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?
 A: This post does not derive the result using only Taylor series, but it only uses the fundamental property of complex numbers, that $z = a+bi$.
Euler derived a awesome formula
$$\frac{\Gamma (s)}{n|p|^s}\sin{(\alpha s)}=\int_0^{\infty}u^{ns-1}\exp{(-\Re(p)u^n)}\sin{(\Im(p)u^n)}du$$ 
Where $tan(\alpha)=\frac{\Im(p)}{\Re(p)}$
Here is an informal proof of the formula https://goo.gl/JKfEO8.
Now use $n=1$, $p=0+i$, $s=0$, $\alpha = \frac{\pi}{2}$. You will also need Eulers Reflection Formula for the Gamma function 
$$\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin{(\pi s)}}$$
If you have trouble solving the integral using the formula you can also watch this https://goo.gl/2WPRQH.
A: HINT: Notice, $$\lim_{t\to \infty}\int_{0}^{t}\frac{1}{t}\left(t-\frac{t^3}{3!}+\frac{t^5}{5!}+\dots \right)dt$$
$$=\lim_{t\to \infty}\int_{0}^{t}\frac{1}{t}\left(t-\frac{t^3}{3!}+\frac{t^5}{5!}+\dots \right)dt$$
$$=\lim_{t\to \infty}\int_{0}^{t}\left(1-\frac{t^2}{3!}+\frac{t^4}{5!}+\dots \right)dt$$
$$=\lim_{t\to \infty}\left(t-\frac{t^3}{3\cdot3!}+\frac{t^5}{5\cdot 5!}+\dots \right)_{0}^{t}$$
$$=\lim_{t\to \infty}\left(t-\frac{t^3}{3\cdot3!}+\frac{t^5}{5\cdot 5!}+\dots \right)$$
