Evaluate $\int_0^{\frac{\pi}{2}} \sqrt[5]{\tan(x)} \frac{\ln(\csc^2(x))}{\sin(2x)}dx$

Question

I came across this integral

$$\int_0^{\frac{\pi}{2}} \sqrt[5]{\tan(x)} \frac{\ln(\csc^2(x))}{\sin(2x)}dx = 5\pi\phi$$ here Is there any integral for the Golden Ratio?.

This looks well beyond my abilities to integrate and I don't really have anything particularly intelligent to contribute towards this computation.

I am particularly interested in how the term $\sqrt[5]{\tan(x)}$ should be handled because it is the most mysterious part of the integrand to me. I normally wouldn't ask this kind of question, but I can't help but feel like this is reasonably doable and not a "Cleo"-type problem.

While playing around on a CAS, I did notice that

$$\int_0^{\frac{\pi}{2}}\sqrt[5]{\tan(x)} \frac{\ln(\sin(x))}{\sin(2x)}dx = \frac{5\pi}{2-2\phi}$$

but I don't think that gives me any intuition on the behavior of the integrand.

Answer 1

Not an answer

Simplify the integrand using trigonometric identities

• $\csc^2(x) = \frac{1}{\sin^2(x)}$, so $\ln(\csc^2(x)) = \ln\left(\frac{1}{\sin^2(x)}\right) = -2\ln(\sin(x))$
• $\sin(2x) = 2\sin(x)\cos(x)$

Substitute these into the integrand $$\sqrt[5]{\tan(x)} \cdot \frac{-2\ln(\sin(x))}{2\sin(x)\cos(x)} = -\sqrt[5]{\tan(x)} \cdot \frac{\ln(\sin(x))}{\sin(x)\cos(x)}$$ Rewrite everything in terms of $\sin$ and $\cos$ $$-\frac{\sin^{1/5}(x)}{\cos^{1/5}(x)} \cdot \frac{\ln(\sin(x))}{\sin(x)\cos(x)} = -\frac{\ln(\sin(x))}{\sin^{4/5}(x)\cos^{6/5}(x)}$$

Let $u = \sin(x)$. Then:

• $du = \cos(x) \, dx$
• $\cos(x) = \sqrt{1 - u^2}$
• When $x = 0$, $u = 0$
• When $x = \frac{\pi}{2}$, $u = 1$

Thus, the integral becomes $$I = -\int_0^1 \frac{\ln(u)}{u^{4/5}(1 - u^2)^{3/5}} \, du$$ Use the following $$\frac{\partial}{\partial a} B(a, b) = \int_0^1 t^{a-1}(1 - t)^{b-1} \ln(t) \, dt$$

This matches the form of the integral if we set

• $a - 1 = -\frac{4}{5} \Rightarrow a = \frac{1}{5}$
• $b - 1 = -\frac{3}{5} \Rightarrow b = \frac{2}{5}$

Thus, the integral becomes $$\int_0^1 \frac{\ln(u)}{u^{4/5}(1 - u^2)^{3/5}} \, du = \frac{\partial}{\partial a} B(a, b) \bigg|_{a = \frac{1}{5}, b = \frac{2}{5}}$$ We know that $B(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)}$

Taking the derivative with respect to $a$ $$\frac{\partial}{\partial a} B(a, b) = B(a, b) \left(\psi(a) - \psi(a + b)\right)$$ where $\psi(x) = \frac{\Gamma'(x)}{\Gamma(x)}$ is the $\textbf{digamma function}$.

Substituting $a = \frac{1}{5}$ and $b = \frac{2}{5}$ $$\frac{\partial}{\partial a} B(a, b) \bigg|_{a = \frac{1}{5}, b = \frac{2}{5}} = B\left(\frac{1}{5}, \frac{2}{5}\right) \left(\psi\left(\frac{1}{5}\right) - \psi\left(\frac{1}{5} + \frac{2}{5}\right)\right)$$ $$\implies\frac{\partial}{\partial a} B(a, b) \bigg|_{a = \frac{1}{5}, b = \frac{2}{5}} = B\left(\frac{1}{5}, \frac{2}{5}\right) \left(\psi\left(\frac{1}{5}\right) - \psi\left(\frac{3}{5}\right)\right)$$

