Is this a proof that $\operatorname{Si}(x)$ is well-defined, continuous and bounded?

Question

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?)

Answer 1

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 $$ \lim_{x\to\infty}\int_0^x\frac{\sin t}{t}\,dt=\frac{\pi}{2} $$

Answer 2

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?