11

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?

River Li
  • 37,323
Ramanasa
  • 440
  • 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. – blub Aug 11 '18 at 03:26
  • I tried to but could not proceed more. It seems there needs some more manipulations. – Ramanasa Aug 11 '18 at 03:34
  • I'm currently writing a sketch of some thoughts. – blub Aug 11 '18 at 03:34
  • If you can deal with power series, then write both $\operatorname{Si}(x)$ and $\arctan (x)$ as power series and note their difference when $x>0.$ – Allawonder Aug 11 '18 at 03:43
  • @Allawonder Power series seems to be a very nice approach. – blub Aug 11 '18 at 03:50
  • 2
    @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 $$ \mathcal{L}\left(\text{Si}(x)\right)(s) = \frac{1}{s}\arctan\frac{1}{s}.$$ – Jack D'Aurizio Aug 11 '18 at 03:52
  • @JackD'Aurizio That's indeed very interesting, I've just skipped to consider the radius of convergence for $\arctan$. – blub Aug 11 '18 at 03:54
  • I also tried a power series method but found difficulty as the power series of $arctan(t)$ converges only in $|t|\le1$. – Ramanasa Aug 11 '18 at 04:00
  • @JackD'Aurizio Oh, i haven't realized that two functions have such an interesing relation. It seems that they are related in some profound manner. – Ramanasa Aug 11 '18 at 04:09
  • @Ramanasa: Indeed I found a way to exploit such relation. My previous answer was a mess, but the actual one is pretty straightforward to follow. – Jack D'Aurizio Aug 11 '18 at 04:45
  • Another approach is to note that $\arctan$ is bounded. This is well known. Since we also know its upper bound, one may try to show instead that $\text{Si}$ exceeds $π/2$ whenever $x>0.$ – Allawonder Aug 11 '18 at 04:51
  • @Allawonder: $\lim_{s\to +\infty}\text{Si}(s)=\frac{\pi}{2}$ and $\frac{\sin s}{s}$ has plenty of simple real roots, so as $s\to +\infty$ the value of $\text{Si}(s)$ oscillates around $\frac{\pi}{2}$. – Jack D'Aurizio Aug 11 '18 at 04:55
  • @JackD'Aurizio Wow. This then is a very interesting problem. – Allawonder Aug 11 '18 at 04:58
  • @Allawonder: on the other hand $\text{Si}(s)\approx \frac{\pi}{2}-\frac{\cos s}{s}$ and $\arctan(s)\approx \frac{\pi}{2}-\frac{1}{s}$ when $s$ is large, so it is not extremely surprising that $\text{Si}(s)>\arctan(s)$ holds for any $s>0$. – Jack D'Aurizio Aug 11 '18 at 04:59
  • @JackD'Aurizio Oh, maybe. But I was referring to the problem of proving the result as interesting, not necessarily to the result itself. – Allawonder Aug 11 '18 at 05:02

4 Answers4

8

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 $$ \text{Si}(x) > \frac{\pi}{2}-\sqrt{2}\,e^x\,\Gamma(0,x).\tag{4}$$

Jack D'Aurizio
  • 353,855
  • 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 $Si(x)=L^{-1}\left(\frac1{s}\arctan\frac{1}{s}\right)(x)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}{e^{sx}\frac1{s}\arctan\frac{1}{s}ds}$? – Ramanasa Aug 11 '18 at 05:09
  • @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 $(1)$. – Jack D'Aurizio Aug 11 '18 at 05:14
  • The magic I am invoking to get rid of the annoying oscillations is the self-adjointness of the Laplace transform: https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_integrals_over_the_positive_real_axis – Jack D'Aurizio Aug 11 '18 at 05:15
  • 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. – Jack D'Aurizio Aug 11 '18 at 05:19
  • 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. – Ramanasa Aug 11 '18 at 05:46
  • @Ramanasa: again, $$\int_{0}^{+\infty} f(x)g(x),dx = \int_{0}^{+\infty}\left(\mathcal{L}f\right)(s)\left(\mathcal{L}^{-1}g\right)(s),ds,$$ or, in other terms, $\langle a,\mathcal{L} b\rangle = \langle \mathcal{L} a,b\rangle$, which is the self-adjointness. – Jack D'Aurizio Aug 11 '18 at 05:47
  • OH, I did not know that there was such a formula. Thanks for referring that. Now I would continue to follow your method. – Ramanasa Aug 11 '18 at 05:50
  • Following your definition, \begin{align} S(x,s)&:=\frac{1}{e^{sx}\sqrt{1+s^2}},\ T(x,s)&:=\frac{1+2sx}{(1+s^2)(1+2sx+x^2+s^2x^2)}. \end{align} But taking $s\rightarrow0+$, $$ S(x,0+)=1> T(x,0+)=\frac{1}{1+x^2}, $$ implying that $$ S(x,s)\le T(x,s) $$ doesn't hold for sufficiently small $s>0$.

    So it might need some modification.

