Is this a proof that $\operatorname{Si}(x)$ is well-defined, continuous and bounded? Let $\operatorname{Si}: \mathbb R \to \mathbb R$, where $\operatorname{Si}(x):=\int^{x}_{0}\frac{\sin(t)}{t}\,dt$
Prove that $\operatorname{Si}$ is well-defined (i), continuous (ii) and bounded (iii). 
My ideas:
for (i): we know that $\sin(t)=\sum^{\infty}_{k=0}(-1)^{k}\frac{t^{2k+1}}{(2k+1)!}$ and therefore $\frac{\sin(t)}{t}= \sum^{\infty}_{k=0}(-1)^{k}\frac{t^{2k}}{(2k+1)!}$, so the series is defined for all $t \in \mathbb R$, since the series converges. By extension that means that $\int^{x}_{0}\frac{\sin(t)}{t}dt$ is defined $\forall x \in \mathbb R$. (Is my reasoning sound here?)
for (ii):
we would need to prove both that $\lim_{\epsilon \to \infty} \int^{\epsilon}_{0}\frac{\sin(x)}{x}$, as well as $\lim_{\epsilon \to 0}\int^{x}_{\epsilon}\frac{\sin(x)}{x}$. Tried using partial integration here and haven't got further. 
for (iii): Follows immediately out of (ii), perhaps even out of (i)?! 
Alternatively looking at $\frac{\sin(t)}{t}$ as a sequence we know that $\Bigl|\frac{\sin(t)}{t}\Bigr|\leq\frac{1}{t}$ and $\frac{1}{t}\to0, n \to \infty,\:$
so $\frac{\sin(t)}{t}$ is bounded and therefore $\int^{x}_{0}\frac{\sin(x)}{x}$ is also bounded. (Is this reasoning correct?)
 A: HINT:
To show that $\text{Si}(x)$ is bounded, we fix $x\in [n\pi,(n+1)\pi]$ and write
$$\begin{align}
\int_0^x \frac{\sin(t)}{t}\,dt&=\sum_{k=0}^{n-1} \int_{k\pi}^{(k+1)\pi}\frac{\sin(t)}{t}\,dt+\int_{n\pi}^x \frac{\sin(t)}{t}\,dt\\\\
&=\sum_{k=0}^{n-1} \int_0^\pi \frac{(-1)^k\sin(t)}{t+k\pi}\,dt+\int_0^{x-n\pi}\frac{(-1)^n\sin(t)}{t+n\pi}\,dt
\end{align}$$
Can you finish?
A: The proof that $\operatorname{Si}$ is well defined is sound: power series can be integrated term by term.
It is also true that (iii) follows from (i) and (ii) and it's sufficient to show that
$$
\lim_{x\to\infty}\operatorname{Si}(x)
$$
is finite, because the function is odd, so its limit at $-\infty$ would be finite as well. The reason is that the extended real line is compact, so a continuous function on the real line is bounded.
Your proof for the limit is unfortunately wrong. Instead, consider
$$
\int_0^x\frac{\sin t}{t}\,dt=
\int_0^1\frac{\sin t}{t}\,dt+
\int_1^x\frac{\sin t}{t}\,dt
$$
and you only need to prove the limit of the second summand is finite. With integration by parts,
$$
\int_1^x\frac{\sin t}{t}\,dt=
\Bigl[-\frac{\cos t}{t}\Bigr]_1^x-
\int_1^x\frac{\cos t}{t^2}\,dt
$$
Now the first summand has limit $\cos 1$ and
$$
\left|\frac{\cos t}{t^2}\right|\le\frac{1}{t^2}
$$
so easily
$$
\lim_{x\to\infty}\int_1^x\left|\frac{\cos t}{t^2}\right|\,dt
$$
is finite, which implies finiteness of
$$
\lim_{x\to\infty}\int_1^x\frac{\cos t}{t^2}\,dt
$$
Dirichlet proved that
$$
\lim_{x\to\infty}\int_0^x\frac{\sin t}{t}\,dt=\frac{\pi}{2}
$$
