14
$\begingroup$

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 $g(z)$:

enter image description here

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).

$\endgroup$
4
  • 6
    $\begingroup$ 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 $1.40$. $\endgroup$
    – ACL
    Mar 10, 2020 at 22:23
  • $\begingroup$ Did you worked on the first and second derivative of $f(x)$? $\endgroup$
    – Shahrooz
    Mar 10, 2020 at 22:35
  • 1
    $\begingroup$ It would be better to suppress the erroneous "trick" to not mislead the subsequent readers, probably ? $\endgroup$ Mar 11, 2020 at 6:29
  • 2
    $\begingroup$ Thanks to @ACL and @ VladimirDotsenko for noting that the trick is erroneous (now removed) $\endgroup$ Mar 11, 2020 at 8:25

3 Answers 3

37
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ I think you switched integration variables from $t$ to $x$, which is somewhat confusing. $\endgroup$
    – user44191
    Mar 10, 2020 at 23:28
  • 1
    $\begingroup$ 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? $\endgroup$ Mar 11, 2020 at 2:30
  • 1
    $\begingroup$ @IosifPinelis hm, it looks that I need that $e$ is irrational, not $e/\pi$. $\endgroup$ Mar 11, 2020 at 6:49
  • 1
    $\begingroup$ @TonyHuynh : Thank you for the information. $\endgroup$ Mar 11, 2020 at 13:03
  • 1
    $\begingroup$ @FedorPetrov: Right, my bad. $\endgroup$ Mar 11, 2020 at 13:04
13
$\begingroup$

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 $k=0,1,\dots$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.