# 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).

• Your initial trick does not seem to work because $\sin(e^t)$ does not have constant sign. The proof of the upper bound on Stackexchange gives $\cos(e^x)+1-(cos(e^{x+1}+1)/e$, which is $(\cos(e^x)-e^{-1}\cos(e^x))+(1-1/e)$. The first term is bounded above by $\sqrt{1+1/e^2}\approx 1.07$, and the second is $\approx 0.63$. This gives an upper bound of $1.70$. Not yet $1.40$.
– ACL
Mar 10 '20 at 22:23
• Did you worked on the first and second derivative of $f(x)$? Mar 10 '20 at 22:35
• It would be better to suppress the erroneous "trick" to not mislead the subsequent readers, probably ? Mar 11 '20 at 6:29
• Thanks to @ACL and @ VladimirDotsenko for noting that the trick is erroneous (now removed) Mar 11 '20 at 8:25

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 $$em!/2=\text{integer}+1/2+\text{small}$$.

• I think you switched integration variables from $t$ to $x$, which is somewhat confusing. Mar 10 '20 at 23:28
• Very nice! Integration by parts, of course, to alleviate the oscillations. Somehow, I had not noticed your answer before completing mine. By the way, we both seem to be using that $e/\pi$ is irrational -- do you know if that is so? Mar 11 '20 at 2:30
• @IosifPinelis hm, it looks that I need that $e$ is irrational, not $e/\pi$. Mar 11 '20 at 6:49
• @TonyHuynh : Thank you for the information. Mar 11 '20 at 13:03
• @FedorPetrov: Right, my bad. Mar 11 '20 at 13:04

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 $$k=0,1,\dots$$.

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.