Five, Phi and Pi in one integral = −5ϕπ

[Tony1]

$$\int_{0}^{\pi\over 2}{\ln(\sin^2 x)\over \sin(2x)}\cdot \sqrt[5]{\tan(x)}\mathrm dx=-5\phi \pi$$

$\phi$ is the golden ratio

Spoiler

Make $u=\tan x$

$${1\over 2}\int_{0}^{\infty}u^{-{4\over 5}}\ln\left({1+u^2\over u^2}\right)\mathrm du$$

hmmm, too complicate to continue.

Any help, please. Thank you!

 

[Theia]

Is there an error in the result? I'm getting

\(\displaystyle I = \int_{0}^{\pi\over 2}{\ln(\sin^2 x)\over \sin(2x)}\cdot \sqrt[5]{\tan(x)}\mathrm dx = -\frac{5 \pi}{\phi}\).

 

[Theia]

Is there an error in the result? I'm getting

\(\displaystyle I = \int_{0}^{\pi\over 2}{\ln(\sin^2 x)\over \sin(2x)}\cdot \sqrt[5]{\tan(x)}\mathrm dx = -\frac{5 \pi}{\phi}\).

Spoiler

Taking your substitution \(\displaystyle u = \tan x\), one obtains

\(\displaystyle I = -\frac{1}{2}\int _0^{\infty}u^{-4/5}\ln \left( \frac{1+u^2}{u^2} \right) \mathrm d u\).

Now taking \(\displaystyle \frac{1+u^2}{u^2} = q\), the integral becomes

\(\displaystyle I = -\frac{1}{4}\int _1 ^{\infty} \ln q \ (q - 1)^{-11/10} \mathrm dq.\)

Let's then integrate by parts (\(\displaystyle \mathrm d f = \ln q\)) and one obtains

\(\displaystyle I = -\frac{5}{2}\int _1 ^{\infty} q^{-1} \ (q - 1)^{-1/10} \mathrm dq.\)

One more substitution: \(\displaystyle 1 - \dfrac{1}{q} = p\) and one obtains

\(\displaystyle I = -\frac{5}{2}\int _0 ^1 p^{-1/10} \ (1 - p)^{-9/10} \mathrm dp.\)

By definition of the Beta function one reads this as

\(\displaystyle I =-\frac{5}{2} \beta \left( \frac{9}{10}, \frac{1}{10} \right) = -\frac{5}{2} \frac{\pi}{\sin \tfrac{\pi}{10}} = -\frac{5\pi}{\phi}\).

 

[Tony1]

Spoiler

Taking your substitution \(\displaystyle u = \tan x\), one obtains

\(\displaystyle I = -\frac{1}{2}\int _0^{\infty}u^{-4/5}\ln \left( \frac{1+u^2}{u^2} \right) \mathrm d u\).

Now taking \(\displaystyle \frac{1+u^2}{u^2} = q\), the integral becomes

\(\displaystyle I = -\frac{1}{4}\int _1 ^{\infty} \ln q \ (q - 1)^{-11/10} \mathrm dq.\)

Let's then integrate by parts (\(\displaystyle \mathrm d f = \ln q\)) and one obtains

\(\displaystyle I = -\frac{5}{2}\int _1 ^{\infty} q^{-1} \ (q - 1)^{-1/10} \mathrm dq.\)

One more substitution: \(\displaystyle 1 - \dfrac{1}{q} = p\) and one obtains

\(\displaystyle I = -\frac{5}{2}\int _0 ^1 p^{-1/10} \ (1 - p)^{-9/10} \mathrm dp.\)

By definition of the Beta function one reads this as

\(\displaystyle I =-\frac{5}{2} \beta \left( \frac{9}{10}, \frac{1}{10} \right) = -\frac{5}{2} \frac{\pi}{\sin \tfrac{\pi}{10}} = -\frac{5\pi}{\phi}\).

Note $$\sin(\pi/10)={1\over 2\phi}$$

$$-{5\over 2}\cdot {\pi\over \sin(\pi/10)}=-5\pi\phi$$