Upper and lower bounds on sine integral?

Question

Define the sine integral $\operatorname{Si}(x)$ by $$ \def\Si{\operatorname{Si}} \Si(x) = \int^x_0 \frac{\sin t}{t} \, dt $$

I want to establish upper and lower bounds on $\frac{1}{\pi} \Si(\pi x)$. The answer here proves a closely-related upper bound. But heuristically, in my particular case, tighter upper and lower bounds appear to be given by $$ \frac{1}{2} - \frac{1}{3 \pi x} \leq \frac{1}{\pi} \Si(\pi x) \leq \frac{1}{2} + \frac{1}{3 \pi x} $$

Mathematica suggests that the inequalities hold out at least as far as $x = 1{,}000{,}000{,}000$, and plots look reassuring:

Are these bounds valid, and if so can they be proved?

Answer 1

We can define sharper bounds from the asymptotics. $$\frac 1 \pi \text{Si}(\pi x)=\frac 12 +\frac {\cos(\pi x)}\pi\sum_{n=0}^\infty (-1)^{n+1}\,\frac {(2n)!}{(\pi x)^{2n+1 }}+\frac {\sin(\pi x)}\pi\sum_{n=0}^\infty (-1)^{n+1}\,\frac {(2n+1)!}{(\pi x)^{2n+2 }} $$

Since the maxima take place at odd values of $x$ and the minima at even values of $x$, we do not care about the second summation and then the lower bound is $$\frac{1}{2}-\frac{1}{\pi ^2 x}+\frac{2}{\pi ^4 x^3}-\frac{24}{\pi ^6 x^5}+O\left(\frac{1}{x^7}\right)$$ and the upper bound is $$\frac{1}{2}+\frac{1}{\pi ^2 x}-\frac{2}{\pi ^4 x^3}+\frac{24}{\pi ^6 x^5}+O\left(\frac{1}{x^7}\right)$$

You had almost the first order since $3\sim\pi$ .

Edit

If you prefer more compact forms, using Padé approximants with $t=\pi x$ $$\frac 12 - \frac 1 \pi \frac{t^2+10}{t \left(t^2+12\right)}<\frac 1 \pi \text{Si}(\pi x) < \frac 12 + \frac 1 \pi \frac{t^2+10}{t \left(t^2+12\right)}$$

$$\frac 12 - \frac 1 \pi \frac{t \left(5 t^2+158\right)}{5 t^4+168 t^2+216}<\frac 1 \pi \text{Si}(\pi x) < \frac 12 +\frac 1 \pi \frac{t \left(5 t^2+158\right)}{5 t^4+168 t^2+216}$$ whose errors are $\frac{432}{\pi ^7 x^7}$ and $\frac{85824}{5 \pi ^9 x^9}$.

Answer 2

Partial answer. By $(6.2.19)$, we have $$ \frac{1}{\pi }\operatorname{Si}(\pi x) = \frac{1}{2} - \frac{1}{\pi }\left( {\operatorname{f}(\pi x)\cos (\pi x) + \operatorname{g}(\pi x)\sin (\pi x)} \right), $$ where, by $\S6.12.\text{ii}$, $$ 0 < \operatorname{f}(z) < \frac{1}{z},\quad 0 < \operatorname{g}(z) < \frac{1}{{z^2 }} $$ for any $z>0$. Consequently, $$ \frac{1}{2} - \frac{1}{{\pi ^2 x}} - \frac{1}{{\pi (\pi x)^2 }} < \frac{1}{\pi }\operatorname{Si}(\pi x) < \frac{1}{2} + \frac{1}{{\pi ^2 x}} + \frac{1}{{\pi (\pi x)^2 }} $$ for $x>0$. If $x > \frac{3}{{\pi ^2 - 3\pi }} = 6.744203419 \ldots$, then $$ \frac{1}{{\pi ^2 x}} + \frac{1}{{\pi (\pi x)^2 }} < \frac{1}{{3\pi x}}. $$ This implies $$ \frac{1}{2} - \frac{1}{{3\pi x}} < \frac{1}{\pi }\operatorname{Si}(\pi x) < \frac{1}{2} + \frac{1}{{3\pi x}}, $$ for $x > \frac{3}{{\pi ^2 - 3\pi }} = 6.744203419 \ldots$.

Answer 3

I'm reasonably confident that the answer is contained within the Digital Library of Mathematical Functions. The relevant section seems to be this one, although it requires a bit of cross-referencing to work out exactly what it says.