$Si(x)\leq Si(\pi)$ for every $x>0$ I need to show that $Si(x)\leq Si(\pi)$ for every $x>0$ where $Si(x)=\int_{0}^{x} \frac {\sin (t)}{t} dt$. I see it graphically but i can't prove it.
Thank you!
 A: As $Si'(x) = \frac{\sin x}{x}$, each local maximum occurs for some values of $x$ for which $\sin x = 0$, i.e., $x = k\pi$ for positive integers $k$, as $x > 0$. When $k$ is odd we have local maxima and when $k$ is even we have local minima. So if we can show that $$Si((2j+1)\pi) \leq Si(\pi) \ \ \ \text{ for all non-negative integers } j$$
we're done. It's probably worth drawing a diagram at this stage to see what we're doing. The proof that follows now is just formalizing a fairly intuitive idea: the area below the $x$-axis over an interval $[(2j+1)\pi, (2j+2)\pi]$ is always of greater magnitude that the area above the axis in the next interval $[(2j+2)\pi, (2j+3)\pi]$.
Let's use induction. The result is trivially true for $j = 0$. For arbitrary non-negative $j$,
$$Si((2j+3)\pi) = \int_0^{(2j+1)\pi} \frac{\sin x}{x} dx + \int_{(2j+1)\pi}^{(2j+3)\pi} \frac{\sin x}{x} dx = Si((2j+1)\pi) + \int_{(2j+1)\pi}^{(2j+3)\pi} \frac{\sin x}{x} dx $$
Now this last integral is negative as
$$\int_{(2j+1)\pi}^{(2j+3)\pi} \frac{\sin x}{x} dx = \int_{(2j+1)\pi}^{(2j+2)\pi} \frac{\sin x}{x} dx +\int_{(2j+2)\pi}^{(2j+3)\pi} \frac{\sin x}{x} dx$$ $$ =  \int_{(2j+1)\pi}^{(2j+2)\pi} \frac{\sin x}{x} dx +\int_{(2j+1)\pi}^{(2j+2)\pi} \frac{\sin (x+\pi)}{(x+\pi)} dx$$
$$ =  \int_{(2j+1)\pi}^{(2j+2)\pi} \frac{\sin x}{x} dx \ - \int_{(2j+1)\pi}^{(2j+2)\pi} \frac{\sin (x)}{(x+\pi)} dx$$
$$ <  \int_{(2j+1)\pi}^{(2j+2)\pi} \frac{\sin x}{x+\pi} dx \ - \int_{(2j+1)\pi}^{(2j+2)\pi} \frac{\sin (x)}{x+\pi} dx \ \ - (*)$$
as for $ x \in [(2j+1)\pi, (2j+2)\pi]$, $\sin x$ is negative and hence $\frac{\sin x}{x} < \frac{\sin x}{x+\pi}$. Finally, the expression (*) is zero.
Hence $Si((2j+3)\pi) < Si((2j+1)\pi)$ for arbitrary $j \geq 0$ and therefore $Si((2j+1)\pi) \leq Si(\pi)$ for all $j \geq 0$.
