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 Math SE a few days ago: How to prove $e^x\left|\int_x^{x+1}\sin(e^t) \,\mathrm d t\right|\le 1.4$?
A few days ago, somebody asked How to prove $ \mathrm{e}^x\left|\int_x^{x+1}\sin\mathrm e^t \mathrm d t\right|\leqslant 2$? on Math StackExchange.
However, this bound does not appear to be sharp so I was wondering how to find the maxima/minima of $$f(x)=e^x\int_x^{x+1}\sin(e^t) \,\mathrm d t$$
or at least how 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 $g(z)$:

Also, notice that $g$ is analytic and $g'(z)=\sin (e z)-\sin (z)+\text{Si}(e z)-\text{Si}(z)$ which might help for the search of critical points (although I don't think that $g'(z)=0$ has closed form solutions).
 A: Integrate by parts:
\begin{align}
\int_x^{x+1}\sin(e^t)dt
& =\int_x^{x+1}e^{-t}d(-\cos(e^t)) \\
& =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-\int_x^{x+1}e^{-t}\cos e^{t}dt\\
& =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-\int_x^{x+1}e^{-2t}d\sin e^{t}\\
& =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-e^{-2(x+1)}\sin e^{x+1}\\
& \hphantom{={}}+e^{-2x}\sin e^x+2\int_x^{x+1}e^{-2t}\sin e^tdt.\end{align}
From here we see that $e^x \int_x^{x+1}\sin(e^t)dt$ is bounded by $1+1/e+O(e^{-x})$ and $1+1/e\approx 1.368$ can not be improved, since both $\cos e^x$ and $-\cos e^{x+1}$ may be almost equal to 1:
if $e^x=2\pi n$ for large integer $n$, then $e^{x+1}=2\pi e n$, we want this to be close to $\pi+2\pi k$, i.e., we want $en$ to be close to $\frac12+k$.
This is possible since $e$ is irrational. Moreover, $e$ is so special number that you may find explicit $n$ for which $en$ is nearly half-integer: $n=m!/2$ for large even $m$ works. Indeed, $e=\sum_{i=0}^{m-1}1/i!+1/m!+o(1/m!)$ yields $em!/2=\text{integer}+1/2+\text{small}$.
A: This can be done with help of Maple in such a way. First, we find the estimated expression explicitly by
 a := (exp(x)*int(sin(exp(t)), t = x .. x + 1) assuming x::real;

$ {{\rm e}^{x}} \left( -{\it Si} \left( {{\rm e}^{x}} \right) +{\it Si}
 \left( {{\rm e}^{x+1}} \right)  \right) 
$
In fact, the integral is reduced to another integrals. Next, the asymptotics of $a$ is found. Maple is not able to find this asymptotics directly so the change $x=\log y$ should be used:
asympt(simplify(eval(a, x = log(y))), y, 2);

$-{\frac {\cos \left( y{\rm e} \right) }{{\rm e}}}+\cos \left( y
 \right) +O \left( {y}^{-1} \right)
 $
Now we return to $x$ by
eval(%, y = exp(x));

$-{\frac {\cos \left( {{\rm e}^{x}}{\rm e} \right) }{{\rm e}}}+\cos
 \left( {{\rm e}^{x}} \right) +O \left(  \left( {{\rm e}^{x}} \right) 
^{-1} \right) $
The rest is as in the Fedor Petrov's answer.
A: Here is a method that will allow one to find the exact upper and lower bounds on $g(z)$ over $z>0$ with any degree of accuracy. 
Take any real $z>0$. Since 
\begin{equation*}
 \frac1y=\int_0^\infty dt\,e^{-y t}
\end{equation*}
for any real $y>0$, we have
\begin{align*}
 \frac{g(z)}z
 &=\int_z^{e z} dy\, \frac{\sin y}y \\ 
 &=\int_0^\infty dt\,\int_z^{e z} dy\,e^{-y t}\sin y \\ 
 &=\int_0^\infty dt\,
\Big( \frac{e^{-t z} (\cos z+t \sin z)}{t^2+1}
-\frac{e^{-e t z} (\cos ez+t \sin ez)}{t^2+1}\Big) \\ 
 &=I_1(z) \cos z+I_2(z)\sin z -I_1(ez) \cos ez-I_2(ez)\sin ez,  \tag{1}
\end{align*}
where 
\begin{align*}
 I_1(z)&:=\int_0^\infty dt\,\frac{e^{-t z}}{t^2+1}, \\ 
 I_2(z)&:=\int_0^\infty dt\,\frac{e^{-t z}t}{t^2+1}. 
\end{align*}
Next, letting $c_1$ and $c_2$ denote functions with values in $(0,1)$, we have 
\begin{align*}
 I_1(z)&=\frac1z\,\int_0^\infty du\,\frac{e^{-u}}{1+u^2/z^2} \\ 
 &=\frac1z\,\int_0^\infty du\,e^{-u} 
 -\frac1z\,\int_0^\infty du\,\frac{u^2e^{-u}}{z^2+u^2} \\ 
 &=\frac1z-\frac{2c_1(z)}{z^3};
\end{align*}
at the last step here, we used the inequality $z^2+u^2>z^2$ for $u>0$;
similarly, 
\begin{align*}
 I_2(z)&=\frac1{z^2}-\frac{3c_2(z)}{z^4}. 
\end{align*}
So, by (1), 
\begin{equation*}
 g=h+r,
\end{equation*}
where 
\begin{equation*}
 h(z):=\cos z-\tfrac1e\,\cos ez
\end{equation*}
and
\begin{equation*}
 r(z):=-\frac{2c_1(z)}{z^2}\, \cos z-\frac{3c_2(z)}{z^3}\,\sin z 
 +\frac{2c_1(ez)}{e^3z^2}\, \cos ez+\frac{3c_2(2z)}{e^4z^3}\,\sin ez 
\end{equation*}
is the "remainder", 
so that 
\begin{equation*}
 |r(z)|<\frac{2.1}{z^2}+\frac{3.1}{z^3},
\end{equation*}
which can be made however small if $z$ is large enough. 
On the other hand, since $e$ is irrational, we will have 
\begin{equation*}
 \sup_{z>0}h(z)=-\inf_{z>0}h(z)=1+1/e=1.367\dots 
\end{equation*}
(which is somewhat close to your value $1.4$). 
So, to compute $\sup_{z>0}g(z)$ and $\inf_{z>0}g(z)$ with any degree of accuracy, it suffices to be able to compute $\sup_{z\in(0,a]}g(z)$ and $\inf_{z\in(0,a]}g(z)$ with any degree of accuracy for any given real $a>0$, which can be done by (say) the interval arithmetic method, using the formula $g(z)=z(\text{Si}(e z)-\text{Si}(z))$ and the monotonicity of the function $\text{Si}$ on each of the intervals of the form $[k\pi,(k+1)\pi]$ for $k=0,1,\dots$. 
