Cómo probar que$\int_{0}^{\pi/2}{\arctan(2\cos^2 x)\over \cos^2 x}\mathrm dx={\pi\over \sqrt{\phi}}?$

Question

Dado que:

Dónde $\phi={\sqrt{5}+1\over 2}$

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$$\int_{0}^{\pi/2}{\arctan(2\cos^2 x)\over \cos^2 x}\mathrm dx={\pi\over \sqrt{\phi}}\tag1$

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$t=2\cos^2 x\implies dt=-4\sin x\cos xdx$

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O lo dejamos en términos de$${1\over 4}\int_{0}^{1}{\arctan t\over t\sqrt{t-t^2}}\mathrm dt\tag2$

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Answer 1

En$(2)$, deberías haber obtenido

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Al permitir$$I=\int_0^2\frac{\arctan(t)}{t\sqrt{2t-t^2}}~\mathrm dt$, obtenemos

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Y por integración por partes,

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que es la parte real de

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Y tomando un contorno semicircular y el teorema del residuo,

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Answer 2

Deje$$I(a)=\int_{0}^{\pi/2}{\arctan(a\cos^2 x)\over \cos^2 x}\mathrm dx.$ $ Then$I(0)=0,I(2)=I$ y \begin{eqnarray} I'(a)&=&\int_{0}^{\pi/2}{1\over{1+a^2\cos^4 x}}\mathrm dx\\ &=&\int_{0}^{\pi/2}{\sec^2x\over{a^2+\sec^4 x}}\sec^2x\mathrm dx\\ &=&\int_{0}^{\infty}{1+u^2\over{a^2+(1+u^2)^2}}\mathrm du\\ &=&\int_{0}^{\infty}{1+u^2\over{(u^2+1+ai)(u^2+1-ai)}}\mathrm du\\ &=&\frac12\int_{0}^{\infty}\left({1\over{u^2+1+ai}}+{1\over{u^2+1-ai}}\right)\mathrm du\\ &=&\frac12\bigg[\frac{1}{\sqrt{1+ai}}\arctan\frac{u}{\sqrt{1+ai}}+\frac{1}{\sqrt{1-ai}}\arctan\frac{u}{\sqrt{1-ai}}\bigg]\bigg|_{0}^{\infty}\\ &=&\frac{\pi}{4}\bigg(\frac{1}{\sqrt{1-ai}}+\frac{1}{\sqrt{1+ai}}\bigg) \end {eqnarray} Entonces \begin{eqnarray} I(2)&=&\frac{\pi}{4}\int_{0}^{2}\bigg(\frac{1}{\sqrt{1-ai}}+\frac{1}{\sqrt{1+ai}}\bigg)\\ &=&\frac{\pi}{4}\bigg(\frac{\sqrt{1+ai}}{i}+\frac{\sqrt{1-ai}}{-i}\bigg)\bigg|_0^2\\ &=&\frac{\pi}{4}\frac{\sqrt{1+2i}-\sqrt{1-2i}}{i}\\ &=&\frac{\sqrt{2\sqrt5-2}}{4}\pi=\frac{\pi}{\sqrt\phi} \end {eqnarray}