How to find this Fourier Transform without the use of integration? If I have $\mathcal{F}(f(t))=F(\omega)$ with $$f(t)= \begin{cases} 1, & 0 \leq t < \pi\\ -1,&- \pi \leq t <0 \\ 0 & \text{otherwise}\end{cases}$$
I have found 
$$F(\omega)= \begin{cases} \frac{2i}{\omega}(\cos(\omega \pi)-1) & \omega \neq 0 \\ 0 & \omega = 0 \end{cases}$$
I do not know how to find a $g(t)=f(t)\sin(\alpha t)$ without the use of integration.
Any help would be appreciated. Thanks in advance.
 A: General Fourier Transform, by formal definition:
$$
\mathcal{F}_{(a,b)}(f(t)) = \sqrt{\dfrac{|b|}{(2\pi)^{1-a}}} \int\limits_{-\infty}^{+\infty} f(t)e^{i b \omega t} dt
$$
As generally known, the pair of values $(a,b)$ are chosen depending on the context of use of the Fourier transform:


*

*$(0,+1)$: Default; Modern Physics.

*$(+1,-1)$: Pure Mathematics; Systems Engineering

*$(-1,+1)$: Classical Physics

*$(0,-2\pi)$: Signal Processing
I will assume you are using the Fourier transform in a context of pure mathematics or systems engineering. Thus:
$$
\mathcal{F}_{(+1,-1)}(f(t)) = \mathcal{F}(f(t)) = F(\omega) \quad\Rightarrow\quad F(\omega)= \int\limits_{-\infty}^{+\infty} f(t)e^{-i \omega t} dt
$$

Since the Fourier transform is defined by an integral, it's impossible to evaluate the fourier transform without the use of integration. Instead of it, you can build a table of Fourier transforms to solve your problem, with the appropriate functions. The question now is...
What are the appropriate functions to build a table of Fourier transforms useful?
You should choose your table of transforms according to the most common functions in your work context. It's something intuitive. There isn't much science in it.
The real trick is to build that table of transforms, multiplying each function by the unit step function. The reason for doing this is very interesting: you can build any function at defined intervals with known functions, with the combination of those known functions multiplied by the unit step function (shifted in time and/or multiplied by a factor, if necessary). 
Note: I will use the following notation to simplify fourier transform.
$$
\mathcal{F}(f(t)) = F(\omega) \quad \Leftrightarrow\quad f(t) \xrightarrow{\mathcal{F}} F(\omega)
$$
And remember these properties of the Fourier transform:
$$
\begin{align}
\mbox{Shifted i}&\mbox{n Time} & f(t\pm a) &\xrightarrow{\mathcal{F}} F(\omega) e^{\pm ia\omega} \\
\mbox{Superposition}&\mbox{ Principle} & af_1(t) + bf_2(t)&\xrightarrow{\mathcal{F}} aF_1(\omega) + bF_2(\omega)
\end{align}
$$

First example: with unit step function, where $\delta(\omega)$ is the Dirac delta function (proof here):
$$
u(t) = \left\{ \begin{matrix} 1, & t > 0 \\ 0, & t < 0 \end{matrix} \right. \quad \Rightarrow \quad u(t) \xrightarrow{\mathcal{F}} \dfrac{1}{i\omega} + \pi\delta(\omega) = U(\omega)
$$
Defining $f_1(t)$ using the unit step function, sweeping from left to right:
$$
f_1(t) = \left\{ \begin{matrix} +1, & 0 < t < +\pi \\ -1, & -\pi < t < 0, \\ 0 & \mbox{otherwise} \end{matrix} \right. \quad \Rightarrow \quad f_1(t) = -u(t+\pi)  +2u(t) -u(t-\pi)
$$
Applying Fourier transform and its properties (and exp form of $\cos(\pi\omega)$):
$$
\begin{align}
f_1(t) &\xrightarrow{\mathcal{F}} -U(\omega)e^{i\pi\omega} + 2 U(\omega) - U(\omega)e^{-i\pi\omega} \\
f_1(t) &\xrightarrow{\mathcal{F}} - \left( \dfrac{1}{i\omega} + \pi\delta(\omega)\right)e^{i\pi\omega} + 2 \left(\dfrac{1}{i\omega} + \pi\delta(\omega)\right) - \left( \dfrac{1}{i\omega} + \pi\delta(\omega)\right)e^{-i\pi\omega} \\
f_1(t) &\xrightarrow{\mathcal{F}} \dfrac{2i}{\omega} \left(\cos(\pi\omega) - 1\right) \\
\end{align}
$$
Since:
$$
\lim\limits_{\omega\to 0} 2\pi\delta(\omega) -\left(\pi\delta(\omega)e^{i\pi\omega} + \pi\delta(\omega)e^{-i\pi\omega} \right) = \lim\limits_{\omega\to 0} 2\pi\delta(\omega)\left( 1-\cos(\pi\omega) \right) = 0\\
\therefore\quad \forall\omega\in\mathbb{R}:\quad 2\pi\delta(\omega) -\left(\pi\delta(\omega)e^{i\pi\omega} + \pi\delta(\omega)e^{-i\pi\omega} \right) = 0
$$

