How can the convolution theorem be used to evaluate this integral?

[leopard]

Homework Statement

Let a function f on (-$$\infty$$, $$\infty$$) be defined as

f(x) = cos x, if |x|<1;
f(x) = 0, otherwise

Find the Fourier transform of f and then evaluate the integral

$$\int ^{\infty}_{\infty} \frac{sin 2w}{w} cosw dw$$

2. The attempt at a solution

I calculate the Fourier transform: $$\frac{1}{\sqrt{2 \pi}}(\frac{sin(1-w)}{1-w} + \frac{sin(1+w)}{1+w})$$

This is the correct answer.

Now, how can this be used to calculate the integral?

 

[mighty2000]

To calculate the integral, you can use the convolution theorem, which states that the Fourier transform of a convolution of two functions is equal to the product of their individual Fourier transforms. In this case, the integral can be written as the convolution of the Fourier transform of f(x) and the function g(w) = sin(2w)/w. So, we can rewrite the integral as:

\int ^{\infty}_{\infty} \frac{sin 2w}{w} cosw dw = \frac{1}{\sqrt{2 \pi}} \int ^{\infty}_{\infty} (\frac{sin(1-w)}{1-w} + \frac{sin(1+w)}{1+w}) g(w) dw

Using the convolution theorem, we can rewrite this as:

\frac{1}{\sqrt{2 \pi}} \int ^{\infty}_{\infty} F(s) G(w-s) ds

Where F(s) and G(s) are the Fourier transforms of f(x) and g(w), respectively. Plugging in the values we calculated for the Fourier transform of f(x), we get:

\frac{1}{\sqrt{2 \pi}} \int ^{\infty}_{\infty} (\frac{sin(1-w)}{1-w} + \frac{sin(1+w)}{1+w}) (\frac{sin(2s)}{s}) ds

This integral can be evaluated using standard techniques for evaluating integrals involving trigonometric functions. Once you have the value of the integral, you can substitute it back into the original expression to get the final answer.