Question

`\rm\int_(0)^(\pi\over 2) \sqrt[5]{\tan(x)}\cdot{\ln(\csc^2(x))\over \sin(2x)} dx \\`​

Answer 1

Substitute
`y=\csc(x)` to rewrite the integral as

`\displaystyle I = \int_0^(\pi/2) \sqrt[5]{\tan(x)} (\ln(\csc^2(x)))/(\sin(2x)) \, dx \\\\ ~~~~ = \int_1^\infty (y \ln(y))/(\left(y^2-1\right)^(11/10)) \, dy`

Substitute
`z=\frac1y` to rewrite again as

`\displaystyle I = -\int_0^1 (\ln(z))/(z^(4/5) \left(1-z^2\right)^(11/10)) \, dz`

Substitute
`z=\sqrt t` to rewrite to

`\displaystyle I = -\frac14 \int_0^1 (\ln(t))/(t^(9/10) (1-t)^(11/10)) \, dt`

One last substitution of
`u=1-t` to rewrite

`\displaystyle I = -\frac14 \int_0^1 (\ln(1-u))/(u^(11/10) (1-u)^(9/10)) \, du`

To summarize, substitute
`\csc(x) = \frac1{√(1-u)}`.

Now write the power series of
`\ln(1-u)` and evaluate the subsequent beta integral.

`\displaystyle I = \frac14 \int_0^1 (du)/(u^(11/10) (1-u)^(9/10)) \sum_(n=1)^\infty \frac{u^n}n \\\\ ~~~~ = \frac14 \sum_(n=1)^\infty \frac1n \int_0^1 u^(n-11/10) (1-u)^(-9/10) \, du \\\\ ~~~~ = \frac14 \sum_(n=1)^\infty \frac{\mathrm{B}\left(n-\frac1{10}, \frac1{10}\right)}n \\\\ ~~~~ = \frac14 \sum_(n=1)^\infty \frac{\Gamma\left(n-\frac1{10}\right) \Gamma\left(\frac1{10}\right)}{\Gamma(n+1)} \\\\ ~~~~ = \frac14 \Gamma\left(\frac1{10}\right) \sum_(n=1)^\infty \frac{\Gamma\left(n-\frac1{10}\right)}{\Gamma(n+1)}`

A lemma:

`\displaystyle \sum_(n=1)^\infty (\Gamma\left(n-\frac1k\right))/(\Gamma(n+1)) = k \Gamma\left(1-\frac1k\right)`

Recall the binomial series,

`\displaystyle (1+x)^\alpha = \sum_(n=0)^\infty \binom \alpha n x^n = \sum_(n=0)^\infty (\Gamma(n-\alpha))/(\Gamma(-\alpha) \Gamma(n+1)) (-x)^n`

Let
`x=-1`, so the left side vanishes. This means

`\displaystyle \sum_(n=1)^\infty (\Gamma(n-\alpha))/(\Gamma(-\alpha) \Gamma(n+1)) = -1 \\\\ ~~~~ \implies \sum_(n=1)^\infty (\Gamma(n-\alpha))/(\Gamma(n+1)) = -\Gamma(-\alpha)`

Let
`\alpha=\frac1k` and use the identity
`x\Gamma(x)=\Gamma(x+1)` to write

`\displaystyle \sum_(n=1)^\infty (\Gamma(n-\alpha))/(\Gamma(n+1)) = -\Gamma\left(-\frac1k\right) = k \Gamma\left(1 - \frac1k\right)`

Let
`k=10`, so our integral is

`\displaystyle I = \frac14 \Gamma\left(\frac1{10}\right) \cdot 10 \Gamma\left(\frac9{10}\right)`

Recall the reflection formula,
`\Gamma(x)\Gamma(1-x) = \frac\pi{\sin(\pi x)}`.

`\displaystyle I = \frac52 \cdot \frac\pi{\sin\left(\frac\pi{10}\right)}`

In an earlier question [28756378] we found the exact value

`\sin\left(\frac\pi{10}\right) = \frac{\sqrt5-1}4`

So we ultimately find that

`\displaystyle I = \frac52 \cdot (4\pi)/(\sqrt5-1) = \frac{5\pi(1+\sqrt5)}2 = \boxed{5\pi\phi}`

(Try saying that 5 times fast!)