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The existence of the Fourier Transform integral is conditioned for some functions. An important example is the sinc (sinus cardinalis) function, which although does not satisfy certain conditions, it nonetheless can be transformed.

To give an example of a case where the sinc function does not satisfy a condition, I would like to solve the following integral, but so far I have not been successful in doing so.

How can I solve this sinus cardinalis integral? Would you be so kind as to help me? Thank you so much!

$$\int_{-\infty}^{\infty} \left|2Af_0 \frac{\sin(2\pi f_0 t)}{2\pi f_0 t}\right|\,dt$$

MarianD
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Daniel Rodriguez
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1 Answers1

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Assuming that $A$ and $f_0$ are nonzero constants, $\int_{-\infty}^{\infty} \left|2Af_0 \frac{\sin(2\pi f_0 t)}{2\pi f_0 t}\right|\,dt$ diverges. This integral would be a constant multiple of $\int_{-\infty}^{\infty} \left|\frac{\sin(t)}{ t}\right|\,dt$ if it converged, so it’s enough to show that $\int_{-\infty}^{\infty} \left|\frac{\sin(2\pi t)}{ t}\right|\,dt$ diverges.

$\displaystyle\int_{-\infty}^{\infty} \left|\frac{\sin(2\pi t)}{ t}\right|\,dt=\sum_{k=-\infty}^\infty\int_k^{k+1}\left|\frac{\sin(2\pi t)}{ t}\right|\,dt$

$\displaystyle\phantom{\int_{-\infty}^{\infty} \left|\frac{\sin(2\pi t)}{ t}\right|\,dt}\ge\sum_{k=-\infty}^\infty\int_{k+\frac{1}{4}-\frac{1}{10}}^{k+\frac{1}{4}+\frac{1}{10}}\left|\frac{\sin(2\pi t)}{ t}\right|\,dt$

$\displaystyle\phantom{\int_{-\infty}^{\infty} \left|\frac{\sin(2\pi t)}{ t}\right|\,dt}\ge\sum_{k=-\infty}^\infty\frac{1}{5}\min_{k+\frac{1}{4}-\frac{1}{10}\le t \le k+\frac{1}{4}+\frac{1}{10}} \left|\frac{\sin(2\pi t)}{ t}\right|$

$\displaystyle\phantom{\int_{-\infty}^{\infty} \left|\frac{\sin(2\pi t)}{ t}\right|\,dt}\ge\sum_{k=0}^\infty\frac{1}{5}\frac{1/2}{k+\frac{1}{2}}>\frac{1}{10}\sum_{k=1}^\infty\frac{1}{k}$.

The harmonic series $\displaystyle\sum_{k=1}^\infty\frac{1}{k}$ diverges, so the integral diverges.

Steve Kass
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