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I want to find the inverse Fourier transforms of:

$$u(\nu + 1) \ \exp(-\nu)$$

Attempt:

So the inverse Fourier transform is given by:

$$\int^\infty_{-\infty} u(\nu + 1) \ e^{-\nu} e^{j2 \pi t} \ d\nu = \int^\infty_{-\infty} u(\nu + 1) \ e^{\nu (j2 \pi t -1)} \ d\nu$$

I know that $FT \Big[ u(t) \Big]= \frac{1}{2} \left( \delta(\nu) - \frac{j}{\pi \nu} \right).$ So do we need to use this property or somehow proceed with direct integration? Also what can I do about the $+1$ in the argument of the Heaviside unit step function?

Any explanation on how to solve this is greatly appreciated.

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  • $\begingroup$ The plus one simply shifts when the Heaviside function turns on by one unit to the left like in normal functional translation, it helps to use the definition of the Heaviside step function as it restricts your domain of integration. That is it is zero for a large part of this integration domain. $\endgroup$ – Triatticus Mar 27 '16 at 15:09
  • $\begingroup$ I used integration by parts to get: $\Big[ u(\nu +1) \frac{e^{\nu (j 2 \pi t -1)}}{(j 2 \pi t -1)} \Big]^{\infty}_{-\infty} - \int^{\infty}_{-\infty} \frac{e^{\nu(j 2 \pi t -1)}}{ j 2 \pi t -1} . \delta (\nu +1) \ d \nu.$ Using the sifting property of the delta function for the second part the expression becomes: $\frac{1}{ j 2 \pi t -1} - \frac{e^{-(j 2 \pi t -1)}}{ j 2 \pi t -1}$. Is this correct? And is it possible to simplify this further? $\endgroup$ – Merin Mar 28 '16 at 3:22
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    $\begingroup$ You don't need integration by parts, the Heaviside function is 1 when the argument is larger than or equal to zero ie that $\nu \geq -1$, hence you can integrate from -1 to infinity against the function 1. Alternatively the Heaviside function is zero when the argument is less than zero, so the rest of the integration reason is unnecessary $\endgroup$ – Triatticus Mar 28 '16 at 3:37
  • $\begingroup$ Thank you very much for the hint. This is what I got: $\int^{\infty}_{-1} e^{\nu(j 2 \pi t -1)} d \nu = \frac{1}{j 2 \pi t -1} \Big[ e^{\nu(j 2 \pi t -1)} \Big]^{\infty}_{-1} = \frac{- e^{-\nu(j 2 \pi t -1)}}{j 2 \pi t -1}.$ Is that the right idea now? $\endgroup$ – Merin Mar 28 '16 at 4:48
  • $\begingroup$ Ill just write up the rest of the hint as an answer, apologies for the lateness of my responses as I have been on mobile all day. $\endgroup$ – Triatticus Mar 28 '16 at 6:54
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We have most of the problem done through comments, just some minor tweaks:

It is easier to see the behavior of the result if we instead represent the integral as

$$ \int_{-\infty}^{\infty} u(\nu + 1) e^{- \nu} e^{2 \pi j \nu t} \mathrm{d}\nu = \int_{-\infty}^{\infty} u(\nu + 1) e^{- \nu (1 - 2 \pi j t)} \mathrm{d}\nu $$ As I mentioned before, the heaviside step function turns on to a value of $1$ after its argument is larger than or equal to zero, this means $\nu \geq -1$ otherwise the integral is zero thus: $$ \int_{-\infty}^{\infty} u(\nu + 1) e^{- \nu (1 - 2 \pi j t)} \mathrm{d}\nu = \int_{-1}^{\infty} e^{- \nu (1 - 2 \pi j t)} \mathrm{d}\nu = \left.\frac{-1}{1 - 2 \pi j t}e^{- \nu (1 - 2 \pi j t)} \right|^{\infty}_{-1} $$ We see that at infinity this exponential term goes to zero so we are left with: $$ -\left( \frac{-1}{1-2 \pi j t} e^{1 - 2 \pi j t} \right) \quad \text{ or } \quad \frac{1}{1-2 \pi j t} e^{1 - 2 \pi j t} $$

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  • $\begingroup$ Thank you very much for the explanation, it makes perfect sense now. $\endgroup$ – Merin Mar 28 '16 at 10:35
  • $\begingroup$ Sure no problem! $\endgroup$ – Triatticus Mar 28 '16 at 12:34

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