I want to try to compute the Dirichlet Integral $$\int_0^\infty \frac{\sin x }{x}dx$$ via substituting the $\sin x $ with the Taylor Series expansion of it, giving $$\int_0^\infty \frac{\sum_{n=0}^\infty (-1)^n\frac {x^{2n+1}}{(2n+1)!}} xdx$$ Now you can factorize one $x$ of the sum and cancel it with the denominator, giving $$\int_0^\infty {\sum_{n=0}^\infty (-1)^n\frac {x^{2n}}{(2n+1)!}} dx$$ Using the power rule of integration we get $$ {\sum_{n=0}^\infty \frac {(-1)^nx^{2n+1}}{(2n+1)(2n+1)!}} \Bigg\rvert_0^\infty$$ By now plugging in the limits of the integral, we get $$\lim_{x \to \infty}{\sum_{n=0}^\infty \frac {(-1)^nx^{2n+1}}{(2n+1)(2n+1)!}}\quad - \quad 0 $$
And that's the part where I've got the problem. I don't know how to continue with computing this sum and then the limit nor if it's even possible in the first place. When asking WolframAlpha about the limit, it says that it's actually $\frac {\pi}{2}$ which is the solution of the original integral but I don't know how you can come up with the solution.
While I was doing some research for tackeling how to solve such limits/series I found out that $$\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{2n+1}=\arctan x$$ but that only holds for $-1 \le x \le 1$. $\quad$ I thought that it would be quite useful since it's almost the formular above and as $x$ approaches infinity for $\arctan x$ it approaches $\frac {\pi}{2}$ . Since $x$ approaches infinity (which clearly is not in the intervall from $-1$ to $1$) it's not that useful though. I also found out that $$-\frac \pi 2 -\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)x^{2n+1}}=\arctan x $$ and that holds for $x \ge 1$
Is it possible to get this 2nd $\arctan x$ formula with somehow rewriting the series? Or is there any other trick how to compute this limit of the series? Btw, I don't have profound knowledge about series or Taylor Series in particular yet.