I'm finding some bounds for the Si function defined as $$ \operatorname{Si}(x) := \int_0^x\frac{\sin t}{t}dt. $$ I observed from WolframAlpha that the inequality $$ \operatorname{Si}(x)>\arctan(x) $$ holds for $x>0$.

I tried to show this analytically but failed and could not find any references regarding this. Could someone help me with this?

  • $\begingroup$ I don't know if this helps but maybe try to express $\arctan(x)$ as the integral from $0$ to $x$ of its derivative $\frac{1}{t^2+1}$ and then form one integral. $\endgroup$ – blub Aug 11 '18 at 3:26
  • $\begingroup$ I tried to but could not proceed more. It seems there needs some more manipulations. $\endgroup$ – Ramanasa Aug 11 '18 at 3:34
  • $\begingroup$ I'm currently writing a sketch of some thoughts. $\endgroup$ – blub Aug 11 '18 at 3:34
  • $\begingroup$ If you can deal with power series, then write both $\operatorname{Si}(x)$ and $\arctan (x)$ as power series and note their difference when $x>0.$ $\endgroup$ – Allawonder Aug 11 '18 at 3:43
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    $\begingroup$ @zzuussee: but a faulty one, since $\arctan(x)$ is not an entire function. On the other hand an interesting relation between $\arctan$ and $\text{Si}$ is $$ \mathcal{L}\left(\text{Si}(x)\right)(s) = \frac{1}{s}\arctan\frac{1}{s}.$$ $\endgroup$ – Jack D'Aurizio Aug 11 '18 at 3:52

All right, I realized that representing $\arctan(x)$ through the integral of an oscillating function is not a good idea. Better to represent both $\arctan(x)$ and $\text{Si}(x)$ as integrals of monotonic and easily-comparable functions. So here it is a polished version of the previous answer. We may safely assume $x>1$ since power series easily prove the statement for $x\in[0,1]$. By the Laplace transform and the Cauchy-Schwarz inequality

$$ \text{Si}(x)=\frac{\pi}{2}-\int_{0}^{+\infty}\frac{\cos(x)+s\sin(x)}{(1+s^2)e^{sx}}\,ds\geq \frac{\pi}{2}-\int_{0}^{+\infty}\frac{ds}{e^{sx}\sqrt{1+s^2}}. \tag{1}$$ By the very definition of $\arctan$ we have $\arctan(x)=\left(\int_{0}^{+\infty}\frac{1}{1+s^2}-\frac{1}{1+(s+x)^2}\right)\,dx$, hence through $\arctan x=\frac{\pi}{2}-\arctan\frac{1}{x}$ we get the following integral representation: $$ \arctan(x) = \frac{\pi}{2}-\int_{0}^{+\infty}\frac{1+2sx}{(1+s^2)(1+2sx+x^2+s^2 x^2)}\,ds. \tag{2}$$ For the sake of brevity, let us denote as $S(x,s)$ and $T(x,s)$ the integrand functions appearing in the RHSs of $(1)$ and $(2)$. If we manage to prove $S(x,s)\leq T(x,s)$ for any $x>1$ and any $s>0$ we are done. But the Padé approximants for the exponential function reveal that this is a pretty loose inequality, so we are good to go: $$ \forall x>0,\qquad \text{Si}(x)>\arctan(x).\tag{3} $$

A strange-looking consequence of $(1)$ and the AM-QM inequality is also $$ \text{Si}(x) > \frac{\pi}{2}-\sqrt{2}\,e^x\,\Gamma(0,x).\tag{4}$$

  • $\begingroup$ It's a really cool way. I did not know I needed such an advanced method to do this. I'm catching up your idea of inverse laplace transform. But could you explain more on the first formula expressing $Si(x)$? Am i able to reach your expression by just calculating $Si(x)=L^{-1}\left(\frac1{s}\arctan\frac{1}{s}\right)(x)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}{e^{sx}\frac1{s}\arctan\frac{1}{s}ds}$? $\endgroup$ – Ramanasa Aug 11 '18 at 5:09
  • $\begingroup$ @Ramanasa: $$ \text{Si}(x) = \int_{0}^{+\infty}\frac{1}{t}\cdot\sin(t)\mathbb{1}_{(0,x)}(t)\,dt $$ if we apply $\mathcal{L}^{-1}$ to $\frac{1}{t}$ and $\mathcal{L}$ to $\sin(t)\mathbb{1}_{(0,x)}(t)$, then recall that $\int_{0}^{+\infty}\frac{\sin t}{t}\,dt=\frac{\pi}{2}$, we reach the integral representation at the beginning of $(1)$. $\endgroup$ – Jack D'Aurizio Aug 11 '18 at 5:14
  • $\begingroup$ The magic I am invoking to get rid of the annoying oscillations is the self-adjointness of the Laplace transform: en.wikipedia.org/wiki/… $\endgroup$ – Jack D'Aurizio Aug 11 '18 at 5:15
  • $\begingroup$ The idea of combining the Laplace (or Mellin) transform with standard inequalities is indeed very powerful, have a look at this fantastic answer by Marko Riedl on a different topic, for instance. $\endgroup$ – Jack D'Aurizio Aug 11 '18 at 5:19
  • $\begingroup$ Thank you for your kind explanation. But still the part (1) doesn't make sense for me. In the expression $$ \operatorname{Si}(x)=\int_0^{\infty} \frac{1}{t}\cdot\sin t \cdot1_{(0,\infty)}dt $$ what exactly mean to apply $L^{-1}$ to $\frac1t$ and $L$ to $\sin t \cdot 1_{(0,\infty)}$? I do not know what property of laplace transform you used here. $\endgroup$ – Ramanasa Aug 11 '18 at 5:46

$$\text{Si}'(x) = \frac{\sin x}{x} = \frac{\sum_k (-1)^kx^{2k+1}/(2k+1)!}{x} = \sum_k (-1)^kx^{2k}/(2k+1)!$$ $$\arctan'(x) = \frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum_k (-x^2)^k = \sum_k (-1)^kx^{2k}$$

$$\text{Si}(x) = \sum_k \frac{(-1)^kx^{2k+1}/(2k+1)}{(2k+1)!}$$ $$\arctan(x) = \sum_k \frac{(-1)^kx^{2k+1}/(2k+1)}{1}$$

$$\frac{1}{(2k+1)!} \geq \frac11$$

This would give the opposite of your inequality... but we must take into account the $(-1)^k$.

  • 3
    $\begingroup$ While $\text{Si}(x)$ is an entire function, the radius of convergence of the Maclaurin series of $\arctan(x)$ is just $1$, so these series do not allow to compare $\text{Si}(x)$ and $\arctan(x)$ outside $|x|\leq 1$. $\endgroup$ – Jack D'Aurizio Aug 11 '18 at 3:50
  • $\begingroup$ That is true. This answer is incomplete anyway. $\endgroup$ – mr_e_man Aug 11 '18 at 3:52

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