MHB How Do You Prove a Specific Fourier Transform Property?

[mathmari]

Hey!

Could you give me a hint how to prove the following property of the Fourier transform, when $F[f(x)]=\widetilde{f}(x)$, where $F[f(x)]$ is the Fourier transform of $f(x)$?

$$F[ \widetilde{f}(x) ]= \frac{f(-k)}{2 \pi}$$

We know that: $ \widetilde{f}(k)=\int_{- \infty}^{+ \infty}{ {f}(x) e^{-i k x}}dx$ and $f(x)=\frac{1}{2 \pi}\int_{- \infty}^{+ \infty}{ \widetilde{f}(k) e^{i k x}}dx$.

I thought the following:

$F[\widetilde{f}(x)]=\int_{-\infty}^{+\infty}{\widetilde{f}(x)e^{-ikx}}dx=2 \pi \frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\widetilde{f}(x)e^{ix(-k)}}dx=2 \pi f(-k)$

But then $2 \pi$ would be at the numerator..
Have I done something wrong or can we not prove this in this way?

Then an other idea is:
$\widetilde{f}(x)=\int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{-ikx}}dk \Rightarrow F[\widetilde{f}(x)]=F[\int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{-ikx}}dk]=\int_{-\infty}^{+\infty}{\int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{-2ikx}}}dkdx$
How could we continue?

Or is there an other way to prove the property?? (Worried)

 

[I like Serena]

Hey!

Could you give me a hint how to prove the following property of the Fourier transform, when $F[f(x)]=\widetilde{f}(x)$, where $F[f(x)]$ is the Fourier transform of $f(x)$?

$$F[ \widetilde{f}(x) ]= \frac{f(-k)}{2 \pi}$$

We know that: $ \widetilde{f}(k)=\int_{- \infty}^{+ \infty}{ {f}(x) e^{-i k x}}dx$ and $f(x)=\frac{1}{2 \pi}\int_{- \infty}^{+ \infty}{ \widetilde{f}(k) e^{i k x}}dx$.

I thought the following:

$F[\widetilde{f}(x)]=\int_{-\infty}^{+\infty}{\widetilde{f}(x)e^{-ikx}}dx=2 \pi \frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\widetilde{f}(x)e^{ix(-k)}}dx=2 \pi f(-k)$

But then $2 \pi$ would be at the numerator..
Have I done something wrong or can we not prove this in this way?

Hi! :)

That is the correct way to prove it and your result is the right result.
None of the other common versions of the Fourier transform will give the result in your problem statement.
So it appears that your problem statement is wrong.

You can see the same result in this wiki table in the row labeled 105.
Your version of the Fourier transform is the last Fourier column in the table.

Then an other idea is:
$\widetilde{f}(x)=\int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{-ikx}}dk \Rightarrow F[\widetilde{f}(x)]=F[\int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{-ikx}}dk]=\int_{-\infty}^{+\infty}{\int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{-2ikx}}}dkdx$
How could we continue?

This is not correct.
It should be:
\begin{aligned}\widetilde{f}(x)&=\int_{-\infty}^{+\infty}{f(k')e^{-ik'x}}dk' \\
\Rightarrow F[\widetilde{f}(x)]
&=F[\int_{-\infty}^{+\infty} f(k')e^{-ik'x} dk'] \\
&=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(k')e^{-ik'x} e^{-ikx} dk'dx \\
&=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(k')e^{-i(k'+k)x} dk'dx
\end{aligned}

Anyway, I do not see how you can continue with that.

 

[mathmari]

Hi! :)

That is the correct way to prove it and your result is the right result.
None of the other common versions of the Fourier transform will give the result in your problem statement.
So it appears that your problem statement is wrong.

You can see the same result in this wiki table in the row labeled 105.
Your version of the Fourier transform is the last Fourier column in the table.

This is not correct.
It should be:
\begin{aligned}\widetilde{f}(x)&=\int_{-\infty}^{+\infty}{f(k')e^{-ik'x}}dk' \\
\Rightarrow F[\widetilde{f}(x)]
&=F[\int_{-\infty}^{+\infty} f(k')e^{-ik'x} dk'] \\
&=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(k')e^{-ik'x} e^{-ikx} dk'dx \\
&=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(k')e^{-i(k'+k)x} dk'dx
\end{aligned}

Anyway, I do not see how you can continue with that.

Ok! (Smile)
Thank you very much! (Smirk)

 

[I like Serena]

I realized that actually we can continue from here using the Dirac delta function:
\begin{aligned}
F[\widetilde{f}(x)]
&=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(k')e^{-i(k'+k)x} dk'dx \\
&=\int_{-\infty}^{+\infty} dk' f(k') \int_{-\infty}^{+\infty} e^{-i(k'+k)x} dx \\
&=\int_{-\infty}^{+\infty} dk' f(k')\ 2\pi\ \delta(k'+k) \\
&= 2\pi f(-k)
\end{aligned}

 

[mathmari]

I realized that actually we can continue from here using the Dirac delta function:
\begin{aligned}
F[\widetilde{f}(x)]
&=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(k')e^{-i(k'+k)x} dk'dx \\
&=\int_{-\infty}^{+\infty} dk' f(k') \int_{-\infty}^{+\infty} e^{-i(k'+k)x} dx \\
&=\int_{-\infty}^{+\infty} dk' f(k')\ 2\pi\ \delta(k'+k) \\
&= 2\pi f(-k)
\end{aligned}

Ok! (flower)
Thanks a lot! (Smirk)