Supposedly I have to use the bonnet theorem and Riemann Stieltjes integration to prove this inequality:
$$\left|\int^{x+1}_x\sin(e^t)dt\right| \le \frac{2}{e^x}$$
With $x>0$, I am struggling to find what would be the point $c\in\left(x,x+1\right)$ in which the function increases or decreases. The function is positive when $2\pi n<x<2\pi n+\pi$ and negative when $2\pi n-\pi<x<2\pi n$, but in those intervals the function increases and decreases. I've also tried the first and second derivative test, both periodic answers, and I have doubts that in such a little interval there's room for the function to increase and decrease so I can split the integral like so, $$\int^{b}_af(x)d\alpha(x)=f(a)\int^{c}_a f(x)d\alpha(x)+f(b)\int^{b}_c f(x)d\alpha(x)=f(a)\left(\alpha(c)-\alpha(a) \right)+f(b)\left(\alpha(b)-\alpha(c) \right)$$
Can't figure out how to split the interval, any pointer will be appreciated.
[adendum]
This is a partial result that I got, I'll added just for context because I think it amounts to nothing:
$$\left|F(x)\right|=\left|\int^{x+1}_x\sin(e^t)dt\right|=\int^{x+1}_x\left|\sin(e^t)\right|dt=\int^{x+1}_x\left|\sum\limits^{\infty}_{n=0}\frac{(-1)^n(e^t)^{1+2n}}{(2n)!}\right|dt=\\\int^{x+1}_x\sum\limits^{\infty}_{n=0}\frac{\left|(-1)^n(e^t)^{1+2n}\right|}{\left|(2n)!\right|}dt=\int^{x+1}_x\sum\limits^{\infty}_{n=0}\frac{(e^t)^{1+2n}}{(2n)!}dt=\int^{x+1}_xe^t\cosh{e^t}dt=\\\sinh{e^{x+1}}-\sinh{e^{x}}=\frac{1}{2}\left(e^{-e^x}-e^{e^x}-e^{-e^{x+1}}+e^{e^{x+1}}\right)$$
Kind regards
Making the variable change $u = e^t$ we have $du = e^t \,dt = u \, dt$ and
$$\int^{x+1}_x\sin(e^t)\,dt = \int^{e^{x+1}}_{e^x} \frac{\sin u}{u} \,du $$
Since $g:u \mapsto \frac{1}{u}$ is a nonnegative decreasing function, by Bonnet's mean value theorem there exists $\xi \in [e^x, e^{x+1}]$ such that
$$\int^{e^{x+1}}_{e^x} \frac{\sin u}{u} \,du = \int^{e^{x+1}}_{e^x} \sin u \,g(u) \,du = \frac{1}{g(e^x+)}\int_{e^x}^\xi \sin u \, du= \frac{1}{e^x}\int_{e^x}^\xi \sin u \, du= \frac{\cos e^x - \cos \xi}{e^x}$$
Note that this form of Bonnet's theorem is found here under the section entitled "Second mean value theorem for definite integrals".
Thus,
$$\left|\int^{x+1}_x\sin(e^t)\,dt\right| = \left|\frac{\cos e^x - \cos \xi}{e^x} \right| \leqslant \frac{2}{e^x}$$