Problem :
Let $x\geq 2$ : $$\int_{0}^{x}\frac{\sin\left(t\right)}{t}dt>\frac{\frac{\pi}{2}x-1}{x}+\frac{\left(1-e^{-x}\right)}{x^{3}}>\arctan(x)$$
I can show one of the side let introduce :
$$f(x)=\frac{\frac{\pi}{2}x-1}{x}+\frac{\left(1-e^{-x}\right)}{x^{3}}-\arctan(x)$$
Then :
$$f'(x)=(-1/(x^{2}+1)+e^{-x}(x+3)-2)/x^{4}$$
We have :
$$e^{-2}(2+3)<2$$
So the derivative is clearly negative and it limits is zero so we have the inequality .
For the other side I have tried power series without any effect .Also in the second reference there is a link with exponential integral so perhaps we can use complex analysis .
Using @JackD'aurizio answer we can compare :
$$\frac{\cos\left(a\right)+x\sin\left(a\right)}{\left(1+x^{2}\right)e^{ax}},\frac{f'\left(xa\right)}{a}$$
Where :
$$f\left(x\right)=\frac{1}{x}-\frac{\left(1-e^{-x}\right)}{x^{3}}$$
How to show it ?
Reference :
Prove that $\int_0^x \frac{\sin t}{t} dt > \arctan x $ for $x>0$
