Fourier Transform of Heaviside I need help with a Fourier Transform.
I know Fourier Transform is defined by:
$$F(\omega)=\int_{-\infty}^{\infty} f(t).e^{-i\omega t}\, dt$$
where $F(\omega)$ is the transform of $f(t)$.
Now, I need to calculate the Fourier Transform of:
$$u(t+\pi) - u(t-\pi)$$
where $u$ is the Heaviside function.
With that, I have to calculate this:
$$\frac{2}{\pi}\int_{0}^{\infty}\frac{\sin{(a\pi)}}{a} \cos{(at)} \,da$$
Any help?
 A: Split the second integral into two pieces,
$$
I(t)=\int_{0}^{\infty}\frac{\sin{\pi a}}{a}\cos(a t)da=\int_{0}^{\infty}\frac{1}{2a}\left(\sin(a(t+\pi))-\sin(a(t-\pi))\right)da 
$$
Due to the eveness of the integrand we get
$$
4 I(t)=\int_{-\infty}^{\infty}\frac{1}{a}\sin(a(t+\pi))da-\int_{-\infty}^{\infty}\frac{1}{a}\sin(a(t-\pi))da=\\\ \underbrace{\Im\int_{-\infty}^{\infty}\frac{1}{a}e^{ia(t+\pi)}da}_{I_1}-\underbrace{\Im\int_{-\infty}^{\infty}\frac{1}{a}e^{ia(t-\pi)}da}_{I_2}
$$
We can now apply residue theorem. There are two things we have to worry about:
-in which part of the Complex plane our integral converges
-How to avoid the singularity at $0$
We solve the second problem by adding a small semicircle at zero to avoid the divergence.  
Now, lets's take $t+\pi>0$ for the moment then we have to close the contour in the upper half plane to calculate $I_1$. The result is
$$
I_1= \pi i
$$
If $t+\pi<0$ we have to close in the lhp. we get
$$I_1=-\pi i$$
put together both cases yields
$$
I_1=\pi i \text{sign}(t+\pi)
$$
A similiar reasoning for $I_2$ gives
$$
I_2=i\pi\text{sign}(t-\pi)
$$
Collecting everything and taking imaginary parts completes our calculation
$$
I=\frac{\pi}{4}(\text{sign}(t+\pi)-\text{sign}(t-\pi))
$$
Feel free to ask, if anything is unclear or look at this question of you
A: The function $f(t)=u(t+\pi)-u(t-\pi)$ is $1$ for $-\pi < t < \pi$ and is zero outside $[-\pi,\pi]$. So the Fourier transform of this function is
$$
    \sqrt{2\pi}\hat{f}(s)=\left.\int_{-\pi}^{\pi}e^{-its}dt = \frac{e^{-its}}{-is}\right|_{t=-\pi}^{\pi}=
             \frac{e^{\pi is}-e^{-\pi is}}{is}=2\frac{\sin(\pi s)}{s}.
$$
Therefore, the inverse transform of the forward transform of $f$ is
$$
\begin{align}
      \frac{f(x+0)+f(x-0)}{2} & =\lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_{-R}^{R}2\frac{\sin(\pi  s)}{s}e^{ixs}ds \\
       & = \lim_{R\rightarrow\infty}\frac{1}{\pi}\int_{0}^{R}\frac{\sin(\pi s)}{s}(e^{ixs}+e^{-ixs})ds \\
       & = \lim_{R\rightarrow\infty}\frac{2}{\pi}\int_{0}^{R}\frac{\sin(\pi s)}{s}\cos(xs)ds \\
       & = \frac{2}{\pi}\int_{0}^{\infty}\frac{\sin(\pi s)}{s}\cos(xs)ds.
\end{align}
$$
A: Instead of adding a small semicircle, we can move the contour so that it avoids the singularity.
Since $\frac{\sin(a\pi)}a\cos(at)$ is even in $a$,
$$
\begin{align}
\int_0^\infty\frac{\sin(a\pi)}a\cos(at)\,\mathrm{d}a
&=\frac12\int_{-\infty}^\infty\frac{\sin(a\pi)}a\cos(at)\,\mathrm{d}a\\
&=\frac14\int_{-\infty}^\infty\frac1a\left[\vphantom{\frac12}\sin(a(\pi+t))+\sin(a(\pi-t))\right]\,\mathrm{d}a\tag{1}
\end{align}
$$
Since $\frac{\sin(a\lambda)}a$ is odd in $\lambda$, we can assume that $\lambda\gt0$ and account for sign later.
Let $\gamma^+=[-R-i,R-i]\cup-i+Re^{i\pi[0,1]}$ and $\gamma^-=[-R-i,R-i]\cup-i+Re^{i\pi[0,-1]}$ as $R\to\infty$, then
$$
\begin{align}
\frac14\int_{-\infty}^\infty\frac{\sin(a\lambda)}a\,\mathrm{d}a
&=\frac14\int_{-\infty-i}^{\infty-i}\frac{\sin(a\lambda)}a\,\mathrm{d}a\\
&=\frac1{8i}\int_{-\infty-i}^{\infty-i}\frac1a e^{ia\lambda}\,\mathrm{d}a
-\frac1{8i}\int_{-\infty-i}^{\infty-i}\frac1a e^{-ia\lambda}\,\mathrm{d}a\\
&=\frac1{8i}\int_{\gamma^+}\frac1z e^{iz\lambda}\,\mathrm{d}z
-\frac1{8i}\int_{\gamma^-}\frac1z e^{-iz\lambda}\,\mathrm{d}z\\
&=\frac1{8i}(2\pi i)-\frac1{8i}(0)\\
&=\frac\pi4\tag{2}
\end{align}
$$
Accounting for the sign of $\lambda$, we get
$$
\frac14\int_{-\infty}^\infty\frac{\sin(a\lambda)}a\,\mathrm{d}a
=\frac\pi4\mathrm{sgn}(\lambda)\tag{3}
$$
Using $(3)$ in $(1)$ gives
$$
\int_0^\infty\frac{\sin(a\pi)}a\cos(at)\,\mathrm{d}a
=\frac\pi4\left[\vphantom{\frac\pi4}\mathrm{sgn}(\pi+t)+\mathrm{sgn}(\pi-t)\right]\tag{4}
$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Note that the Heaviside function $\ds{\,{\rm u}\pars{t}}$ can be written as
$$
\,{\rm u}\pars{t}
=-\int_{-\infty}^{\infty}
\frac{\expo{-\ic\omega t}}{\omega + \ic 0^{+}}\,\frac{\dd\omega}{2\pi\ic}
$$
such that
\begin{align}&\,{\rm u}\pars{t + \pi} - \,{\rm u}\pars{t - \pi}
=\int_{-\infty}^{\infty}\frac{1}{2\pi\ic}
\frac{\expo{\ic\omega\pi} - \expo{-\ic\omega\pi}}{\omega + \ic 0^{+}}\,
\expo{-\ic\omega t}\,\dd\omega
=\int_{-\infty}^{\infty}\frac{1}{\pi}
\frac{\sin\pars{\pi\omega}}{\omega + \ic 0^{+}}\,
\expo{-\ic\omega t}\,\dd\omega
\\[5mm]&=\int_{-\infty}^{\infty}
\frac{\sin\pars{\pi\omega}}{\pi\omega}\,\expo{-\ic\omega t}\,\dd\omega
\\[1cm]&\frac{\sin\pars{\pi\omega}}{\pi\omega}\ \mbox{is the}\
{\it\mbox{Fourier Transform}}\ \mbox{of}\ 
\,{\rm u}\pars{t + \pi} - \,{\rm u}\pars{t - \pi}. 
\end{align}

