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: enter image description here

  • 7
    $\begingroup$ Would the downvoter care to explain the downvote ? $\endgroup$ Mar 8, 2020 at 17:25
  • $\begingroup$ I don't know if it helps anybody but the x-value of the critical points of the function $f(x)$ must satisfy the equation $\sin(e^{x+1})-\sin(e^x) = \operatorname{Si}(e^{x})-\operatorname{Si}(e^{x+1})$, again where $\operatorname{Si}(x)$ denotes the Sine Integral. Wolfram|Alpha gives numerical approximations $\endgroup$ Mar 9, 2020 at 12:40
  • 6
    $\begingroup$ $g(z)$ is certainly not periodic. Nice problem, by the way! $\endgroup$
    – Dirk
    Mar 9, 2020 at 16:50

2 Answers 2


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.


[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


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