Proving $-2 < \int_a^{+\infty} \frac{\sin(t)}{t}\,\mathrm dt < 2$ for $a>0$ If true, I'd like to show that $$-2 < \int_a^{+\infty} \frac{\sin(t)}{t}\,\mathrm{d}t < 2 \quad \forall a > 0$$ knowing that this integral exists. I think that is true and that we could even find a better bound. Any advice?
I have no problem showing that this integral is bounded (since it is continuous with regard to $a$ and has a finite limit in $+\infty$) but can't manage to show that bounds $-2$ and $2$ are right. The problem looks a bit like the sum $\sin(k)$ being bounded, but I don't know how to deal with the division by $k$.
I tried to split the integral in different part between $n\pi$ and $(n+2)\pi$ but it led to nothing good or simple.
Thank you!
 A: $\int_{0}^{a}\frac{\sin(x)}{x}\,dx = \text{Si}(a)$ is a differentiable function with stationary points at $\pm\pi,\pm 2\pi,\pm 3\pi,\ldots$ by the fundamental theorem of Calculus. It is straightforward to check by Leibniz criterion that $\text{Si}(\pi)$ is an absolute maximum and $0$ is an absolute minimum over $[0,+\infty)$. It follows that for any $a\geq 0$ we have
$$ \frac{\pi}{2}-\text{Si}(\pi)\leq \int_{a}^{+\infty}\frac{\sin x}{x}\,dx \leq \frac{\pi}{2} $$
implying
$$ -\frac{1}{3}\leq \int_{a}^{+\infty}\frac{\sin x}{x}\,dx \leq \frac{5}{3}. $$

Late addendum. It might be interesting to notice that the constant $\text{Si}(\pi)$ (involved, for instance, in Gibbs' phenomenon for the sawtooth wave) can be numerically approximated through the Cauchy-Schwarz inequality. Indeed, by the Laplace transform
$$\text{Si}(\pi)=\frac{\pi}{2}+\int_{0}^{+\infty}\frac{e^{-\pi x}}{1+x^2}\,dx $$
and the functions $(1+x)e^{-\pi x}$ and $\frac{1}{(1+x)(1+x^2)}$ have a very similar behaviour in a right neighbourhood of the origin. It follows that
$$\text{Si}(\pi)\leq \frac{\pi}{2}+\sqrt{\int_{0}^{+\infty}(1+x)^2 e^{-2\pi x}\,dx \int_{0}^{+\infty}\frac{dx}{(1+x)^2(1+x^2)^2}}=\frac{\pi}{2}+\sqrt{\frac{2\pi^2+2\pi+1}{32\pi^2}}$$
and the inequality is pretty tight. It becomes even tighter if the previous $(1+x)$ factor is replaced by $\left(1+\frac{\pi}{2}x\right)$.
A: HINT: Use the fact that the sequence
$$
a_n=\int_{(n-1)\pi}^{n\pi} \frac{\sin t}{t}\, dt, \qquad n\geq 1,
$$
is decreasing in absolute value. Then use the fact that
$$
\int_0^\infty \frac{\sin t}{t}\, dt=\sum_{n=1}^\infty (-1)^{n+1}|a_n|,
$$
together with the estimate for alternating series with decreasing coefficients
$$
\sum_{n=1}^{2m}(-1)^{n+1}|a_n|\leq \sum_{n=0}^\infty  (-1)^{n+1}|a_n|\leq \sum_{n=1}^{2m-1} (-1)^{n+1}|a_n|,
$$
valid for any integer $m\geq 1$.
