Quantcast
Channel: Eat, Study, Sleep, Repeat.
Viewing all articles
Browse latest Browse all 12

Fourier Transform Solution for the Dirichlet Integral (sin(x)/x)

$
0
0

In mathematics, there are several integrals known as the Dirichlet integral, after the German mathematician Peter Gustav Lejeune Dirichlet.

One of those is

\int_0^{\infty} \frac{\sin x}{x}dx

In this post I will present my solution to this integral, using Fourier transforms and their properties. I will also show how to calculate the integrals for

\int_0^{\infty} (\frac{\sin x}{x})^n dx

for n = 2,3,4

In later posts, so as to keep individual solutions separate, and keep posts at a decent length.

Before I begin, I would like to mention that this is not meant as a very rigorous mathematical solution, so some details and steps are skipped. Also, some assumptions are made that are not explained, but they can be proven to be valid.

 The Integral \int_0^{\infty} \frac{\sin x}{x}dx

First, let’s have a look at what this function actually looks like: Sin(x)/x

Although the integral is only defined from zero to infinity, I wanted to plot this in order to show that the function is indeed even (symmetric about the y-axis). In order to solve this, we need to consider the following function: f(x) = h(1-|x|) where h(1-|x|) represent the Heaviside step function. This gives us a function that is explicitly:

f(x) =  \left\{  	\begin{array}{ll}  		1  & \mbox{if } |x| \leq 1 \\  		0 & \mbox{if } |x| > 1  	\end{array}  \right.

Heaviside Step Function

Now this being said, let’s take the Fourier transform of this function. F(\xi) =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{i\xi x} dx

However, considering our function, this can be simply written as:

F(\xi) =\frac{1}{\sqrt{2\pi}} \int_{-1}^{1} f(x) e^{i\xi x} dx = \frac{1}{\sqrt{2\pi}}[\frac{e^{i\xi}}{i\xi}- \frac{e^{-i\xi}}{i\xi}]_{-1}^1

Therefore, the Fourier transform is:

F(\xi) = \frac{2}{\sqrt{2\pi}\xi}\frac{e^{i\xi}-e^{-i\xi}}{2i} = \sqrt{\frac{2}{\pi}}\frac{\sin\xi}{\xi}

Interesting result, is it not?

Now we only need to use the inversion theorem, to get that:

f(x) =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{i\xi x}\sqrt{\frac{2}{\pi}}\frac{\sin\xi}{\xi}d\xi

Considering that the integration here is by \xi , we can treat x as a parameter and take x=0

f(0) =\frac{1}{\pi} \int_{-\infty}^{\infty}\frac{\sin\xi}{\xi}d\xi

But we already know that f(0) = 1, so we immediately get that

\int_{-\infty}^{\infty}\frac{\sin\xi}{\xi}d\xi = \pi

Considering the fact that \frac{\sin{\xi}}{\xi} is even in \xi, we have:

\int_0^{\infty}\frac{\sin\xi}{\xi}d\xi = \frac{1}{2}\int_{-\infty}^{\infty}\frac{\sin\xi}{\xi}d\xi = \frac{\pi}{2}

The conditions required for the functions involved, so that we can use the inversion theorem can be found here.


Filed under: Engineering School, Tips and Tricks Tagged: dirichlet integral, fourier transform, sinx/x

Viewing all articles
Browse latest Browse all 12

Latest Images

Trending Articles



Latest Images