Supremum of a sine integral

Question

Let $M_T=\int\limits_{0}^{T}\frac{\sin(t)}{t}dt$ be a sine integral. Why is $2\displaystyle\sup_{T}M_T < \infty$?

Answer 1

$$\int_{0}^{T}\frac{\sin\left(t\right)}{t}dt=\overset{\infty}{\underset{n=0}{\sum}}\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}\int_{0}^{T}t^{2n}dt=\overset{\infty}{\underset{n=0}{\sum}}\frac{\left(-1\right)^{n}T^{2n+1}}{\left(2n+1\right)!\left(2n+1\right)}<\infty$$ for the Leibniz's criterion, $\forall T\in\mathbb{R}^{+}$. Furthermore note that $$\int_{0}^{\infty}\frac{\sin\left(t\right)}{t}dt=\int_{0}^{\infty}\mathfrak{L}\left\{ \sin\left(t\right)\right\} \left(s\right)ds=\int_{0}^{\infty}\frac{1}{s^{2}+1}ds=\frac{\pi}{2}$$where $\mathfrak{L}$ is the Laplace transform.

Answer 2

For any $k\geq 1$ we have: $$a_k\triangleq \left|\int_{k\pi}^{(k+1)\pi}\frac{\sin t}{t}\,dt\right|\leq\log\left(1+\frac{1}{k}\right)$$ and $a_k>a_{k+1}$, since $|\sin(t+\pi)|=|\sin t|$ but $\frac{1}{t+\pi}<\frac{1}{t}$.

Leibniz' criterion hence gives that $\frac{\sin t}{t}$ is a Riemann integrable function over $\mathbb{R}^+$, and: $$ M_T \leq M_{\pi} $$ holds for any $T\in\mathbb{R}^+$.