Chapter 8: Problem 12

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

Expert verified

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.