How do I calculate Fourier Transform of $\frac{\sin t}{t}$?

Question

How do I calculate Fourier Transform of $\,\dfrac{\sin t}{t}$?

I tried to calculate transfrom from $\sin(t)$ and $\frac{1}{t}$ and than multiply but I don't get correct solution.

Answer 1

The sinc function is not in $L^1({\mathbb R})$, hence you cannot compute its Fourier transform directly, using the formula $$\hat f(\xi)=\int_{-\infty}^\infty f(x)e^{-2\pi i\xi x}\>dx\ ,\tag{1}$$ or similar. But sinc is in $L^2({\mathbb R})$, hence by a fundamental theorem of Fourier analysis it has a Fourier transform $g\in L^2({\mathbb R})$, and conversely: sinc is the Fourier transform of some $g\in L^2({\mathbb R})$. Fortunately in this example the $g$ is in $L^1$, so that we can directly verify that $\hat g={\rm sinc}$. As sinc is an even function it then follows by the $L^2$ inversion theorem that $\widehat{\rm sinc}=g$.

This $g$ is a rectangle function: $$g(\xi)=b\quad (-a\leq \xi\leq a), \qquad g(\xi)=0 \quad({\rm otherwise})\ ,\tag{2}$$ whereby the values $a$, $b$ depend on your exact specification of the Fourier transform.

The space $L^p({\mathbb R})$ consists of all functions $f:\>{\mathbb R}\to{\mathbb R}$ having finite integral $\int_{-\infty}^\infty|f(t)|^p\>dt$. Since $\int_{-\infty}^\infty\bigl|{\rm sinc}(t)\bigl|^2\>dt$ is finite we know that ${\rm sinc}\in L^2({\mathbb R})$. The function $g$ in $(2)$ is in $L^1({\mathbb R})$, hence it is allowed to compute its Fourier transform in the usual way $(1)$. It turns out that $\hat g={\rm sinc}$, up to scaling constants.

Answer 2

The reason that your approach failed is that $\frac{1}{t}$ does not have a Fourier Transform.

One possible calculation of the Fourier Transform of $\frac{\sin t}{t}$ is as follows. It's not very rigorous, but it gets the job done and can be made rigorous, if one so wishes, using distributions.

\begin{align} \hat{f}(\omega)&=\mathcal{F}\left[\frac{\sin t}{t}\right](\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\frac{\sin t}{t}e^{i\omega t}\,dt\\ \hat{f}'(\omega)&= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\frac{\sin t}{t}\frac{\partial}{\partial\omega}e^{i\omega t}\,dt\\ &=\frac{i}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\sin t\,e^{i\omega t}\,dt\\ &=\frac{1}{2\sqrt{2\pi}}\left[\int_{-\infty}^\infty e^{i(\omega+1)t}\,dt - \int_{-\infty}^\infty e^{i(\omega-1)t}\,dt\right]\\ &=\sqrt{\frac{\pi}{2}}\left[\delta(\omega + 1)+ \delta(\omega - 1)\right] \end{align}

Now, we know that

$$\hat{f}(0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\frac{\sin t}{t}\,dt = \sqrt{\frac{\pi}{2}}$$

and thus, we arrive at $\hat{f}(\omega)$ by integrating $\hat{f}'(\omega)$. Thus, we have:

\begin{align} \hat{f}(\omega) &= \sqrt{\frac{\pi}{2}}\left[H(\omega + 1) -H(\omega-1)\right]\\ &=\sqrt{\frac{\pi}{2}}H(1-|\omega|) \end{align}

where $H$ is the Heaviside step function.

Answer 3

So I'll first note the definition of the sinc() function

$$\mathrm{sinc}\,\,{t} \equiv \dfrac{\sin{\pi t}}{\pi t}$$

and an integral property of the sinc() function that I'll use below

$$\int_{-\infty}^{\infty} \mathrm{sinc}\,\,{at} \,\, \mathrm{d}t = \dfrac{1}{|a|}$$

Here is one method to compute the Fourier Transform of $\dfrac{\sin{t}}{t}$:

$$\begin{align*}F(s) &= \int_{-\infty}^{\infty} \dfrac{\sin{t}}{t} e^{-2\pi i s t}\mathrm{d}t\\ &=\int_{-\infty}^{\infty} \dfrac{\sin{t}\cos{2\pi st}}{t}\mathrm{d}t -i\int_{-\infty}^{\infty} \dfrac{\sin{t}\sin{2\pi st}}{t}\mathrm{d}t\\ &=\int_{-\infty}^{\infty} \dfrac{\sin{t}\cos{2\pi st}}{t}\mathrm{d}t -i0\\ &=\int_{-\infty}^{\infty} \dfrac{\sin(t+2\pi s t)}{2t}+\dfrac{\sin(t-2\pi s t)}{2t}\mathrm{d}t\\ &=\int_{-\infty}^{\infty} \dfrac{\sin\left[\left(\frac{1}{\pi}+2s\right)\pi t\right]}{2t}+\dfrac{\sin\left[\left(\frac{1}{\pi}-2s\right)\pi t\right]}{2t}\mathrm{d}t\\ &=\int_{-\infty}^{\infty} \dfrac{\left(\frac{1}{\pi}+2 s\right)\pi}{2}\mathrm{sinc}\left[\left(\frac{1}{\pi}+2s\right)t\right]+\dfrac{\left(\frac{1}{\pi}-2 s\right)\pi}{2}\mathrm{sinc}\left[\left(\frac{1}{\pi}-2s\right)t\right]\mathrm{d}t\\ &=\pi\left[\dfrac{1}{2}\dfrac{\frac{1}{\pi}+2s}{\left|\frac{1}{\pi}+2s\right|}+\dfrac{1}{2}\dfrac{\frac{1}{\pi}-2s}{\left|\frac{1}{\pi}-2s\right|}\right]\\ &=\pi\left[\dfrac{1}{2}\dfrac{s+\frac{1}{2\pi}}{\left|s+\frac{1}{2\pi}\right|}-\dfrac{1}{2}\dfrac{s-\frac{1}{2\pi}}{\left|s-\frac{1}{2\pi}\right|}\right]\\ &=\pi\left[\dfrac{1}{2}\mathrm{sgn}\left(s+\frac{1}{2\pi}\right)-\dfrac{1}{2}\mathrm{sgn}\left(s-\frac{1}{2\pi}\right)\right]\\ &= \begin{cases}\pi&\quad |s|< \frac{1}{2\pi}\\ 0&\quad |s| >\frac{1}{2\pi}\end{cases} \end{align*}$$