Proving $-2 < \int_a^{+\infty} \frac{\sin(t)}{t}\,\mathrm dt < 2$ for $a>0$

Question

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!

Answer 1

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 $m\geq 1$.

Answer 2

$\int_{0}^{a}\frac{\sin(x)}{x}\,dx = \text{Si}(a)$ 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 $$ -\frac{1}{3}\leq \int_{a}^{+\infty}\frac{\sin x}{x}\,dx \leq \frac{5}{3}. $$

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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 $\left(1+\frac{\pi}{2}x\right)$.