Find FT of Function: Solution & Explanation

[LCSphysicist]

Homework Statement

I will post a print

Relevant Equations

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I need to find the FT of this function. Here is my attempt:

$$H(f) = \int_{0}^{\infty} ke^{-2 \pi i f t}dt$$

We know that $\delta(t) = \int_{-\infty}^{\infty} e^{2 \pi i f t} df$, the part with sin in this integration vanish, so that, and knowing that cos is a even function, we can write $\delta(t) = \int_{-\infty}^{\infty} e^{2 \pi i f t} df = \int_{-\infty}^{\infty} e^{-2 \pi i f t} df$.

Now, for example, $\int_{-a}^{a}x^2 = 2\int_{0}^{a}x^2$. So we can write the delta above as $\delta(t) = \int_{\infty}^{\infty} e^{-2 \pi i f t} df = 2\int_{0}^{\infty} e^{-2 \pi i f t} df$.

Putting this in the first formula $$H(f) = \int_{0}^{\infty} ke^{-2 \pi i f t}dt$$:

$H(f) = \int_{0}^{\infty} ke^{-2 \pi i f t} dt = k \delta(f) /2$

This make sense or you? Is it right?

 

[BvU]

Is it right?

One possible check is to do the inverse FT. That would give you a constant function, so: no !
Something got lost on the way...

$\$

 

[Orodruin]

Now, for example, $\int_{-a}^{a}x^2 = 2\int_{0}^{a}x^2$. So we can write the delta above as $\delta(t) = \int_{\infty}^{\infty} e^{-2 \pi i f t} df = 2\int_{0}^{\infty} e^{-2 \pi i f t} df$.

Is
$$
\int_{-a}^a x\, dx = 2 \int_0^a x\, dx?
$$

 

[Office_Shredder]

I think your confusion is they $\delta(t)$ is a thing that needs to be integrated again in order to get the function value. So

$$\int_{-\infty}^{\infty}g(x) \int_{-\infty}^{\infty} e^{2\pi i x t} dt dx= g(0)$$

But
$$\int_{-\infty}^{\infty}g(x) e^{2\pi i x t} dx \neq g(0)$$

This latter integral is just a multiple of the inverse Fourier transform of g.