The digamma function has known values at rational points. In particular

• $\psi\left(\frac{1}{5}\right) = -\gamma - \ln(5) - \frac{\pi}{2}\cot\left(\frac{\pi}{5}\right)$
• $\psi\left(\frac{3}{5}\right) = -\gamma - \ln(5) + \frac{\pi}{2}\cot\left(\frac{3\pi}{5}\right)$

where $\gamma$ is the Euler-Mascheroni constant. Using the identity $\cot\left(\frac{\pi}{5}\right) = \sqrt{5 + 2\sqrt{5}}$ and $\cot\left(\frac{3\pi}{5}\right) = -\sqrt{5 - 2\sqrt{5}}$, we can simplify $$\psi\left(\frac{1}{5}\right) - \psi\left(\frac{3}{5}\right) = -\frac{\pi}{2}\cot\left(\frac{\pi}{5}\right) - \frac{\pi}{2}\cot\left(\frac{3\pi}{5}\right)$$ Substituting the values of $\cot$ $$\psi\left(\frac{1}{5}\right) - \psi\left(\frac{3}{5}\right) = -\frac{\pi}{2}\left(\sqrt{5 + 2\sqrt{5}} - \sqrt{5 - 2\sqrt{5}}\right)=-\frac{\pi}{2}\left(\sqrt{10-2\sqrt{5}}\right)$$

By using wolfram, I got $B\left(\frac15,\frac25\right)\approx6.838$

So we have $$I\approx\frac{\pi}{2}\left(\sqrt{10-2\sqrt{5}}\right)\cdot 6.838\approx25.24$$ where as $$5\pi\phi\approx25.41$$

This is the best I was able to do.

Answer 2

\begin{aligned}\int_{0}^{\frac{\pi}{2}}{\sqrt[5]{\tan{x}}\frac{\ln{\left(\csc^{2}{x}\right)}}{\sin{\left(2x\right)}}\,\mathrm{d}x}&=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\sqrt[5]{\tan{x}}\frac{\ln{\left(1+\frac{1}{\tan^{2}{x}}\right)}}{\tan{x}}\sec^{2}{x}\,\mathrm{d}x}\\ &=\frac{1}{2}\int_{0}^{+\infty}{\sqrt[5]{y}\frac{\ln{\left(1+\frac{1}{y^{2}}\right)}}{y}\,\mathrm{d}y}\end{aligned}

We made a substitution $ \left\lbrace\begin{aligned}y&=\tan{x}\\ \mathrm{d}y&=\sec^{2}{x}\,\mathrm{d}x\end{aligned}\right. $, and we'll be making another one $ \left\lbrace\begin{aligned}y&=x^{-5}\\ \mathrm{d}y&=-\frac{5}{x^{6}}\end{aligned}\right.$ :

\begin{aligned}\int_{0}^{\frac{\pi}{2}}{\sqrt[5]{\tan{x}}\frac{\ln{\left(\csc^{2}{x}\right)}}{\sin{\left(2x\right)}}\,\mathrm{d}x}&=\frac{5}{2}\int_{0}^{+\infty}{\frac{\ln{\left(1+x^{10}\right)}}{x^{2}}\,\mathrm{d}x}\\ &=\frac{5}{2}\left[-\frac{\ln{\left(1+x^{10}\right)}}{x}\right]_{0}^{+\infty}+\frac{5}{2}\int_{0}^{+\infty}{\frac{10x^{9}}{x\left(1+x^{10}\right)}\,\mathrm{d}x}\\ &=25\int_{0}^{+\infty}{\frac{x^{8}}{1+x^{10}}\,\mathrm{d}x}\end{aligned}

Now setting $ u=\frac{1}{1+x^{10}} $, we end up with : \begin{aligned}\int_{0}^{\frac{\pi}{2}}{\sqrt[5]{\tan{x}}\frac{\ln{\left(\csc^{2}{x}\right)}}{\sin{\left(2x\right)}}\,\mathrm{d}x} &=\frac{5}{2}\int_{0}^{1}{u^{-\frac{9}{10}}\left(1-u\right)^{\frac{9}{10}-1}\,\mathrm{d}u}\\&=\frac{5}{2}\beta\left(1-\frac{9}{10},\frac{9}{10}\right)\\ &=\frac{5}{2}\Gamma\left(1-\frac{9}{10}\right)\Gamma\left(\frac{9}{10}\right)\\ &=\frac{5}{2}\pi\csc{\left(\frac{9}{10}\pi\right)}\\ &=\frac{5}{2}\pi\sec{\left(\frac{2\pi}{5}\right)}\\ &=5\pi\varphi \end{aligned}

