# 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:

• Would the downvoter care to explain the downvote ? Mar 8, 2020 at 17:25
• 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 Mar 9, 2020 at 12:40
• $g(z)$ is certainly not periodic. Nice problem, by the way!
– Dirk
Mar 9, 2020 at 16:50

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.
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