$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
$$\frac{\sin x}{x} = 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n+1)!}$$
Using the d'Alembert test and $t=x^2$ the radius of convergence is $\infty$ so for every x this sum equals the function.
So I have $$\int_0^\infty\frac{\sin x}{x}\,dx=\int_0^\infty\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n+1)!}$$
But how do I prove that the integral itself converges?
I've had this idea but i'm not sure if it's formally valid:
$$\displaystyle\int_0^\infty\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n+1)!}= \\ \displaystyle\lim_{M\to\infty}\displaystyle\int_0^M\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n+1)!}= \\ \displaystyle\lim_{M\to\infty}\displaystyle\int_0^M\sum_{n=0}^\infty(-1)^n\frac{t^{n}}{(2n+1)!}= \\ \text{This is a power series so we can integrate each member separably } \\ \displaystyle\lim_{M\to\infty}\sum_{n=0}^\infty\displaystyle\int_0^M(-1)^n\frac{t^{n}}{(2n+1)!}= \\ \displaystyle\lim_{M\to\infty}\sum_{n=0}^\infty\displaystyle\int_0^M(-1)^n\frac{x^{2n}}{(2n+1)!}= \\ (\text{Using Newton-Leibniz formula}) \\ \displaystyle\lim_{M\to\infty}\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\frac{M^{2n+1}}{(2n+1)}-\sum0 $$ And because power series-es have uniform convergence the radius has unchanged so if this sum converges therefore the original $\displaystyle\int_0^\infty\frac{\sin x}{x}dx$ converges.
Also $\sum_{n=0}^\infty\frac{x^n}{n!}$ has infinite radius of convergence, but of course $$ \lim_{x\to\infty}\sum_{n=0}^\infty\frac{x^n}{n!}=\infty $$ It's true that your series has alternating signs, so the example is not of the same kind, but it's too week an argument.
Note also that the integral $$ \int_0^\infty\frac{\sin x}{x}\,dx $$ is only “conditionally convergent”, in the sense that $$ \int_0^\infty\left|\frac{\sin x}{x}\right|\,dx $$ does not converge.
The usual proof is by considering the integral as $$ \int_0^1\frac{\sin x}{x}\,dx+ \int_1^\infty\frac{\sin x}{x}\,dx $$ where the first term poses no problem because the function has a removable singularity at $0$.
The second term is computed by parts: $$ \int_1^t\frac{\sin x}{x}\,dx= \Bigl[\frac{-\cos x}{x}\Bigr]_1^t-\int_1^t\frac{\cos t}{t^2}\,dt $$ Again, the first term poses no issue and $$ \left|\frac{\cos t}{t^2}\right|\le\frac{1}{t^2} $$ so the integral is absolutely convergent. Here you may possibly use power series, but I don't think it's worth the trouble.