Fourier transform of $\frac{\sin x}{x}$ without contour integration

Question

Find the Fourier transform of $$\frac{\sin x}{x}$$

I know that question was asked before here, here on M.SE, but those solutions use contour integration. Can I solve it without that? One possible solution may be,

Take

$$ \mathscr F\{f(x)\}=F(\omega)=\int_{-\infty}^\infty \frac{\sin x}{x}e^{-i\omega x}dx $$

If we use Leibniz integral rule,

$$ \begin{align} \frac{d}{d\omega}F(\omega)&=\int_{-\infty}^\infty \frac{\sin x}{x}\frac{\partial}{\partial \omega}e^{-i\omega x}dx\\ &=-i\int_{-\infty}^\infty \sin xe^{-i\omega x}dx\\ &=-i\int_{-\infty}^\infty\frac{e^{ix}-e^{-ix}}{2i}e^{-i\omega x}dx\\ &=-\frac12\int_{-\infty}^\infty(e^{ix}e^{-i\omega x}-e^{-ix}e^{-i\omega x})dx\\ &=-\frac12\underbrace{\int_{-\infty}^\infty e^{-(i\omega-i)x}-e^{-(i+i\omega)x}dx}_I \end{align} $$

But it seems the integral is problematic for their limit. Any help will be appreciated.

Answer 1

There is a much simpler way by using the Fourier inversion theorem!

Then the result just follows from the fact that $$ {\cal F}^{-1}(\mathbf 1_{[-1,1]})(x) = \frac{1}{2\pi}\int_{-1}^1 e^{i\,\omega\,x}\,\mathrm d \omega = \frac{e^{ix}-e^{-ix}}{2i\pi\,x} = \frac{\sin x}{\pi\,x}. $$ and so $$ {\cal F}\left(\frac{\sin x}{x}\right) = \pi\,{\cal F}\left(\frac{\sin x}{\pi\,x}\right) = \pi\,\mathbf 1_{[-1,1]}(\omega) $$

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Let me add details if you are not familiar with the Fourier transform: the Fourier inversion theorem tells that you can find the function from its Fourier transform by writing $$ f(x) = \frac{1}{2\pi} \int_{\Bbb R} \widehat f(\omega) \, e^{i\omega\,x}\,\mathrm d \omega = \frac{1}{2\pi} \cal F\big(\hat f\big)(-x) $$ where $\widehat f = {\cal F}(f)$. Equivalently, it means that the inverse operation to the Fourier transform is given by $$ {\cal F}^{-1}(g)(x) = \frac{1}{2\pi} \int_{\Bbb R} g(\omega) \, e^{i\omega\,x}\,\mathrm d \omega = \frac{1}{2\pi} \cal F(g)(-x) $$

Answer 2

One "easy" way is to use the convolution theorem. I'll sketch out how I would do it:

First, depending on the definition of the Fourier transform, some will use a multiplicative factor depending if they use angular frequency $\omega$, or $2\pi f$ or even $2\pi k$. Lets stick with $\omega$ and "ignore" multiplicative factors for the moment. You can figure out those yourself at the end.

Now, note that the Fourier transform of $\sin(x)$ is something proportional to $\delta(\omega-1)-\delta(\omega+1)$ where $\delta(\bullet)$ is the Dirac delta. The reason is that $\sin(x)$ is composed by two complex exponentials as: $\sin(x) = \frac{\exp(ix)}{2i}-\frac{\exp(-ix)}{2i}$, and the Fourier transform of a complex exponential $\exp(ix)$ is:

$$ \int_{-\infty}^{\infty}\exp(ix)\exp(-i\omega x)\,\mathrm dx = -\int_{-\infty}^{\infty}\exp(i(\omega-1)x)\,\mathrm dx = 2\pi\delta(\omega-1) $$ similarly for $\exp(-i\omega x)$. More details on the last step can be found here or here for a different approach.

In addition, the Fourier transform of $1/x$ be found as well, resulting in something proportional to $\text{sgn}(\omega)$ where $\text{sgn}(x)$ is the sign function (returns 1 or -1 depending on the sign of the input). For more information on this part, see this.

Now, The Fourier transform of $\sin(x)\frac{1}{x}$ will be the convolution of the Fourier transforms of $\sin(x)$ and $1/x$ by separate. Now, since $\delta(\omega-a)\ast f(\omega) = f(\omega-a)$ for any $a$ and any function $f(\bullet)$, this results in $$ \begin{aligned} \mathcal{F}\left\{\sin(x)\frac{1}{x}\right\} = \mathcal{F}\left\{\sin(x)\right\}\ast\mathcal{F}\left\{\frac{1}{x}\right\} &\propto \left(\delta(\omega-1)-\delta(\omega+1)\right)\ast\left(\text{sgn}(\omega)\right)\\ & =\text{sgn}(\omega-1)-\text{sgn}(\omega+1) \end{aligned} $$

In the following you will find some additional Wolfram Alpha links that walks you through this process:

• FT of $\sin(x)$
• FT of $1/x$
• Convolution of both FTs
• FT of $\sin(x)/x$

I hope this helps!