How do you find Fourier Transform of a non-integrable function? How do you find a Fourier Transform for a non-integrable function?
For $$\frac{\sin(4x)}{\pi x}$$  
According to https://owlcation.com/stem/How-to-Integrate-sinxx-and-cosxx the function is not integrable.
 A: Although it's not $L^1$, it's $L^2$ and we can make sense of its Fourier transform. You can find its formula using the $L^2$-inversion theorem. Hint: pick an arbitrary rectangle pulse; it is $L^1$, find its Fourier transform. You should get something close to $\sin x/x$. After a few adjustments you find your function.
A: You can write it in form of a $\mathrm{sinc}$ function whose Fourier transform is just a rectangular (brickwall) function.
A: The function $f(z)=\frac{\sin(z)}{z}$ has a removable discontinuity and is analytic in the complex plane.  As such, it is integrable on any smooth contour $C$.
While $f(x)$ has no anti-derivative in terms of elementary functions, the integral 
$$F(a,b)=\int_a^b \frac{\sin(x)}{x}\,dx$$
exists for any real-valued $a$ and $b$.
In fact, we have
$$\int_{-\infty}^\infty \frac{\sin(x)}{x}\,dx =\pi$$
The Fourier Transform of the sinc function, is given by 
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{\sin(x)}{x}e^{ikx}\,dx=\begin{cases}\pi&,|k|<1\\\\0&,|k|>1\end{cases}}\tag 1$$

At the request of @user1952009, I thought it would be useful to present a section on the methodology used to find the Fourier Transform of the sinc function.  
The Fourier Transform of $f(x)=\frac{\sin(x)}{x}$ is given by the Cauchy-Principal Value integral
$$\begin{align}
\mathscr{F}\{f\}(k)&=\text{PV}\left(\int_{-\infty}^{\infty}\frac{\sin(x)}{x}e^{ikx}\,dx\right)\\\\
&=\lim_{L\to \infty}\int_{-L}^{L}\frac{\sin(x)}{x}e^{ikx}\,dx\\\\
&=\lim_{L\to \infty}\int_{-L}^L \frac{\sin(x)\cos(kx)}{x}\,dx\\\\
&=\frac12\lim_{L\to \infty}\int_{-L}^L \frac{\sin((k+1)x)-\sin((k-1)x)}{x}\,dx\\\\
&=\frac12\lim_{L\to \infty}\text{Im}\left(\int_{-L}^L \frac{e^{(k+1)x}-e^{(k-1)x}}{x}\,dx\right)\\\\
\end{align}$$
Now, we move to the complex plane and invoke Cauchy's Integral Theorem to evaluate the integral 
$$\begin{align}
0&=\oint_C \frac{e^{itz}}{z}\,dz\\\\
&=\int_{\epsilon\le |x|\le L}\frac{e^{itx}}{x}\,dx+\int_{\text{sgn}(t)\pi}^{0}\frac{e^{it\epsilon e^{i\phi}}}{\epsilon e^{i\phi}}\,i\epsilon e^{i\phi}\,d\phi+\int_0^{\text{sgn}(t)\pi}\frac{e^{itL e^{i\phi}}}{L e^{i\phi}}\,iLe^{i\phi}\,d\phi
\end{align}$$
As $\epsilon\to 0^+$ and $L\to \infty$, we find that (after taking the imaginary part)
$$\lim_{(\epsilon,L)\to (0^+,\infty)}\int_{\epsilon\le |x|\le L}\frac{\sin(tx)}{x}\,dx=\pi\text{sgn}(t)$$
Thus, when $t=k\pm 1$, we have
$$\text{PV}\left(\int_{-\infty}^\infty\frac{\sin(tx)}{x}\,dx\right)=\begin{cases}\pi&,k>\mp 1\\\\-\pi&,k<\mp1\end{cases}$$
Therefore, we find that
$$\text{PV}\left(\int_{-\infty}^\infty \frac{\sin(x)}{x}e^{ikx}\right)=\begin{cases}\pi&,|k|<1\\\\0&,|k|>1\\\\\pi/2&,k=\pm 1\end{cases}$$ 
