Assume that $a>0$ and $r\in[0,1)$. How to prove that the function $$f(p)=\int_{-\pi}^\pi \left (1+r^2+2 r \cos x\right)^{a/2} |(2+a) \cos(x+p)-a r \cos(p)| \, dx$$ attains its maximum for $p=\pi/2$?
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3$\begingroup$ It may help to know how this problem arose. $\endgroup$– Iosif PinelisCommented Jan 30 at 20:52
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2$\begingroup$ Note that $f(\pi/2)=2 r^{-1}[(1+r)^{a+2}-(1-r)^{a+2}]$, $f'(\pi/2)=0$ and $f''(\pi/2)=-f(\pi/2)<0$. In fact, all odd derivatives are zero, while the even ones oscillate between $\pm f(\pi/2)$... $\endgroup$– Fred HuchtCommented Jan 30 at 22:04
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1$\begingroup$ ... the resulting candidate $f(p) = f(\pi/2) \sin p$, has, however, radius of convergence zero due to the branch cuts of the square root. $\endgroup$– Fred HuchtCommented Jan 30 at 22:22
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$\begingroup$ Welcome to MO. This assumes a local maximum for all $a$ or all $r$ at $p = \pi/2$ so maybe differentiate under the integral sign with respect to $a$ or $r$ might help a little bit to show it is at least a local maximum ? $\endgroup$– mickCommented Jan 30 at 22:28
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