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I cant seem to get this right; I end up with $\dfrac{\cos(2 \pi^2 \xi )}{\frac{1}{4}-4\pi^2 \xi^2}$ after ;

$$\frac{1}{2} \left(\frac{e^{i\pi}(\frac{1}{2}+2\pi\xi)}{i(\frac{1}{2}+2\pi\xi)}-\frac{e^{-i\pi}(\frac{1}{2}-2\pi\xi)}{i(\frac{1}{2}-2\pi\xi)} - \frac{e^{-i\pi}(\frac{1}{2}+2\pi\xi)}{i(\frac{1}{2}+2\pi\xi)}+\frac{e^{i\pi}(\frac{1}{2}-2\pi\xi)}{i(\frac{1}{2}-2\pi\xi)}\right)$$

which by $e^\frac{i\pi}{2}=i$ equals

$$\frac{e^{i\pi2\pi\xi}}{\frac{1}{2}+2\pi\xi} + \frac{e^{i\pi2\pi\xi}}{\frac{1}{2}-2\pi\xi} + \frac{e^{-i\pi2\pi\xi}}{\frac{1}{2}+2\pi\xi} + \frac{e^{-i\pi2\pi\xi}}{\frac{1}{2}-2\pi\xi}=\frac{\cos(2 \pi^2 \xi)}{\frac{1}{4}-4\pi^2 \xi^2}$$

My definition of the transform is $\displaystyle \int_{-\infty}^\infty f(x)e^{-i2\pi x \xi} \, dx $

Mark Viola
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user123124
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  • I'm confused (about title + definition of the transform). Is the integral you're trying to compute $\int_{-\pi}^\pi dx \cos(x/2)e^{-2\pi\mathrm{i}x\xi}$? [Your definition of Fourier transform does not have $\mathrm{i}$ in the exponent, and the 'truncation' alluded at in the title is not evident]. If the answer is yes, then your result seems correct. – Pierpaolo Vivo Mar 14 '16 at 18:02
  • Why do you believe that your answer is incorrect? It is indeed correct. – Mark Viola Mar 14 '16 at 18:20
  • @Dr.MV since Im supposed to use the result to compute $ \int_{-\infty}^{\infty} \frac{cos(\pi \xi) }{1-4\xi^2} $ and all constants seem messed up to do this – user123124 Mar 14 '16 at 18:28

1 Answers1

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This is just a confusion of definitions of the fourier transform, not having $2 \pi$ in exponent gives the right transform. Evaluating $cos(\frac{x}{2})$ at $0$ gives the value of the integral via fourier inversion formula.

user123124
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