I know a good number of methods to solve this integral like Feynman Trick, Laplace transform etc.
$$\int_0^{\infty} \dfrac{\sin x}{x} \mathrm{d}x = \dfrac{π}{2} $$
Do y'all know some creative and not-so-common/(exotic?) methods to solve this? I have so much to learn :)
Here is a convoluted process which introduces a few concepts. You can see this integral as the Mellin transform of $\sin(x)$ $$ \mathcal{M}_x[\sin(x)](s) = \int_0^\infty x^{s-1} \sin(x) \; dx $$ by the Ramanujan Master theorem, if a function can be expressed as $$ f(x) = \sum_{k=0}^\infty \frac{(-1)^k}{k!}\phi(k)x^k $$ then the Mellin transform is given by $$ \mathcal{M}_x[\sin(x)](s) = \int_0^\infty x^{s-1} f(x) \; dx = \Gamma(s) \phi(-s) $$ where $\Gamma(s)$ is the gamma function, we have \begin{equation} \sin(x) = \sum_{k=0}^\infty \frac{(-1)^k}{k!}[x \bmod 2=1]x^k \end{equation} where the $[]$ is an Iverson bracket. A suitable function that alternates between $0$ and $1$ to give the correct coefficients is $|\sin(\frac{\pi k}{2})|$ so we could say $$ \mathcal{M}_x[\sin(x)](s) = \int_0^\infty x^{s-1} \sin(x) \; dx = \Gamma(s)|\sin(\frac{\pi s}{2})| $$ you integral is retrieved when $s=0$, $\Gamma(0)|\sin(0)|$ is indeterminate so we can take the limit which is $\pi/2$.
