Prove that $|Si(1/x) - \pi/2| \leq 2x$ for $x>0$

Question

Prove that $|Si(1/x) - \frac \pi 2| \leq 2x$ for $x>0$, where $Si(x)=\int_0^x \frac{\sin t}t \, dt$.

Answer 1

$|Si(x)- \pi /2|=|\int_{x}^{\infty} \frac {sin(t)}{t} dt |\leq| \int_{x}^{\infty} \frac {e^{it}}{t} dt|$

$=|[\frac {e^{it}}{it}]_{x}^{\infty}-\bigg (\int_{x}^{\infty} \frac {e^{it}}{-it^2} dt\bigg ) | $

$=|-[\frac {e^{it}}{ix}]+\bigg (\int_{x}^{\infty} \frac {e^{it}}{it^2} dt\bigg ) | $

$=|\frac 1{x}\bigg (\int_{1}^{\infty} \frac {e^{itx}-e^{ix}}{t^2} dt\bigg ) | $

$=|\frac 1{x}\bigg (\int_{1}^{\infty} \frac {e^{ix(t-1)}-1}{t^2} dt\bigg ) | $

so what it's all comes to; is proving :

$=|\bigg (\int_{1}^{\infty} \frac {e^{ix(t-1)}-1}{t^2} dt\bigg ) | \leq 1$

let $v=x(t-1)$ :

$=|\bigg (\int_{1}^{\infty} \frac {e^{ix(t-1)}-1}{t^2} dt\bigg ) | \leq 1$

is equivalent to :

$=|\bigg (\int_{0}^{\infty} \frac {e^{iv}-1}{(v+x)^2}\times x dv\bigg ) | \leq 1$

$|\bigg (\int_{0}^{\infty} \frac {e^{iv}-1}{(v+x)^2}\times x dv\bigg ) | = |[\frac {i(1-e^{iv})-v}{(v+x)^2}]+\bigg (\int_{0}^{\infty} \frac {i(1-e^{iv})-v}{2(v+x)^3}\times x dv\bigg ) |=|\bigg (\int_{0}^{\infty} \frac {i(1-e^{iv})-v}{2(v+x)^3}\times x dv\bigg ) |$

we use $ |(1-e^{iv})|\leq |v|$

$\leq |\bigg (\int_{0}^{\infty} \frac {2v}{2(v+x)^3}\times x dv\bigg ) |=|\bigg (\int_{0}^{\infty} \frac {vx}{(v+x)^3} dv\bigg ) | = |\bigg (\int_{0}^{\infty} \frac {x}{2(v+x)^2} dv\bigg ) |$

last equality is obtained by integration by parts

therefore

$|\bigg (\int_{0}^{\infty} \frac {x}{2(v+x)^2} dv\bigg ) |\leq \frac 1{2}$

Answer 2

I will prove it for $x = 2n\pi$ where $n \in \mathbb Z$:

First, restate the inequality to be proved as $$|\operatorname{Si}(x) - \frac{\pi}2| < \frac1x$$

Then

$$\int_0^{2n\pi} \frac{\sin t}{t}\,dt = \frac\pi2 - \int_{2n\pi}^\infty \frac{\sin t}{t}\,dt = \sum_{k=n}^\infty \left(\int_{2n\pi}^{(2n+1)\pi}\frac{\sin t}t\,dt + \int_{(2n+1)\pi}^{2(n+1)\pi}\frac{\sin t}t\,dt\right).$$

Now notice that $$\frac{1}{(2n+1)\pi}<\int_{2n\pi}^{(2n+1)\pi}\frac{\sin t}t\,dt < \frac{1}{2n\pi}$$ and $$-\frac{1}{(2n+1)\pi}<\int_{(2n+1)\pi}^{2(n+1)\pi}\frac{\sin t}t\,dt < -\frac{1}{2(n+1)\pi},$$ so $$\sum_{k=n}^\infty \left(\frac{1}{(2n+1)\pi} - \frac{1}{(2n+1)\pi} \right)< \sum_{k=n}^\infty \left(\int_{2n\pi}^{(2n+1)\pi}\frac{\sin t}t\,dt + \int_{(2n+1)\pi}^{2(n+1)\pi}\frac{\sin t}t\,dt\right) < \sum_{k=n}^\infty \left(\frac{1}{2n\pi} - \frac{1}{2(n+1)\pi} \right)$$

The LHS series equals $0$ and the RHS series telescopes to $\frac{1}{2n\pi}$, so $$0 < \int_0^{2n\pi} \frac{\sin t}{t}\,dt < \frac{1}{2n\pi}.$$

A similar argument can be made for odd multiples of $\pi$.

The argument can be made in general by noticing that $[x\pi,(x+2)\pi]$ can be broken up into the two sets $[x\pi, \lceil x \rceil \pi] \cup [(1 + \lceil x \rceil) \pi, (x+2)\pi]$ and $[\lceil x \rceil \pi, (1 + \lceil x \rceil) \pi]$.