Integrate sinx/x using fourier transform?

[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.

 

[nabeel17]

anyone?

 

[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.

 

[nabeel17]

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.

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.

 

[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.

 

[lurflurf]

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