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$$f(t)= \begin{cases} e^{-t} & 0<t<1 \\ 0 & \text{otherwise} \end{cases}$$

How can I solve this function's Fourier transform?

I am stuck at here:

Daniel R - OP \begin{align} F(k)&=\int_{-\infty}^{\infty}f(t)\exp(-2\pi itk)\;\mathrm dt \\ &=\int_{0}^{1}\exp(-t)\exp(-2\pi itk)\;\mathrm dt \\ &=\int_{0}^{1}\exp(-2\pi itk-t)\;\mathrm dt \\ &=\int_{0}^{1}\exp((-2\pi ik-1)t)\;\mathrm dt \\ \end{align}

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    $\begingroup$ Hint: by definition. By the way, are $x$ and $t$ related? $\endgroup$ – TZakrevskiy Nov 26 '14 at 11:26
  • $\begingroup$ I've just realised that i wrote the equation wrongly.Sorry about that. $\endgroup$ – Pyro Nov 26 '14 at 11:28
  • $\begingroup$ I am new to TeX commands. So I could not write where I am stuck at. Therefore I could not identified my problem. $\endgroup$ – Pyro Nov 26 '14 at 13:21
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Applying the defintion of Fourier transform gives you

$$\begin{align} F(k)&=\int_{-\infty}^{\infty}f(t)\exp(-2\pi itk)\;\mathrm dt \\ &=\int_{0}^{1}\exp(-t)\exp(-2\pi itk)\;\mathrm dt \\ &=\int_{0}^{1}\exp(-2\pi itk-t)\;\mathrm dt \\ &=\int_{0}^{1}\exp((-2\pi ik-1)t)\;\mathrm dt \\ \end{align}$$

Can you take it from here?

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Using the definiontion of f, we can plug it in the usual Fourier Transform formula $$\hat{f(\xi)}=\frac{1}{2\pi}\int_{-\infty}^{\infty}{f(t)e^{-2\pi i\xi t}dt}=\frac{1}{2\pi}\int_{0}^{1}{e^{-t(1+2\pi i \xi)}dt}=\frac{1}{2\pi}\frac{1-e}{1+2\pi i \xi}$$

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  • $\begingroup$ Isn't $\frac{1}{2\pi}$ a part of Inverse Fourier Transform? $\endgroup$ – Pyro Nov 26 '14 at 13:24

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