MHB How Do You Solve the Integral from POTW #156?

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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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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

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

Spoiler

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