4

For all $x\in\mathbb{R}\backslash\{0\}$ the cardinal sine function $\text{sinc}(x) = \sin(x)/x$ is trivially bounded by

$$ |\text{sinc}(x)| \le \frac{1}{|x|},$$

since $\sin(x)\le 1$. I am wondering if the sine integral can similarly be bounded by

$$ \left|\text{Si}(x) - \frac{\pi}{2}\right| = \left|\text{si}(x)\right| \le \frac{1}{|x|} \qquad\forall x>0,$$

where $\text{Si}(x) = \int_0^x \text{sinc}(y)\,\mathrm{d}y$ and $\text{si}(x) = -\int_x^\infty \text{sinc}(y)\,\mathrm{d}y$. When plotting the graphs it looks like this might be true.

2 Answers2

5

$$\frac{\pi}{2}-\text{Si}(x) = \int_{x}^{+\infty}\frac{\sin(t)}{t}\,dt = \int_{0}^{+\infty}\frac{\sin(x)\cos(t)+\cos(x)\sin(t)}{x+t}\,dt $$ due to the Laplace transform, can be written as $$ \int_{0}^{+\infty}\frac{\cos(x)+s\sin(x)}{1+s^2}\,e^{-sx}\,ds $$ which by the Cauchy-Schwarz inequality is bounded by $$ \int_{0}^{+\infty}\frac{e^{-sx}}{\sqrt{1+s^2}}\,ds < \int_{0}^{+\infty}e^{-sx}\,ds=\frac{1}{x},$$ hence your conjecture is correct.

Jack D'Aurizio
  • 361,689
  • Nice proof. Can you provide a reference (preferably a book) for this and other inequalities related to the trigonometric integral functions and the exponential integral? –  Oct 15 '18 at 15:59
  • @Chris: I am not sure I know one. Actually I just applied a useful integral transform (the Laplace transform) and a useful algebraic inequality (CS). On the other hand you may find multiple instances of such combination in my notes. – Jack D'Aurizio Oct 15 '18 at 16:09
1

I could not understand the above proof, so make a new one based on change of variables (first step) integration by parts (2nd step) as \begin{align} \left| \int_x^\infty \frac{\sin(t)}{t} dt \right| & = \left| \int_1^\infty \frac{\sin(xt)}{t} dt \right| \\ & = \left| \int_1^\infty \frac{\Im \exp(\Im x t)}{t} dt \right| \\ & = \frac{1}{x}\left| -e^{\Im } +\int_1^{\infty} \frac{e^{\Im t}}{t^2} dt \right| \\ & \leq \frac{2}{x} \end{align}

Hadi Daneshmand
  • 11