Prove that $\int_0^x \frac{\sin t}{t} dt > \arctan x $ for $x>0$

Question

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?

Answer 1

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 $k>1$.

Answer 2

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: $$ \forall x>0,\qquad \text{Si}(x)>\arctan(x).\tag{3} $$

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

Answer 3

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

Answer 4

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) :

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