hand, we can move on to the integral of x²ⁿexp(-ax²). The key to solving this integral is to express both this integral and the previous one as a function of 'a':

At first glance, this might seem unusual since 'a' is merely a constant. However, differentiating I₀(a) with respect to 'a' will clarify this approach. We can derive I₂(a) by calculating -d/da I₀(a).

To find I₄(a), we can differentiate I₂(a) and reverse the sign, leading us to the relationship I₄(a) = -d/da I₂(a). In terms of I₀(a), this can be represented as d²/da² I₀(a).

A pattern emerges: each subsequent integral (I₂(a), I₄(a), I₆(a), …) is simply the prior integral differentiated with respect to 'a', with the sign inverted. This recursive property allows us to express the 2nth integral in a concise manner.

Using the power rule, each differentiation of (sqrt{pi/a}) results in multiplying by the power of 'a' (i.e., -1/2, -3/2, -5/2, etc.) while reducing the power of 'a' by one each time. This can be neatly summarized as follows:

Here, the double factorial, !!, signifies a product over all integers of the same parity (either odd or even). For example, 9!! = 9*7*5*3*1. This can also be validated through induction.

In conclusion, the strength of this method lies in our decision to represent the integral as a function of 'a' and differentiate accordingly. This technique, known as differentiation under the integral sign or Feynman's trick, serves as a powerful tool for simplifying seemingly complex integrals.

## Chapter 2: Exploring Additional Integrals

### Section 2.1: The Integral of x^(2n+1)exp(-ax²)

Next, we will consider the integral of x^(2n+1)exp(-ax²). Note that the limits of integration are 0 to ∞ since this integral represents an odd function, and integrating over -∞ to ∞ will yield zero.

To solve this integral, we can make the straightforward substitution of u = x², which will eliminate 'x' from the equation. This yields a simpler form; however, the resulting integral remains challenging.

A more straightforward approach involves using differentiation under the integral sign. We can differentiate the familiar integral of exp(-au) with respect to 'a':

Repeating this process will increment the exponent of 'u' while alternating the sign. Ultimately, we can derive the following relationship:

The repeated application of (-d/da) on 1/a results in n!/a^(n+1). By combining all this information, we can solve the original integral, leading us to the desired result.

### Section 2.2: The Integral of exp(-ax² + bx + c)

Finally, we will examine the integral of exp(-ax² + bx + c). This scenario generalizes the Gaussian integral by allowing the exponent to be any quadratic function (with a remaining positive).

The first step in solving this integral is to factor out eᶜ, as it is simply a constant. However, the challenge arises from the additional bx term in the exponent. Converting to polar coordinates would complicate matters further, resulting in an expression like -ar² + br(sinθ + cosθ), which cannot be resolved through u-substitution.

By completing the square, we can simplify the integral considerably:

Notice that the resulting integral is essentially the first Gaussian integral we solved, just translated horizontally! Because we are integrating across the entire space, the definite integral's value remains invariant under horizontal translations, allowing us to conclude that:

Thank you for reading.

In this video, the Gaussian Integral is solved using polar coordinates, illustrating the foundational method in a visual format.

This video demonstrates a unique and engaging way to solve the Gaussian Integral, showcasing alternative techniques and insights.