# Integrate sinx/x using fourier transform?

1. Nov 18, 2013

### nabeel17

I learned how to integrate it using the complex plane and semi circle contours but I was wondering if there is a way using fourier transforms. I know that the fourier transform of the rectangle wave form is the sinc function so I was thinking maybe i could do an inverse fourier on sinc x and get back the rectangle function and integrate that? Or something along those lines.

2. Nov 24, 2013

### nabeel17

anyone?

3. Nov 25, 2013

### lurflurf

Sure just write the integral as the Fourier transform or inverse Fourier transform of 1/x. Of course if you can take the transform you already know the integral.

4. Nov 29, 2013

### nabeel17

Why 1/x? why does that help? I was thinking to change sinx/x by taking the fourier transform of it which is just the box function and see if that would get me anywhere.

5. Nov 29, 2013

### lurflurf

I will assume the following
$$\int_{-\infty}^\infty \! \frac{\sin(t)}{t} \, \mathrm{d}t\\ \mathrm{sinc}(t)=\lim_{x\rightarrow t} \frac{\sin(x)}{x}\\ \mathcal{F} \{ \mathrm{f}(t) \}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \! \mathrm{f}(t)e^{\imath \, \omega \, t} \, \mathrm{d}t$$
What you suggest works fine
we know (as you say)
$$\mathcal{F}\left\{\frac{1}{2}\sqrt{\frac{\pi}{2}}(\mathrm{sgn}(1-t)+\mathrm{sgn}(1+t))\right\}=\mathrm{sinc}(\omega)\\ \text{then}\\ \left. \int_{-\infty}^\infty \! \frac{\sin(t)}{t} \, \mathrm{d}t=\sqrt{2 \pi} \mathcal{F} ^{-1} \{ \mathrm{sinc}(\omega) \} \right|_{t=0}$$$$=\frac{\pi}{2}((\mathrm{sgn}(1-0)+\mathrm{sgn}(1+0)))=\pi$$
This does not really help us though as to do that we already know the integral.

6. Nov 29, 2013

### lurflurf

1/x seems simpler to me
$$\mathcal{F} \{ t^{-1} \}=\imath \, \sqrt{\frac{\pi}{2}}\mathrm{sgn}(t)\\ \text{so}\\ \left. \int_{-\infty}^\infty \! \frac{\sin(t)}{t} \, \mathrm{d}t=-\imath \, \sqrt{2 \pi} \mathcal{F} \{ t^{-1} \}\right|_{\omega=0}=\pi \, \mathrm{sgn}(0)=\pi$$

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