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Hopefully this a short question.

$u(t)$ is the Unit step function or the Heaviside function. $a$>0

Solving using the basic definition, $$F[x(t)]=\int_{-\infty}^{\infty}e^{-at}u(t)e^{-iwt}$$ we get $$F[x(t)]=\frac{1}{iw+a}$$ If we were to start from the Fourier transform of $u(t)$ $$F[x(t)]=\frac{1}{iw}+\pi\delta(w)$$ and use the frequency shifting property of $F[e^{-at}x(t)]=X(iw+a)$, ($X(iw)$ is the FT of $x(t)$)

we get $$F[x(t)]=\frac{1}{iw+a}+\pi\delta(w+a)$$

I first assumed the discrepancy arose due to unwillingness to deal with delta functions in a common transform. but that does not really explain why i haven't seen it literally anywhere.

If there is a alternative explanation, please share.

Animesh Shukla
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  • The frequency shift property involves a complex exponential, not an exponential with a real exponent. Your claim of $F[e^{at}x(t)]=X(iw-a)$ is an incorrect statement and not the frequency shift property.

    Also this integral $F[x(t)]=\int_{-\infty}^{\infty}e^{at}u(t)e^{-iwt} dt$ does not exist for $a > 0$, so the transform doesn't exist as stated after that.

    Please carefully examine your question and correct the errors.

     – Andy Walls Nov 21 '24 at 04:01
  • @AndyWalls you are correct about $a$. I have changed changed it to -a and specified $a>0$. As for the frequency shifting property, all real numbers are complex numbers with imaginary part zero. If we check this property on Wikipedia, we can just write $e^{-at}$ as $e^{i(ia)t}$ and put that in that formula to arrive at what i have stated in my question. – Animesh Shukla Nov 21 '24 at 07:30

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