How do you find Fourier Transform of the non-integrable function $\frac{\sin(4x)}{\pi x}$?

Question

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.

Answer 1

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.

Answer 2

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

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{\sin(x)}{x}e^{ikx}\,dx=\begin{cases}\pi&,|k|<1\\\\0&,|k|>1\end{cases}}\tag 1$$

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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

$$\text{PV}\left(\int_{-\infty}^\infty \frac{\sin(x)}{x}e^{ikx}\right)=\begin{cases}\pi&,|k|<1\\\\0&,|k|>1\\\\\pi/2&,k=\pm 1\end{cases}$$

Answer 3

You can write it in form of a $\mathrm{sinc}$ function whose Fourier transform is just a rectangular (brickwall) function.