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For every positive integer $n$, define $\displaystyle a_n = \int_{0}^{n\pi} \frac{\sin{x}}{1+x}$. Prove that $a_2 < a_4 < \cdots < a_5 < a_3 < a_1$.

This doesn't look easily integrated, so there must be some other trick we can use to simplify the integral or to compare it to other values of $n$. Maybe we can say $\displaystyle \int_{0}^{2k\pi} \frac{\sin{x}}{1+x} < \displaystyle \int_{0}^{2(k+1)\pi} \frac{\sin{x}}{1+x}$ and try to prove that first.

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For the even terms, note that

$$\begin{align} a_{2n+2}-a_{2n}&=\int_{2n\pi}^{2n\pi +2\pi}\frac{\sin(x)}{x+1}\,dx\\\\ &=\int_0^{2\pi}\frac{\sin(x)}{x+2n\pi +1}\,dx\\\\ &=\int_0^\pi \sin(x)\left(\frac{1}{x+2n\pi +1}-\frac{1}{x+2\pi+2n\pi +1}\right)\,dx\\\\ &=\int_0^\pi \sin(x)\left(\frac{2\pi}{(x+2n\pi +1)(x+2\pi+2n\pi +1)}\right)\,dx\\\\ &>0 \end{align}$$

Proceed along a similar line for the odd terms.

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What you really want is to prove $$ \displaystyle \int_{0}^{2(k+1)\pi} \frac{\sin{x}}{1+x} - \int_{0}^{2k\pi} \frac{\sin{x}}{1+x} = \int_{2k\pi}^{2(k+1)\pi} \frac{\sin{x}}{1+x} = \int_{0}^{2 \pi} \frac{\sin{x}}{1+x+2\pi k} > 0 $$ which is a complete cycle of $\sin x$. Knowing that $\frac{1}{1+x}$ is a decreasing function on $x$, you can see by intuition that this integral is positive.

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    $\begingroup$ That last equality is not correct. The denominator should be $x+2k\pi +1$. ;-)) -Mark $\endgroup$ – Mark Viola Apr 1 '16 at 22:09
  • $\begingroup$ Anyway, separating the last integral in one on $[0,\pi]$, and another on $[\pi,2\pi]$, and then translating the second and using $\sin(\pi+x)=-\sin(x)$, it's easy to check your intuition. $\endgroup$ – Jean-Claude Arbaut Apr 1 '16 at 22:11
  • $\begingroup$ @Dr.MV: oh, I see $\endgroup$ – user164550 Apr 1 '16 at 22:14
  • $\begingroup$ @Jean-ClaudeArbaut: thanks $\endgroup$ – user164550 Apr 1 '16 at 22:15
  • $\begingroup$ @Jean-ClaudeArbaut I believe that my answer addresses your well written comments. $\endgroup$ – Mark Viola Apr 1 '16 at 22:21

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