Prove that $\int_0^x \frac{\sin t}{t} dt > \arctan x $ for $x>0$. 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?
 A: 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
$$ \text{Si}(x) > \frac{\pi}{2}-\sqrt{2}\,e^x\,\Gamma(0,x).\tag{4}$$
A: $$\text{Si}'(x) = \frac{\sin x}{x} = \frac{\sum_k (-1)^kx^{2k+1}/(2k+1)!}{x} = \sum_k (-1)^kx^{2k}/(2k+1)!$$
$$\arctan'(x) = \frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum_k (-x^2)^k = \sum_k (-1)^kx^{2k}$$
$$\text{Si}(x) = \sum_k \frac{(-1)^kx^{2k+1}/(2k+1)}{(2k+1)!}$$
$$\arctan(x) = \sum_k \frac{(-1)^kx^{2k+1}/(2k+1)}{1}$$
$$\frac{1}{(2k+1)!} \geq \frac11$$
This would give the opposite of your inequality... but we must take into account the $(-1)^k$.
