calculus - Can one solve $int_{0}^{infty} frac{sin(x)}{x} dx$ *from its Taylor series antiderivative directly*?

This question was inspired by this question:

Evaluating the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$?

Well, can anyone prove this without using Residue theory. I actually thought of doing this:
$$\int_{0}^{\infty} \frac{\sin x}{x} \, dx = \lim_{t \to \infty} \int_{0}^{t} \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \, dt$$
but I don't see how $\pi$ comes here, since we need the answer to be equal to $\frac{\pi}{2}$.

Answers were given to the stated question -- how to prove without using Residue theory. Yet the quote suggests an obvious follow-up question: can you prove the integral from the Taylor series expansion directly, somehow?