Prove that $\int_0^\infty \frac{\sin x}{x}\,dx$ converges using power series $$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
$$\frac{\sin x}{x} = 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n+1)!}$$
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:
$$\displaystyle\int_0^\infty\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n+1)!}=
\\
\displaystyle\lim_{M\to\infty}\displaystyle\int_0^M\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n+1)!}=
\\
\displaystyle\lim_{M\to\infty}\displaystyle\int_0^M\sum_{n=0}^\infty(-1)^n\frac{t^{n}}{(2n+1)!}=
\\
\text{This is a power series so we can integrate each member separably }
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\displaystyle\lim_{M\to\infty}\sum_{n=0}^\infty\displaystyle\int_0^M(-1)^n\frac{t^{n}}{(2n+1)!}=
\\
\displaystyle\lim_{M\to\infty}\sum_{n=0}^\infty\displaystyle\int_0^M(-1)^n\frac{x^{2n}}{(2n+1)!}=
\\
(\text{Using Newton-Leibniz formula})
\\
\displaystyle\lim_{M\to\infty}\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\frac{M^{2n+1}}{(2n+1)}-\sum0
$$
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.
 A: 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.
A: You could try Ramanujan's master theorem,

$$\int_0^\infty y^{s-1}\sum_{k\ge0}\frac{(-y)^k}{k!}\phi(k)dy=\Gamma(s)\phi(-s).$$

With $y=x^2$, our integral is$$\frac12\int_0^\infty y^{-1/2}\sum_{k\ge0}\frac{(-y)^k}{(2k+1)!}dy=\frac12\Gamma\left(\frac12\right)\phi\left(-\frac12\right)$$with the analytic continuation$$\phi(k)=\frac{k!}{(2k+1)!}=\frac{\Gamma(k+1)}{\Gamma(2k+2)}$$so our result is$$\frac{\Gamma^2\left(\frac12\right)}{2\Gamma\left(1\right)}=\frac{\pi}{2}.$$
