I can't quite get an inverse Fourier Transform to match up with a statement in my textbook. At one point, my textbook writes:

"If $g$ is a function that is one on the interval $(- \pi, \pi]$ and zero otherwise, then $\mathcal{F}^{-1}g(\xi) = \sqrt{2 \pi} \text{sinc}(\xi)$"

I tried to calculate to verify this statement, but do not get the same answer. The inverse Fourier transform is defined as:

$$\mathcal{F}^{-1}[g](x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} g(\lambda) e^{i \lambda x} d \lambda$$

In this case we have $g = 1$ in the interval $[- \pi, \pi]$. So my calculation will be:

$$\mathcal{F}^{-1}g(\xi) = \frac{1}{\sqrt{2 \pi}} \int_{- \pi}^{\pi} e^{i \lambda \xi} d \lambda$$

However, when I calculate this, I get the answer:

$$\frac{\sqrt{\frac{2}{\pi}} \sin(\pi \xi)}{\xi}$$

which does not correlate with the book. Wolframalpha also gives this answer. So is the book wrong here? I would really apprecaite some input here!


$\operatorname{sinc}(x)$ is defined in two different ways; one of them is $$\operatorname{sinc}(x)=\begin{cases}\frac{\sin(\pi x)}{\pi x},&x \neq 0,\\1,&x = 0.\end{cases}$$ So, multiply your answer by $\frac{\pi}{\pi}$, simplify, use the definition above, and enjoy!

  • 1
    $\begingroup$ Ah! So easy. Duh! Thank you so much. Here I am going crazy over something so obvious. Really appreciate it :) $\endgroup$ – Kristian May 8 '13 at 11:58

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