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Post Closed as "Not suitable for this site" by Will Jagy, Stefan Kohl♦, Tilman, Ryan Budney, S. Carnahan♦
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Marjo
Marjo
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How to prove that the function $$f(r)=(1 - r^2) \int_0^{2\pi}|Re[\frac{e^{i a} (2 - e^{-i s} r)}{(-e^{i s} + r)^2}]|ds\le 8$$$$f(r)=(1 - r^2) \int_0^{2\pi}|Re[\frac{e^{i a} (2 - e^{-i s} r)}{(-e^{i s} + r)^2}]|ds$$ for real $a$ and $r\in[0,1]$ attains its maximum for $r=0$ with $f(0)=8$.

How to prove that the function $$f(r)=(1 - r^2) \int_0^{2\pi}|Re[\frac{e^{i a} (2 - e^{-i s} r)}{(-e^{i s} + r)^2}]|ds\le 8$$ for real $a$ and $r\in[0,1]$ attains its maximum for $r=0$.

How to prove that the function $$f(r)=(1 - r^2) \int_0^{2\pi}|Re[\frac{e^{i a} (2 - e^{-i s} r)}{(-e^{i s} + r)^2}]|ds$$ for real $a$ and $r\in[0,1]$ attains its maximum for $r=0$ with $f(0)=8$.

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Marjo
Marjo
  • 11
  • 3

Inequality for an integral

How to prove that the function $$f(r)=(1 - r^2) \int_0^{2\pi}|Re[\frac{e^{i a} (2 - e^{-i s} r)}{(-e^{i s} + r)^2}]|ds\le 8$$ for real $a$ and $r\in[0,1]$ attains its maximum for $r=0$.