# Fourier Transform of $\exp(-t)$

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

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}

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

\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}
Using the definiontion of f, we can plug it in the usual Fourier Transform formula $$\hat{f(\xi)}=\frac{1}{2\pi}\int_{-\infty}^{\infty}{f(t)e^{-2\pi i\xi t}dt}=\frac{1}{2\pi}\int_{0}^{1}{e^{-t(1+2\pi i \xi)}dt}=\frac{1}{2\pi}\frac{1-e}{1+2\pi i \xi}$$
• Isn't $\frac{1}{2\pi}$ a part of Inverse Fourier Transform? – Pyro Nov 26 '14 at 13:24