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While finding the Fourier Transform of the unit step function $u(t)$ , I came across the following integral:

$$\int_{0}^{\infty}e^{-i\omega t}dt = \left[-\frac{e^{-i\omega t}}{i \omega}\right]_{0}^{\infty} $$

The textbook says that the integral will not converge.Can anyone explain the reason why so?

shilov
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hacker804
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  • The issue is that the Fourier Transform of the unit step function is NOT a function. It is not defined by the integral as written here. Rather, it is a Tempered Distribution and defined as a functional. See THIS ANSWER for a rigorous development. – Mark Viola Apr 22 '21 at 02:17

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Note that for a finite upper limit $b$, we have (assuming $w \neq 0$) $$\begin{aligned} \int_0^{b}e^{-iwt} dt &= \left.-\frac{e^{-iwt}}{iw}\right|_{t=0}^{t=b} \\ & = \frac{1-e^{-iwb}}{iw} \\ &= e^{-iwb/2}\left(\frac{e^{iwb/2} - e^{-iwb/2}}{iw}\right) \\ &= 2e^{-iwb/2}\frac{\sin(wb/2)}{w} \\ \end{aligned}$$ Taking the absolute value, we obtain $$\left|\int_0^b e^{-iwt} dt\right| = \left|\frac{2}{w}\right|\cdot\left|\sin(wb/2)\right|$$ For fixed $w$ and $b \to \infty$, the first factor on the right hand side is constant, but the second factor oscillates in the range $[0,1]$. Therefore, $$\lim_{b \to \infty}\left|\int_0^b e^{-iwt}dt\right|$$ does not converge. Consequently, neither does $$\lim_{b \to \infty}\int_0^b e^{-iwt}dt$$ which means that the improper integral $$\int_0^{\infty}e^{-iwt}dt$$ does not exist.

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The complex exponential function is periodic/oscillatory. The integral won't converge for more or less the same type of reasoning that, for example, $\displaystyle \lim_{x\to+\infty} \sin x$ doesn't exist.