Then,
\begin{align}&\color{#66f}{\large%
\frac{2}{\pi}\int_{0}^{\infty}\frac{\sin\pars{a\pi}}{a}\cos\pars{at}\,\dd a}
=\int_{-\infty}^{\infty}\frac{\sin\pars{\pi a}}{\pi a}\cos\pars{at}\,\dd a
=\Re\int_{-\infty}^{\infty}\frac{\sin\pars{\pi a}}{\pi a}\expo{-\ic at}\,\dd a
\\[5mm]&=\Re\int_{-\infty}^{\infty}\braces{%
\int_{-\infty}^{\infty}\bracks{\,{\rm u}\pars{x + \pi} - \,{\rm u}\pars{x - \pi}}
\expo{\ic ax}\,\frac{\dd x}{2\pi}}\expo{-\ic at}\,\dd a
\\[5mm]&=\Re\int_{-\infty}^{\infty}\bracks{%
\,{\rm u}\pars{x + \pi} - \,{\rm u}\pars{x - \pi}}\ \overbrace{%
\int_{-\infty}^{\infty}\expo{\ic\pars{x - t}a}\,\frac{\dd a}{2\pi}}
^{\dsc{\delta\pars{x - t}}}\ \,\dd x
=\color{#66f}{\large\,{\rm u}\pars{t + \pi} - \,{\rm u}\pars{t - \pi}}
\end{align}

However, $\ds{\,{\rm u}\pars{x} = \frac{\,{\rm sgn}\pars{x} + 1 }{2}}$
such that
\begin{align}&\color{#66f}{\large%
\frac{2}{\pi}\int_{0}^{\infty}\frac{\sin\pars{a\pi}}{a}\cos\pars{at}\,\dd a}
=\color{#66f}{\large%
\frac{\,{\rm sgn}\pars{t + \pi} - \,{\rm sgn}\pars{t - \pi}}{2}}
\end{align}


Note that Wikipedia
  use the symbol $\ds{\,{\rm H}}$ for the Heaviside Step Function while the
  current use is $\ds{\Theta}$. Even $\ds{\tt Mathematica}$ calls it
  $\ds{\tt HeavisideTheta}$ and $\ds{\Theta\pars{x}}$ in its documentation.

