- #1
xoureo
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Homework Statement
If [tex]F(k)=TF\{f(x)\},k\neq 0[/tex] where TF is the Fourier transform ,and
[tex]F(0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(u)du\neq 0[/tex] ,
show that
[tex]TF\{\int_{-\infty}^{x}f(u)du\}=-i \frac{F(k)}{k} +\pi F(0)\delta(k)[/tex]
Homework Equations
The Attempt at a Solution
I attempt the following:
[tex]TF\{\int_{-\infty}^{x}f(u)du\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \{\int_{-\infty}^{x}f(u)du\}e^{-ikx}dx[/tex]
and , integration by parts, with
[tex]w=\int_{-\infty}^{x}f(u)du[/tex]
[tex]dw=f(x)dx[/tex]
give
[tex]\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[/tex]
The last term is
[tex] -\frac{i}{k}F(k)[/tex]
What should i do with the first?. I do this, but it probably be bad:
[tex]\frac{i}{k} e^{-ikx}\}|_{-\infty}^{\infty}=\int_{-\infty}^{\infty}e^{-ikx}dx = 2\pi\delta(k)[/tex]
and
[tex]\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) [/tex]
Then,
[tex]
\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)
[/tex]
where is an extra 2.
I tried to do this problem using convolution, but again i can't achieve the desired result:
[tex](f*g)(x)=\int_{-\infty}^{\infty}f(u)g(x-u)du[/tex]
If [tex]g(x-u)[/tex] is the Heaviside function, and with [tex]x-u>0[/tex] , the convolution is
[tex](f*g)(x)=\int_{-\infty}^{\infty}f(u)g(x-u)du=\int_{-\infty}^{x}f(u)du[/tex]
Then,
[tex]TF\{\int_{-\infty}^{x}f(u)du\}=TF\{(f*g)(x)\}=\sqrt{2\pi}F(k)G(k)[/tex]
But, the Fourier transform of the Heaviside function is:
[tex]TF\{g(x)\}=\frac{-i}{k\sqrt{2\pi}}+\sqrt{\frac{\pi}{2}}\delta(k)[/tex]
Hence,
[tex]TF\{\int_{-\infty}^{x}f(u)du\}=\frac{-iF(k)}{k}+\pi F(k)\delta(k)[/tex]
But, in the last term appears [tex]F(k)[/tex] , and i want to get [tex]F(0)[/tex]. What is wrong??
Thanks