What is the theory of distribution that makes possible to calculate Fourier transform of the Sine function? I am an Engineering student.
Sine wave function is a power signal present from $-\infty$ to $+\infty$.
I have have read that because of distribution theory,the Fourier transform of the Sine function  is possible. But what is distribution theory behind it?
Note : Please try to understand that I am talking about function that tends from $-\infty$ to $+\infty$ , not 0 to $+\infty$
 A: Ok let me see if I can't answer your question from a mathematical perspective in a way that might be helpful for an engineer. The first thing you need to understand is that $\varphi_{n}(x)=\frac{1}{\sqrt{2\pi}}e^{ikx} , \, k \in \mathbb{Z}$ forms an orthonormal sequence in $L^{2}([-\pi,\pi]))$. Now, anytime you take the inner product of two orthonormal functions such as 
$$
\langle \varphi_{n},\varphi_{m}\rangle=\int^{\pi}_{-\pi} e^{i(m-n)x} \, dx=\delta_{mn}=\begin{cases} 0 & \text{if } m \neq n \\1 & \text{if }m=n \end{cases}
$$
for example with $n \neq m$
$$
\langle\varphi_{n},\varphi_{m}\rangle=\frac{1}{2\pi}\int^{\pi}_{-\pi}e^{i(m-n)x} \, dx
=0$$
and 
$$
\langle \varphi_{n},\varphi_{n}\rangle=\frac{1}{2\pi}\int^{\pi}_{-\pi}e^{i(n-n)x}\, dx=1
$$
Now IMO, the Fourier transform is best understood as a linear operator $F:L^{2}(\mathbb{R}) \to L^{2}(\mathbb{R})$ given by
$$
Ff=\int^{\infty}_{-\infty} f(x)e^{-i 2 \pi \omega x} dx
$$
which more generally belongs to the class of integral transforms of the form
$$
Tf=\int^{\infty}_{-\infty}K(x,\omega)f(x) dx
$$
Hopefully by now you can see that the Fourier transform looks a lot like an inner product of an orthonormal basis function in $L^{2}(\mathbb{R})$ and a function $f \in L^{2}(\mathbb{R})$. For an engineer, you might interpret the inner product as a sort of correlation between the given function and the basis function. However, note that the Fourier transform gives does not give us a number like the inner product but it does give us a function $\hat{f}(\omega)$ and for any $\omega_{0} \in \mathbb{R}$, we have 
$$
\hat{f}(\omega_{0})=\lambda_{0}
$$
and this value $\lambda_{0}$ scales our basis function $e^{i 2\pi \omega_{0} x}$ so that we have $\lambda_{0} e^{-i 2 \pi \omega_{0} x}$ so that we have 
$$
\lambda_{0}e^{ik \omega_{0}x}=\lambda_{0} \cos(\omega_{0} x)+i\lambda_{0} \sin(\omega_{0}x)
$$
which tells us the amplitude (which can be interpreted as the contribution) of our basis function at a given frequency. Now, if we only had countably many values of $\lambda_{0}$, we could write our original function $f$ as 
$$
f=\sum^{\infty}_{k=-\infty} \lambda_{k}e^{-i 2 \pi \omega_{k}x}
$$
where 
$$\lambda_{k}=\int^{\infty}_{-\infty} f(x)e^{i 2\pi \omega_{k}} \, dx$$
but since $\omega_{0} \in \mathbb{R}$ for most functions (think about how many terms you'd need to write a constant function or any non-periodic function in terms of sines and cosines) then we have that
$$
f(x)=\int^{\infty}_{-\infty} \hat{f}(\omega) e^{i 2 \pi \omega x} \, dx
$$
So now to finally answer your question, the connection between distribution theory and the Fourier transform of a sine function has to do with something called the uncertainty principle or Gabor limit. This just means that there is a tradeoff between time localization and frequency localization. This is to say that the more that one zooms in to a specific value of time $f(t_{0})$, the more spread out the the values of $\hat{f}(\omega)$. This is formalized in the following statement where $D[f]=\int^{\infty}_{\infty} x^{2} f(x) \, dx$
$$
