Fourier transform of t, 1/t and t^n

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SUMMARY

The Fourier transforms of the functions t, 1/t, and t^n can be derived using distribution theory, as these functions do not meet the integrability conditions required for standard Fourier transforms. The Fourier transform of t is represented as the derivative of the delta function, specifically √(2π)δ'(ω), while the transform of 1/t is proportional to the sign function, sgn(ω). This approach utilizes the definition of the Fourier transform, given by the integral involving e^(-iωt) and the function f(t).

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I would like to know how one finds the Fourier transforms of

t,

\frac{1}{t}

and

{t}^{n}

with the definition of the Fourier transform as

\mathscr{F}\{f(t)\}=\mathcal{F}\{f(t)\}=\frac{1}{ \sqrt{2\pi} }\int\limits_{-\infty}^{\infty}{e}^{-i\omega t}f(t)\mbox{d}t

I have tried the definition of a Fourier transform and I got some weird limits. Laplace transforms are so much easier!

Thanks in advance.
 
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In the usual definition of Fourier transform, f(t) is usually presumed to be integrable, or square integrable. None of your functions satisfy this requirement.
 
The functions do, however, have Fourier transforms in terms of distributions. Consider

\sqrt{2\pi}\delta(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty dt~e^{i\omega t}.

Now, take a derivative of both sides with respect to the frequency:

\sqrt{2\pi}\delta'(\omega) = \frac{i}{\sqrt{2\pi}} \int_{-\infty}^\infty dt~t e^{i\omega t}.

You can keep taking derivatives to get the Fourier transform of tn. For 1/t, the Fourier transform will be proportional to the \mbox{sgn}(\omega) function, where sgn(x) returns the sign of x.
 

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