# An improper integral (Related to the Fourier transform)

1. Jul 21, 2011

### asmani

How to show this?
$$\int_{-\infty}^{+\infty}e^{-i2\pi xs}ds=\delta(x)$$
This is a part of a problem of "Bracewell, R. The Fourier Transform and Its Applications, 3rd ed. New York: McGraw-Hill, pp. 100-101, 1999". This isn't a homework, I found it http://mathworld.wolfram.com/FourierTransform1.html" [Broken]. But I'm not sure even if this is true.

Last edited by a moderator: May 5, 2017
2. Jul 21, 2011

### Mute

Try showing that

$$\int_{-\infty}^{\infty} dx~\delta(x)f(x) = \int_{-\infty}^\infty dx \int_{-\infty}^\infty ds~e^-{2\pi i s x}f(x) = f(0).$$

3. Jul 22, 2011

### Dickfore

$$\int_{-B}^{B}{e^{-i \, 2 \pi \, x \, s} \, ds} =\left.\frac{e^{-i \, 2 \pi \, x \, s}}{-i \, 2 \pi \, x}\right|_{B}^{B} = \frac{\sin{(\pi \, B \, x)}}{\pi \, x}$$

Next, consider the function:
$$\mathrm{sinc}(x) \equiv \frac{\sin{(\pi \, x)}}{\pi \, x}$$
and evaluate the definite integral (by contour integration)
$$\int_{-\infty}^{\infty}{\mathrm{sinc}(x) \, dx} = ?$$
Then, notice that the integral I evaluated is:
$$B \, \mathrm{sinc}(B \, x)$$
What happens if you take the limit $B \rightarrow \infty$?

4. Jul 22, 2011

### asmani

I can show that if f is continuous at 0, then:
$$\int_{-\infty}^{+\infty}f(x)\delta(x)dx=f(0)$$
And if f has the Fourier transform, then:
$$\int_{-\infty}^{+\infty}\left [\int_{-\infty}^{+\infty}f(x)e^{-i2\pi xs}dx \right ]ds=f(0)$$
Does it help?

Thanks

Last edited: Jul 22, 2011
5. Jul 22, 2011

### asmani

I can see that if $B \rightarrow \infty$, then:
$$\int_{-\infty}^{\infty}{B\mathrm{sinc}(Bx) \, dx} = 1$$
But $B\mathrm{sinc}(Bx)$ is not equal to 0 for every x≠0, while $\delta (x)$ is.

Thanks

Last edited: Jul 22, 2011
6. Jul 22, 2011

### asmani

Last edited by a moderator: Apr 26, 2017