Calculus 4

For the non-hindrance of calculations we'll denote $s$ as $\omega$ and use $s$ for our calculations in the Laplace transformation.

Now, using the well known formula for Laplace transforms:

$\int_0^\infty \dfrac{f(t)}{t} \,dt =\int_0^\infty \mathcal{L}\{f(t)\}(s) \,ds$

We have

$\begin{aligned} \int_0^{\infty} \dfrac{\sin(\omega t)}{t} \,dt &= \int_0^{\infty} \mathcal{L} \left\{ \sin(\omega t) \right\} (s) \,ds \\ &= \int_0^{\infty} \dfrac{\omega}{s^2 + \omega^2} \,ds \\ &= \dfrac{\pi}{2} \omega \sqrt{\dfrac{1}{\omega^2}} \end{aligned}$

Here, it may seem fair to introduce $\omega$ into the root and cancel out the $\sqrt{\dfrac{\omega^2}{\omega^2}}$ term in the final step. But remember that $\omega \in \mathbb{C}$ and so it would be an improper method to do so.

Thus, we derive that

$\int_0^{\infty} \dfrac{\sin(\omega t)}{t} = \dfrac{\pi}{2} \text{ iff } \begin{cases} \mathfrak{I} (\omega^2) \neq 0 \\ \mathfrak{R}(\omega^2) \ge 0 \end{cases}$

Take s=0 We get the Integral as 0 Which is not equal to $\frac { \pi }{ 2 }$