# Property fourier transform

1. Dec 26, 2008

### xoureo

1. The problem statement, all variables and given/known data
If $$F(k)=TF\{f(x)\},k\neq 0$$ where TF is the Fourier transform ,and

$$F(0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(u)du\neq 0$$ ,

show that

$$TF\{\int_{-\infty}^{x}f(u)du\}=-i \frac{F(k)}{k} +\pi F(0)\delta(k)$$

2. Relevant equations

3. The attempt at a solution
I attempt the following:

$$TF\{\int_{-\infty}^{x}f(u)du\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \{\int_{-\infty}^{x}f(u)du\}e^{-ikx}dx$$

and , integration by parts, with

$$w=\int_{-\infty}^{x}f(u)du$$

$$dw=f(x)dx$$

give

$$\frac{1}{\sqrt{2\pi}}\{\int_{-\infty}^{x}f(u)du \frac{i}{k} e^{-ikx}\}|_{-\infty}^{\infty} -\frac{1}{\sqrt{2\pi}} \frac{i}{k}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$

The last term is

$$-\frac{i}{k}F(k)$$

What should i do with the first?. I do this, but it probably be bad:

$$\frac{i}{k} e^{-ikx}\}|_{-\infty}^{\infty}=\int_{-\infty}^{\infty}e^{-ikx}dx = 2\pi\delta(k)$$

and

$$\frac{1}{\sqrt{2\pi}}\left[\lim_{x\rightarrow\infty}\int_{-\infty}^xf(z)dz-\lim_{x\rightarrow -\infty}\int_{-\infty}^xf(z)dz\right] = F(0)$$

Then,

$$\int_{-\infty}^{\infty}\frac{dx}{\sqrt{2\pi}}\int_{-\infty}^xf(z)dze^{-ikx}=-i\frac{F(k)}{k}+2\pi F(0)\delta(k)$$

where is an extra 2.

I tried to do this problem using convolution, but again i cant achieve the desired result:

$$(f*g)(x)=\int_{-\infty}^{\infty}f(u)g(x-u)du$$

If $$g(x-u)$$ is the Heaviside function, and with $$x-u>0$$ , the convolution is

$$(f*g)(x)=\int_{-\infty}^{\infty}f(u)g(x-u)du=\int_{-\infty}^{x}f(u)du$$

Then,

$$TF\{\int_{-\infty}^{x}f(u)du\}=TF\{(f*g)(x)\}=\sqrt{2\pi}F(k)G(k)$$

But, the fourier transform of the Heaviside function is:

$$TF\{g(x)\}=\frac{-i}{k\sqrt{2\pi}}+\sqrt{\frac{\pi}{2}}\delta(k)$$

Hence,

$$TF\{\int_{-\infty}^{x}f(u)du\}=\frac{-iF(k)}{k}+\pi F(k)\delta(k)$$

But, in the last term appears $$F(k)$$ , and i want to get $$F(0)$$. What is wrong??

Thanks

2. Dec 26, 2008

### Dick

There's nothing wrong with your convolution argument. Think what happens if you integrate a test function against F(k)*delta(k). You get the same thing as if you integrate it against F(0)*delta(k) if F is continuous.F(k)*delta(k) and F(0)*delta(k) are the same distribution.

3. Dec 27, 2008

### xoureo

Ok,i understand it now, thanks for the help