     – Ramanasa Aug 11 '18 at 06:33
  • @Ramanasa: you are right, but the approach is salvageable through a step of integration by parts applied to the RHS of $(1)$, leading to a comparison between $\frac{\text{arcsinh(s)}}{x e^{xs}}$ and a rational function. I will fix it later, I am a bit busy at the moment. – Jack D'Aurizio Aug 11 '18 at 06:36
  • Thanks a lot. Again i appreciate for your insight and effort. I would also try to finish it. – Ramanasa Aug 11 '18 at 06:45
  • After several attempts, I found it difficult to control the integrand function and complete the inequality. That might be an another nontrivial problem. So I think the proof is still open, despite of a very good break through. – Ramanasa Aug 15 '18 at 09:08
  • I turned this on to another question. Please look at https://math.stackexchange.com/questions/2884723/prove-int-0-infty-frac1esx-sqrt1s2ds-arctan-left-frac1x-ri. – Ramanasa Aug 16 '18 at 12:54
  • and https://mathoverflow.net/questions/308511/prove-int-0-infty-frac1esx-sqrt1s2ds-arctan-left-frac1x-ri. – Ramanasa Aug 17 '18 at 08:49
4

Remarks: You said "could not find any references regarding this". Here is a reference.

This inequality is proved in this article:

Graham Jameson, and Nick Lord and James McKee, “An inequality for Si(x)”, Math. Gazette 99 (2015). https://www.maths.lancs.ac.uk/jameson/siineqnotes.pdf

River Li
  • 37,323
0

Power series approach, for $x\leq1$:

$$\text{Si}(x)=\sum_k\frac{(-1)^kx^{2k+1}}{(2k+1)!\,(2k+1)}$$ $$\arctan(x)=\sum_k \frac{(-1)^kx^{2k+1}}{(2k+1)}$$ $$\text{Si}(x)-\arctan(x)=\sum_k\frac{(-1)^{k-1}x^{2k+1}}{2k+1}\left(1-\frac1{(2k+1)!}\right)$$ $$=\frac{5}{18}x^3-\frac{119}{600}x^5+\frac{5039}{35280}x^7-\cdots$$ $$\overset?>0$$

Since this is an alternating series and the first term is positive, it suffices to show that the terms are always getting smaller:

$$\frac{x^{2k+1}}{2k+1}\left(1-\frac1{(2k+1)!}\right)\overset?<\frac{x^{2k-1}}{2k-1}\left(1-\frac1{(2k-1)!}\right)$$

We certainly have $x^{2k+1}\leq x^{2k-1}$ and $1-1/(2k+1)!<1$, so what remains is

$$\frac{1}{2k+1}\overset?<\frac{1}{2k-1}\left(1-\frac1{(2k-1)!}\right)$$ $$\frac{2k-1}{2k+1}=1-\frac{2}{2k+1}\overset?<1-\frac{1}{(2k-1)!}$$ $$\frac{2k+1}{2}\overset?<(2k-1)!$$

which is evidently true for $k>1$.

mr_e_man
  • 5,364
-1

There is someting to dig with $\tanh(x)$ because :

For $x>0$ it seems we have :

$$\int_{0}^{x}\frac{\sin\left(t\right)}{t}dt>\arctan\left(x\right)>\tanh\left(x\right)-1+\frac{\frac{\pi}{2}x-1}{x}$$

Improving it we have the following conjecture :

It seems we have for $x\geq 2$ :

$$\int_{0}^{x}\frac{\sin\left(t\right)}{t}dt>\frac{\frac{\pi}{2}x-1}{x}+\frac{\left(1-e^{-x}\right)}{x^{3}}>\arctan(x)$$

As other attempt we can try brute force with :

There exists $a,b,x,n>0$ such that $x\in[a,b],c=-\pi$ then it seems we have :

$$\arctan(x)<\frac{\frac{\pi}{2}x-1}{x}+\frac{\left(1-e^{-nx\sin\left(x\right)}\right)}{x^{3}}<Si(x)$$

Or :

$$\arctan(x)<\frac{\frac{\pi}{2}x-1}{x}+\frac{\left(1-e^{-n\left(x+c\right)\sin\left(x+c\right)}\right)}{\left(x+c\right)^{3}}<Si(x)$$

Using the derivative (without $n$ for the first and with for the second) :

$$f(x)=(x^{2}-3)/x^{4}-\left(e^{-\left(x\cos\left(x\right)\right)}\left(x^{2}\sin(x)-x\cos(x)-3\right)\right)/x^{4},g(x)=\left(1/x^{2}+\left(3\left(e^{\left(nx+n\pi\right)\sin\left(x)\right)}-1\right)\right)/\left(x+\pi\right)^{4}+e^{\left(nx+n\pi\right)\sin x}\right)\left(-\left(\left(n\sin x\right)/(x+\pi)^{3}-\left(n\cos x\right)/(x+\pi)^{2})\right)\right)$$

Miss and Mister cassoulet char
  • 3,511