4
$\begingroup$

I need to use the Fourier Transform to confirm that this result is true:

$$\int_0^\infty2te^{-at}*\cos(t)\,dt=2(a^2-1)/((a^2-1)^2+4a^2)$$

How can I do so? I'm using fourier transform identities such as $x(t)*\cos(w_0t)$ where $x(t)=2*t*e^{-at}$, but the end result contains terms of $\omega$. What am I supposed to do with omega and how do I get rid of it to make it look like that?

  • 1
    I assume the integration is with respect to $t$, where does the $r$ come from? Please consider formatting your question to make it more clear. – 2017-02-19
  • 0
    @Alex I've edited – 2017-02-19
  • 0
    Since you did not accept nor upvoted my answer (which very clearly made use of the Fourier Transform and its properties), I realize that it wasn't helpful to you so I deleted it – 2017-02-21

1 Answers 1

2

Hint. By using a Laplace Transform table one has

$$ \mathscr{L}(\cos t) = \int_0^\infty e^{-at}\cdot\cos(t)\,dt=\dfrac{a}{a^2 + 1} \tag 1$$

differentiating both sides of $(1)$ with respect to $a$ yields

$$\mathscr{L}(t \cos t) = \int_0^\infty t\:e^{-at}\cdot\cos(t)\,dt=\dfrac{a^2-1}{(a^2 + 1)^2} \tag 2$$

which gives the announced result since $(a^2+1)^2=(a^2-1)^2+4a^2$.

  • 0
    Thank you, but I'm looking for a way in fourier transform – 2017-02-19
  • 0
    @Goldname This is a way with Fourier transform since here $\mathscr{L}(t \cos t) =(\mathscr{L}( \cos t) )'$ and in general $\mathscr{L}(t f(t)) =(\mathscr{L}(f(t)) )'$ – 2017-02-19
  • 0
    But why is it when I try to take the fourier transform, it results in omegas? I can't get rid of them. – 2017-02-19
  • 0
    @Goldname Where are your calculations? I'm going to sleep in 5 minutes ;) – 2017-02-19
  • 0
    Well F(2*t*e^(-a*t)) = $2/(a+jw)^2$ so you can see that an omega appears – 2017-02-19