fourier analysis
$$f(t)= \begin{cases}
e^{-t} & 0<t<1 \\
0 & \text{otherwise}
\end{cases}$$
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}
I write the Fourier transform as
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: \frac{\sin{x}}{x} e^{i k x} $$
Consider, rather, the integral
$$ \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{i x}-e^{-i x}}{x} e^{i k x} $$
$$ = \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} - \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{-i (1-k) x}}{x} $$
Consider the following integral corresponding to the first integral:
$$\oint_C dz \: \frac{e^{i (1+k) z}}{z} $$
where $C$ is the contour defined in the illustration below:
This integral is zero because there are no poles contained within the contour. Write the integral over the various pieces of the contour:
$$\int_{C_R} dz \: \frac{e^{i (1+k)z}}{z} + \int_{C_r} dz \: \frac{e^{i (1+k) z}}{z} + \int_{-R}^{-r} dx \: \frac{e^{i (1+k) x}}{x} + \int_{r}^{R} dx \: \frac{e^{i (1+k) x}}{x} $$
Consider the first part of this integral about $C_R$, the large semicircle of radius $R$:
$$\int_{C_R} dz \: \frac{e^{i (1+k)z}}{z} = i \int_0^{\pi} d \theta e^{i (1+k) R (\cos{\theta} + i \sin{\theta})} $$
$$ = i \int_0^{\pi} d \theta e^{i (1+k) R \cos{\theta}} e^{-(1+k) R \sin{\theta}} $$
By Jordan's lemma, this integral vanishes as $R \rightarrow \infty$ when $1+k > 0$. On the other hand,
$$ \int_{C_r} dz \: \frac{e^{i (1+k) z}}{z} = i \int_{\pi}^0 d \phi \: e^{i (1+k) r e^{i \phi}} $$
This integral takes the value $-i \pi$ as $r \rightarrow 0$. We may then say that
$$\begin{align} & \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} = i \pi & 1+k > 0\\ \end{align}$$
When $1+k < 0$, Jordan's lemma does not apply, and we need to use another contour. A contour for which Jordan's lemma does apply is one flipped about the $\Re{z}=x$ axis. By using similar steps as above, it is straightforward to show that
$$\begin{align} & \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} = -i \pi & 1+k < 0\\ \end{align}$$
Using a similar analysis as above, we find that
$$\int_{-\infty}^{\infty} dx \: \frac{e^{-i (1-k) x}}{x} = \begin{cases} -i \pi & 1-k < 0 \\ i \pi & 1-k >0 \\ \end{cases} $$
We may now say that
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: \frac{\sin{x}}{x} e^{i k x} = \begin{cases} \pi & |k| < 1 \\ 0 & |k| > 1 \\ \end{cases} $$
To translate to your definition of the FT, divide the RHS by $\sqrt{2 \pi}$.
$$\hat{f}(k)=\int e^{-2\pi i xk}e^{-|x|}\operatorname{sign}(x)dx=\int_0^\infty e^{-2\pi i xk-x}dx-\int_{-\infty}^0 e^{-2\pi i kx+x}dx=\int_0^\infty e^{-(2\pi k i+1)x}dx -\int_{-\infty}^0 e^{-(2\pi i k-1)x}dx.$$
Can you go on?
Now
$$\int_0^\infty e^{-(2\pi k i+1)x}dx=\frac{1}{2\pi k i+1}, $$
as $|e^{-(2\pi k i+1)x}|=e^{-x}$ and $x\geq 0$, and
$$\int_{-\infty}^0 e^{-(2\pi i k-1)x}dx=-\frac{1}{2\pi k i-1},$$
as $|e^{-(2\pi k i-1)x}|=e^{x}$ and $x\leq 0$. In summary
$$\hat{f}(k)=\frac{1}{2\pi k i+1}+\frac{1}{2\pi k i-1}=-\frac{4\pi ki}{4\pi^2 k^2 +1}$$
Best Answer
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?