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.

Answer 2

You can solve it using Feynman's technique. But it boils down to providing the same calculation steps, with an intuitive explanation. See this video for example

Answer 3

As said by aleden, the case of the indefinite integrals has been settled by Liouville and others (Risch), and the antiderivative of $\dfrac{\sin x}x$ is proven to have no closed-form expression.

But the case of definite integrals, some of which are sporadically solvable (by various methods such as residues) when their definite counterparts are not, remains black magic.

Answer 4

There are loads of ways to evaluate this so-called Dirichlet integral. Just to expand on Andrei's answer, Feynman's trick gets $$\int_0^\infty\frac{\sin x dx}{x}=\int_0^\infty dy\int_0^\infty\sin x \exp -xy dx=\int_0^\infty\frac{dy}{1+y^2}=\frac{\pi}{2}.$$That first $=$ sign is where we use the trick, but I've concealed the details. Let's spell them out:$$\int_0^\infty\frac{\sin x dx}{x}=\left.\int_{0}^{\infty}\frac{\sin x\exp-xydx}{x}\right|_{y=0}=-\int_0^\infty\partial_y\int_{0}^{\infty}\frac{\sin x\exp-xydx}{x},$$since we're trying to evaluate at $y=0$ a function that vanishes at $y=\infty$. But I didn't hide all that just so no one line of LaTeX would be too long. You see, Feynman's trick isn't even needed to obtain the double-integral expression; you can simply use $\frac{1}{x}=\int_0^\infty e^{-xy}dy$. After that, the result of the argument is the same either way; but I prefer this use of $\frac{\Gamma(s)}{x^s}=\int_0^\infty y^{s-1}\exp -xy dy$ because it's probably easier than a double-layered Feynman's trick for the next power of sinc.