Why do we use laplace transform to solve the integral $\int _{0}^{\infty}\frac{\sin x}{x}dx$ ? why is it insolvable if we go the normal way

Question

We use Laplace Transform to solve $\int _{0}^{\infty}\frac{\sin x}{x}dx$ as shown below:
$\DeclareMathOperator{\arccot}{arccot}$
We use the property $L(\frac{\sin x}{x})=\int _{s}^{\infty}\frac{1}{s^{\prime2}+1}ds^\prime=\frac{\pi}{2}-\arctan(s)=\arccot(s)$.

Now we use the laplace transform definition $\int _{0}^{\infty}e^{-sx}\frac{\sin x}{x}dx=\arccot(s)$
and put $s=0$, getting:

$\int _{0}^{\infty}\frac{\sin x}{x}dx=\arccot(0)=\frac{\pi}{2}$

I would like to know why this is the case (if possible in a intuitive way) and why this function is not solvable (the integration continues indefinitely) using normal (Integration by parts) integration methods.

Edit: The edits are in bold (in the original question).

Edit 2: I thank everyone for your answers and comments. I would request you to put up some visual/graphic representation (if possible) to explain this so that I can have an intuitive understanding. For example, The definite integral gives the area under the curve right ? So what changes when we solve it using Laplace (or other methods) and why is it not possible to find the area of the curve using integration by parts?

Answer 1

By Louville's Thoerem, the sinc function has no elementary anti-derivative and so the improper integral has to be calculated through other methods such as Laplace Transform or through complex analytical methods.