D[f]D[\hat{f}]=\int^{\infty}_{\infty} \lvert x^{2} f(x)\rvert^{2} \, dx \int^{\infty}_{\infty} \lvert \omega^{2} \hat{f}(\omega) \rvert^{2} \, d\omega \geq \frac{\lvert \lvert f \rvert \rvert^{4}_{2}}{16 \pi^{2}}$$
where it should be fairly obvious that $D[f]$ is a linear functional that measures the "spread" or "variance" of $f$. Now, imagine that one wants to take the Fourier transform of a distribution $\delta: C^{\infty}_{c} \to \mathbb{R}$ which is clearly a linear functional defined by 
$$
\delta[\varphi]=\varphi[0]
$$
for every $\varphi \in C^{\infty}_{c}$. Now we often define $\delta$ as a distributional derivative of the heaviside function 
$$
H(x)=\begin{cases}0 & \text{if } x<0 \\ 1 & \text{if } x\geq 0 \end{cases}
$$
such that 
$$
\delta[\varphi]=-\int^{\infty}_{-\infty} \varphi'(x) H(x) \, dx
$$
for every $\varphi \in C^{\infty}_{c}$
and then you end up with a Lebesgue-Steiltjes integral 
$$
-\int^{\infty}_{-\infty} \varphi'(x) H(x) \, dx=\int^{\infty}_{-\infty} \varphi(x) \, dH(x)
$$
and I personally prefer this definition so we write
$$
\delta[\varphi]=\int^{\infty}_{-\infty} \varphi(x)\, dH(x)
$$
for all $\varphi \in C^{\infty}_{c}$ which then gives rise to the heuristic definition
$$
\delta(x)=\begin{cases} 0 & \text{if } x \neq 0 \\ \infty & \text{if } x=0 \end{cases}
$$
where $\delta(x)$ is also subject to the constraint that 
$$
\int^{\infty}_{-\infty} \delta(x) \, dx=1
$$
and if we try to take the Fourier transform of the tempered distribution $\delta[\varphi]$, we get
$$
\hat{\delta}(\omega)=\int^{\infty}_{-\infty} \delta(x) e^{-i 2 \pi \omega x} \, dx=1
$$
Now note that that 
$$
\delta(x)=\int^{\infty}_{-\infty} 1(e^{i 2 \pi \omega x}) \, dx 
$$
Now, consider that for $f(x)=\sin(x)=\frac{e^{-ix}-e^{ix}}{2i}$, we have
$$
\int^{\infty}_{-\infty} \frac{e^{-ix}-e^{ix}}{2i} e^{-i 2 \pi \omega x} \, dx\\
=\int_{- \infty}^{\infty} - \frac{\left(\mathrm{e}^{\mathrm{ix}}\!\left(\mathrm{e}^{- 2 \pi i \mathrm{\omega} x}\right) - \mathrm{e}^{- \mathrm{ix}}\!\left(\mathrm{e}^{- 2 \pi i \mathrm{\omega} x}\right)\right)}{2 i} \,d x \\ = \int^{\infty}_{-\infty} \frac{e^{2 \pi i \omega x+ix}-e^{-2 \pi i \omega x-ix}}{2i} \\ = \frac{1}{2 i}\int^{\infty}_{-\infty} e^{ix(2 \pi \omega +1)}-e^{ix(- 2 \pi \omega -1)} \, dx \\ = \frac{1}{2i} \left(\delta(2 \pi \omega +1)-\delta(2 \pi \omega -1)\right) $$
which intuitively makes sense as a function $f(x)=sin(x)$ can be written in terms of a single frequency $f_{0}=\frac{1}{2 \pi}$
A: Devraj - $$ \int_{-\infty}^{\infty} \frac{e^{2 \pi f_{0} i t}-e^{-2 \pi f_{0} i t}}{2i} e^{2 \pi i k t} dt =\frac{1}{2i} \int_{-\infty}^{\infty} -e^{-2\pi i(f-f_{0})t}+e^{2 \pi i(f+f_{0})} dt=\frac{1}{2i}\left[ \delta(f+f_{0})-\delta(f-f_{0}) \right]$$
A: One way to see such a thing is by approximating your pure $\sin x$ function/signal by truncated or otherwise modified functions whose Fourier transforms exist as integrals, and look at the limit of such things. For example, truncating $\sin x$ to be $0$ outside expanding intervals $[-2\pi N,2\pi N]$. We can integrate explicitly 
$$
2i\int_{-N}^N e^{-i xt}\,\sin x\;dx
= \int_{-N}^N e^{ix(1-t)}\,dt - \int_{-N}^N e^{ix(-1-t)}\,dx
= [{e^{ix(1-t)}\over i(1-t)}]^N_{-N} - [{e^{ix(-1-t)}\over i(-1-t)}]^N_{-N}
$$
$$
= {e^{iN(1-t)}\over i(1-t)} - {e^{-iN(1-t)}\over i(1-t)} + \hbox{two more terms}
= {2\sin N(1-t)\over 1-t} + {2\sin N(-1-t)\over -1-t}
$$
$$
= {2\sin N(t-1)\over t-1} + {2\sin N(t+1)\over t+1}
$$
These are essentially "sinc" functions (Google-able), and a further discussion would show that these approximate Dirac deltas as $N\to +\infty$...
