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The exact value is close to zero judging by the graph. By the fundamental theorem of calculus, $\frac{d}{dt}\int_{t}^{t+1}\sin e^x \ dx=\sin e^{t+1}-\sin e^{t}$. The solution will be given when we solve for $t$ such that $\sin e^{t+1}-\sin e^{t}=0$ . I think we have to play around with the periodicity of sine. So if $\sin e^{t+1}=k$, then $e^{t+1}+2\pi n=k$ and we have $e^t=\frac{k-2\pi n}{e}$ but I can't really work with this.

Other considerations: The inside of the integral can be transformed to $\frac{\sin u}{u}$ by making the substitution $u=e^x$.

Answer: judging by the comments I think we an do:

$ee^t=\pi e^t$ So that $e^t=\frac{\pi}{e+1}$ which gives us $t=\ln \frac{\pi}{e+1}$

recmath
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  • There are other possibilities, such as $e\cdot e^t=\pi-e^t$. – André Nicolas May 05 '14 at 22:55
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    The value you're looking for is $\ln\left(\frac{\pi}{e+1}\right)$. I can't make a write-up of it right now but if no one else does I can later. – Eugene Bulkin May 05 '14 at 22:59
  • So either $\frac{e^{t+1} - e^t}{\pi}$ is an even integer or $\frac{e^{t+1}+e^t}{\pi}$ is an odd integer. Then $t = \log(2\pi n) - \log(e-1)$ or $\log((2n+1)\pi) - \log(1+e)$ for some $n \in \mathbb{Z}$. – jdc May 05 '14 at 23:06
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    $$\sin x - \sin y= 2\sin\left(\frac{x-y}2\right)\cos\left(\frac{x+y}2\right)$$ – chubakueno May 05 '14 at 23:30

1 Answers1

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Let $f(t) = \int_{t}^{t+1}\sin e^x \ dx$. As was noted in the comments, $\sin x - \sin y= 2\sin\left(\frac{x-y}2\right)\cos\left(\frac{x+y}2\right)$, so $$f'(t) = \frac{d}{dt}\int_{t}^{t+1}\sin e^x \ dx = \sin e^{t+1}-\sin e^t = \\ 2\sin\left(\frac{e^{t+1}-e^t}{2}\right)\cos\left(\frac{e^{t+1}+e^t}{2}\right) = 2\sin\left(e^t\cdot\frac{e-1}{2}\right)\cos\left(e^t\cdot\frac{e+1}{2}\right)$$

So $f'(t) = 0$ when $e^t\cdot\frac{e-1}{2}=n\pi$ or $e^t\cdot\frac{e+1}{2}=n\pi+\pi/2$, or $t = \ln\left(\frac{2\pi n}{e-1}\right)$ or $t = \ln\left(\frac{\pi(2n+1)}{e+1}\right)$ for any $n$.

Now, we want the value closest to zero as you noted, which gives us $\ln(\pi/(e+1))$. $f''(t) = e^t(e\cos e^{t+1} - \cos e^t)$, so $f'' = \frac{\pi}{e+1}\left(e\cos\left(\pi\cdot\frac{e}{e+1}\right)-\cos\left(\frac{\pi}{e+1}\right)\right)$. You can approximate $e$ with $2$ or $3$, but either way $f'' < 0$, so this is a maximum.

There's some more messy algebra if you wish to prove that this is the absolute maximum, if you so desire.

Eugene Bulkin
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