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Integrate sinx/x using fourier transform?

  1. Nov 18, 2013 #1
    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. jcsd
  3. Nov 24, 2013 #2
    anyone?
     
  4. Nov 25, 2013 #3

    lurflurf

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    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.
     
  5. Nov 29, 2013 #4
    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.
     
  6. Nov 29, 2013 #5

    lurflurf

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    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.
     
  7. Nov 29, 2013 #6

    lurflurf

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    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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