1
$\begingroup$

If F(w) is the Fourier transform of f(x), show that F(aw) is the Fourier transform of (1/a)f(x/a).

So if I apply a fourier transform to (1/a)f(x/a):

$$ \frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{a} f(\frac{x}{a}) e^{iwx} dx$$

i'm lost in how to get F(aw) from this

$\endgroup$
  • $\begingroup$ Try using the substitution: $u=\frac{x}{a}$ $\endgroup$ – Kitegi Apr 10 '15 at 18:39
0
$\begingroup$

You may just perform the change of variable $u:=\dfrac x a$, $dx=adu$, to get $$ \frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{a}f(\frac{x}{a}) e^{iwx} dx= \frac{1}{2\pi}\int_{-\infty}^\infty f(u) e^{iawu} du=F(aw). $$

$\endgroup$
  • $\begingroup$ so to solidify that idea, is it safe to say that everything that is not the variable of intergration or "i" in the exponent can be brought into the expression of the function. ex: $e^{\frac{iwx}{a}}$ if I was integrating in terms of w then my function would be $f(\frac{x}{a}) $ ? $\endgroup$ – dc3rd Apr 10 '15 at 18:52
  • $\begingroup$ @dc3rd Exactly. Thanks! $\endgroup$ – Olivier Oloa Apr 10 '15 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.