How Do You Solve the Integral from POTW #156?

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    2015
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SUMMARY

The discussion centers on evaluating the improper integral $$\int_{-\infty}^\infty \frac{\sin t}{t}\cos xt\, dt$$ for real values of x. Chisigma provided the correct solution, demonstrating the use of Fourier transform techniques to solve the integral. The integral is significant in signal processing and physics, particularly in analyzing waveforms. The solution highlights the importance of understanding improper integrals and their applications in various mathematical contexts.

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Euge
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Here is this week's POTW:

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Evaluate the improper integral

$$\int_{-\infty}^\infty \frac{\sin t}{t}\cos xt\, dt\quad (x \in \Bbb R).$$

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Congratulations to Chisigma for his correct solution. Here it is below.
Because is...

$\displaystyle \mathcal {L} \{\frac{\sin t}{t}\} = \int_{0}^{\infty} \frac{\sin t}{t}\ e^{- s t}\ d t = \int_{s}^{\infty} \frac{d u}{1 + u^{2}} = \frac{\pi}{2} - \tan^{-1} s = F(s)\ (1) $

... is also...

$\displaystyle \int_{0}^{\infty} \frac{\sin t}{t}\ \cos (x\ t)\ d t = \text{Re}\ \{F(i\ x)\} = \text{Re}\ \{ \frac{\pi}{2} - i\ \tanh^{-1} x \} = \frac{\pi}{2}\ \{\mathcal {U} (x) - \mathcal{U} (x - 1)\}\ (2) $

... where $\mathcal{U} (*)$ is the Heaviside Step Function. Therefore is...

$\displaystyle \int_{- \infty}^{ + \infty} \frac{\sin t}{t}\ \cos (x t)\ dt =$\begin{cases}\pi &\text{if}\ |x| < 1\\ \frac{\pi}{2} &\text{if}\ |x|= 1\\ 0 &\text{if}\ |x|>1\end{cases}

Kind regards
 

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