Second example: integrate and include in your table of Fourier transform:
$$
\begin{align}
\forall a\in\mathbb{R}:&
&
e^{-a t}u(t) &\xrightarrow{\mathcal{F}} \dfrac{1}{a+\omega i} = E_{(a)}(\omega)
& 
e^{-ia t}u(t) &\xrightarrow{\mathcal{F}} \dfrac{1}{ai+\omega i} + \pi\delta(\omega + a) = E_{ia}(\omega)
& &
\end{align} 
$$
$$
\therefore \quad E_{ia}(\omega) = E_{(ia)}(\omega) + \pi\delta(\omega + a)
$$
Defining $f_2(t)$:
$$
f_2(t) = f_1(t)\sin(\alpha t) = -u(t+\pi)\sin(\alpha t)  +2u(t)\sin(\alpha t) -u(t-\pi)\sin(\alpha t)
$$
Where it's convenient to define (using exp form of $\sin(\alpha t)$):
$$
\begin{align} 
g(t) &= u(t+a)\sin(\alpha t) \\
&= \dfrac{i}{2}e^{-i\alpha t}u(t+a) - \dfrac{i}{2}e^{i\alpha t}u(t+a) \\
\therefore\quad g(t-a)&= \dfrac{i}{2}e^{-i\alpha t}u(t) - \dfrac{i}{2}e^{i\alpha t}u(t) \\
&= \dfrac{ie^{ia \alpha}}{2}e^{-i\alpha t}u(t) - \dfrac{ie^{-ia \alpha}}{2}e^{i\alpha t}u(t) \\
\end{align} 
$$
Applying Fourier transform and its properties:
$$
\begin{align} 
g(t-a) &\xrightarrow{\mathcal{F}} \dfrac{i}{2}e^{ia \alpha}E_{+ia}(\omega) - \dfrac{i}{2}e^{-ia \alpha}E_{-ia}(\omega) \\
\therefore \quad g(t) &\xrightarrow{\mathcal{F}} \dfrac{i}{2}e^{ia (\alpha+\omega)}E_{+ia}(\omega) - \dfrac{i}{2}e^{-ia (\alpha-\omega)}E_{-ia}(\omega) = G_{a}(\omega)
\end{align}
$$
Thus:
$$
u(t+a)\sin(\alpha t) \xrightarrow{\mathcal{F}} G_{a}(\omega)
$$
And finally:
$$
\begin{align}
f_2(t) &\xrightarrow{\mathcal{F}} -G_{-\pi}(\omega) + 2G_{0}(\omega) - G_{+\pi}(\omega) \\
f_2(t) &\xrightarrow{\mathcal{F}} \dfrac{2\alpha}{\alpha^2 - \omega^2} - \dfrac{2\alpha}{\alpha^2 - \omega^2} \cos(\pi\alpha)\cos(\pi\omega) - \dfrac{2\omega}{\alpha^2 - \omega^2} \sin(\pi\alpha)\sin(\pi\omega)
\end{align}
$$
I leave to you the verification of this solution as homework (use angle sum and difference identities)
A: You can consider the integrals against the $\sin$ and $\cos$ terms separately. Those with $\cos$ vanish because your function is odd. The coefficient of the first $\sin$ term is
$$
          2\frac{1}{\pi}\int_{0}^{\pi}\sin(x)dx
$$
The only terms that survive are those involving
$$
      \sin(x),\sin(3x),\sin(5x),\sin(7x),\cdots
$$
You end up with
$$
       2\frac{1}{\pi}\int_{0}^{\pi}\sin((2n+1)x)dx
        = 2\frac{1}{\pi}\int_{0}^{\pi/(2n+1)}\sin((2n+1)x)dx
$$
The integral over $\sin((2n+1)x$ over $[0,\pi]$ cancels except for $\int_{0}^{\pi/(2n+1)}\sin((2n+1)x)dx$ which is the same as the area under $\sin(x)$ on $[0,\pi]$, but scaled by $\frac{1}{2n+1}$. So the above equals
$$
          \frac{1}{2n+1}2\frac{1}{\pi}\int_{0}^{\pi}\sin(x)dx.
$$
So the Fourier series is
$$
      \left[\sum_{k=0}^{\infty}\frac{1}{2k+1}\sin((2k+1)x)\right]\frac{2}{\pi}\int_{0}^{\pi}\sin(x')dx'.
$$
A: Since
$$\mathcal{F} \left( \sin (\alpha t) \right) = i \pi \left( \delta (\omega + \alpha) - \delta (\omega - \alpha) \right)$$
and multiplication in the time domain is equivalent to convolution in the frequency domain, we have
$$G (\omega) = i \pi F (\omega + \alpha) - i \pi F (\omega - \alpha)$$
