Question

In this problem, you will directly compute the convolution of two Gaussian functions in two steps. a. Use completing the square to evaluate $$ \int_{-\infty}^{\infty} e^{-a t^{2}+\beta t} d t $$ b. Use the result from part a to directly compute the convolution in Example 8.16: $$ (f * g)(x)=e^{-b x^{2}} \int_{-\infty}^{\infty} e^{-(a+b) t^{2}+2 b x t} d t $$

Short answer

The convolution \((f * g)(x)=e^{-b x^{2}} \int_{-\infty}^{\infty} e^{-(a+b) t^{2}+2 b x t} dt\) simplifies to \( e^{\frac{-abx^{2}}{a+b}} * \sqrt{\frac{\pi}{a+b}}\).

Step-by-step solution

Integral Calculation by Completing the Square

To compute the integral of \(\int_{-\infty}^{\infty} e^{-a t^{2}+\beta t} d t\), completing the square in the exponent is a viable approach.An expression of the form \(ax^{2}+bx\) can be rewritten as \(a(x+ \frac{b}{2a})^{2} - \frac{b^{2}}{4a}\), which, in this case, allows us to rewrite \(-at^{2}+\beta t\) as \(-a(t-\frac{\beta}{2a})^{2} + \frac{\beta^{2}}{4a}\) in the exponent of our integral.Now the integral becomes: \(\int_{-\infty}^{\infty} e^{-a(t-\frac{\beta}{2a})^{2} + \frac{\beta^{2}}{4a}} dt\). This integral can now be observed as a Gaussian integral and can be solved.

Gaussian Integral Result

The next step is to solve the integral. In general, a Gaussian integral of the form \(\int_{-\infty}^{\infty} e^{-ax^{2}} dx\) is equal to \(\sqrt{\frac{\pi}{a}}\).The integral \(\int_{-\infty}^{\infty} e^{-a(t-\frac{\beta}{2a})^{2} + \frac{\beta^{2}}{4a}} dt\) therefore equals \(e^{\frac{\beta^{2}}{4a}} \sqrt{\frac{\pi}{a}}\).

Implementation of Integral Result into Convolution

Now, the resulting integral has to be implemented into the convolution \((f * g)(x)=e^{-b x^{2}} \int_{-\infty}^{\infty} e^{-(a+b) t^{2}+2 b x t} dt\).Recognize the similarity between this and the previously solved integral. We can complete the square on the exponent and use our Gaussian integral result. In this case, \(-at^{2}+\beta t\) is replaced by \(-(a+b)t^{2}+2bxt\), hence, \(\beta\) becomes \(2bx\) and \(a\) becomes \(a+b\).By substituting the result of the integral, the convolution becomes: \(e^{-b x^{2}} * e^{\frac{(2bx)^{2}}{4(a+b)}} * \sqrt{\frac{\pi}{(a+b)}}\).

Simplification

Finally, simplify the convolution: \(e^{-b x^{2} + \frac{b^{2}x^{2}}{a+b}} * \sqrt{\frac{\pi}{a+b}}\) to give \( e^{\frac{-abx^{2}}{a+b}} * \sqrt{\frac{\pi}{a+b}}\). After simplification, this is our final solution.

Key concepts

Completing the Square

Completing the square is a mathematical technique often used to simplify the process of solving quadratic equations, as well as to evaluate integrals involving quadratic expressions. The process involves rewriting a quadratic expression in a way that highlights a perfect square trinomial. This can make it much easier to work with, especially in calculus.

When we have an expression of the form \(ax^2 + bx\), our goal is to express it as \(a(x + \, \frac{b}{2a})^2 - \, \frac{b^2}{4a}\). This transformation allows simplifying the evaluation of integrals because it converts the expression to a form that is more readily recognized as a Gaussian function.

For example, in the exercise provided, we have \(-at^2 + \beta t\). By completing the square, we rewrite it as \(-a(t - \frac{\beta}{2a})^2 + \frac{\beta^2}{4a}\). This results in an expression that represents a shifted Gaussian, which is easier to integrate.

Gaussian Integral

The Gaussian integral is a fundamental concept in calculus and probability theory. It involves the integral of the exponential function of a negative quadratic term, typically of the form \(\int_{-\infty}^{\infty} e^{-ax^2} \, dx\). This has a well-known result: \(\sqrt{\frac{\pi}{a}}\).

In the context of our exercise, after completing the square, the exponent of the integral becomes \(-a(t - \frac{\beta}{2a})^2 + \frac{\beta^2}{4a}\). We recognize this as a Gaussian integral because it follows the standard form, which allows us to solve it using the standard Gaussian result. The solution involves multiplying by \(e^{\frac{\beta^2}{4a}}\), which accounts for the shift in the Gaussian center. Thus, the integral evaluates to \(e^{\frac{\beta^2}{4a}} \sqrt{\frac{\pi}{a}}\). This elegant result is crucial in simplifying complex integrals related to Gaussian functions.

Integral Calculation

Integral calculation is a critical aspect of calculus, enabling us to determine areas under curves, among other applications. In this specific exercise, calculating the convolution integral translates to integrating a product of Gaussian functions across a continuous domain.

Let's revisit how this is applied. We started with \((f * g)(x) = e^{-bx^2} \int_{-\infty}^{\infty} e^{-(a+b)t^2 + 2bxt} \, dt\). The core challenge was evaluating the integral involving the quadratic term in the exponent. The solution involved completing the square on \(-(a+b)t^2 + 2bxt\) and recognizing the form as a Gaussian integral.

These transformations allowed us to handle the integral using our earlier results: specifically, we substitute \(a\) with \(a+b\) and \(\beta\) with \(2bx\). Once integrated, the convolution simplifies to \(e^{-b x^2 + \frac{b^2 x^2}{a+b}} \sqrt{\frac{\pi}{a+b}}\). This demonstrates how combining these mathematical tools can simplify and solve what initially appear to be complex convolution tasks.