How to prove $e^x\left|\int_x^{x+1}\sin(e^t) \,\mathrm d t\right|\le 1.4$? Related question asked by me on MathOverflow: How to prove $e^x\left|\int_x^{x+1}\sin(e^t) \,\mathrm d t\right|\le 1.4$?
This is a follow-up question to the question How to prove $ \mathrm{e}^x\left|\int_x^{x+1}\sin\mathrm e^t \mathrm d t\right|\leqslant 2$?, in which a weaker bound is proven using a nice trick.
Now my question is how to maximize and minimize $$f(x)=e^x\int_x^{x+1}\sin(e^t) \,\mathrm d t$$
or at least to prove $-1.4\le f(x)\le 1.4$. 
Some observations, using the substitution $y=e^t$:
$$f(x)=e^x \int_{e^x}^{e^{x+1}} \frac{\sin(y)}y\,\mathrm dy=g(e^x),$$
where I have defined $$g(z)=z \int_z^{e z} \frac{\sin(y)}y\,\mathrm dy = z (\operatorname{Si}(e z)-\operatorname{Si}(z)).$$
($\operatorname{Si}$ is the Sine integral.)
So the question reduces to: What are the maxima/minima of $g(z)$ for $z\geq 0$ ?
Using the series of $\mathrm{Si}(z)$, we get 
$$g(z)=\sum_{k=1}^\infty (-1)^{k-1} \frac{z^{2k}(e^{2k-1}-1)}{(2k-1)!\cdot(2k-1)}$$
and here is a plot of $g(z)$, which seems to be periodic:

 A: By the result of [1], we have
$$\left|\mathrm{Si}(y) - \frac{\pi}{2}\right| \le \frac{1}{y}, \ y > 0.\tag{1}$$
Thus, we have, for $z > 0$,
$$\frac{\pi}{2} - \frac{1}{z} \le \mathrm{Si}(z) \le \frac{\pi}{2} + \frac{1}{z}$$
and
$$\frac{\pi}{2} - \frac{1}{\mathrm{e}z} \le \mathrm{Si}(\mathrm{e}z) \le \frac{\pi}{2} + \frac{1}{\mathrm{e}z}.$$
Thus, we have, for $z > 0$,
$$-1 - \frac{1}{\mathrm{e}} \le z(\mathrm{Si}(\mathrm{e}z) - \mathrm{Si}(z)) \le 1 + \frac{1}{\mathrm{e}}.$$
Also, $1 + \frac{1}{\mathrm{e}} \approx 1.367879441$. We are done.
Reference
[1] Upper bound for the sine integral
Remark: The following stronger inequality holds, which implies (1):
$$\left|\mathrm{Si}(y) - \frac{\pi}{2}\right| \le \arctan \frac{1}{y}, \ y > 0.$$
See: Graham Jameson, Nick Lord and James McKee, An inequality for Si(x), Math. Gazette 99 (2015).
https://www.maths.lancs.ac.uk/jameson/siineqnotes.pdf
A: This question has received some answers on MathOverflow.

*

*Fyodor Petrov solves with integration by parts

*Iosif Pinelis suggests a method that allows arbitrarily accurate bounds with interval arithmetic