Answer 3

You can make it more general $$I_n=\int\sqrt[n]{\tan(x)} \,\,\frac{\log(\csc^2(x))}{\sin(2x)}\,dx$$ $$x=\tan^{-1}(t) \quad \implies \quad I_n=\frac 12 \int t^{\frac{1}{n}-1} \log \left(1+\frac{1}{t^2}\right)\,dt$$ Using integration by parts $$I_n=\frac{1}{2} n\, t^{\frac{1}{n}} \log\left(1+\frac{1}{t^2}\right)+n\int\frac{t^{\frac{1}{n}-1}}{1+t^2}\,dt $$ The last integral express in terms of the Gaussian hypergeometric function $$\int\frac{t^{\frac{1}{n}-1}}{1+t^2}\,dt =n \,t^{\frac{1}{n}} \, _2F_1\left(1,\frac{1}{2 n};1+\frac{1}{2 n};-t^2\right)$$ Integrating between $0$ and $\infty$ gives $$J_n=\int_0^{\frac{\pi}{2}} \sqrt[n]{\tan(x)} \,\,\frac{\log(\csc^2(x))}{\sin(2x)}\,dx=\frac{\pi n }{2} \csc \left(\frac{\pi}{2 n}\right)$$ and then your result for $n=5$.

Edit

Even more general, if $0<\Re\left(\frac{m}{n}\right)<2$, $$K_{m,n}=\int_0^{\frac{\pi}{2}} \sqrt[n]{\tan^m(x)} \,\,\frac{\log(\csc^2(x))}{\sin(2x)}\,dx=\frac{\pi n}{2 m}\,\csc \left(\frac{\pi m}{2 n}\right)$$ and, if $m=\frac n 5$, the same result.

Answer 4

Noticing that when we put $y=\tan x$, we have $$ I=\frac{1}{2} \int_0^{\infty} \sqrt[5]{\tan x} \frac{\ln \left(\csc ^2 x\right) \sec ^2 x}{\tan x} d x= \frac{1}{2} \int_0^{\infty} y^{-\frac{4}{5}} \ln \left(1+\frac{1}{y^2}\right) d y $$ Integration by parts yields $$ I= \frac{5}{2} \int_0^{\infty} \ln \left(1+\frac{1}{y^2}\right) d (y^{\frac{1}{5}}) =5 \int_0^{\infty} \frac{y^{-\frac{4}{5}}}{y^2+1} d y $$ Via putting $z=\frac{1}{y^2+1}$ gives

$$I=\frac{5}{2}B\left(\frac{9}{10} ,\frac{1}{10}\right)=\frac{5\pi}{2}\csc \frac{\pi}{10} =5\pi \phi $$

where the second last step uses the reflection property of Beta function.

Answer 5

$$\begin{align*} I &= \int_0^\tfrac\pi2 \frac{(\tan x)^{1/5}}{\sin(2x)} \log\left(\csc^2x\right) \, dx \\ &= -\frac14 \int_0^1 \frac{\log(1-y)}{y^{11/10} (1-y)^{9/10}} \, dy & \sin x=\sqrt{1-y} \\ &= \sum_{n\ge1} \frac1{4n} \int_0^1 y^{n-\frac{11}{10}}(1-y)^{-\frac9{10}} \, dy \\ &= \sum_{n\ge1} \frac{\operatorname{Beta}\left(n-\frac1{10},\frac1{10}\right)}{4n} \\ &= \frac14 \Gamma\left(\frac1{10}\right) \underbrace{\sum_{n\ge1} \frac{\Gamma\left(n-\frac1{10}\right)}{\Gamma(n+1)}}_{=-\Gamma\left(-\frac1{10}\right)} & (*) \\ &= \frac52 \Gamma\left(\frac1{10}\right)\Gamma\left(\frac9{10}\right) \\ &= \frac{5\pi}{2\sin\frac\pi{10}} \equiv \boxed{5\pi\phi} & \text{reflection formula} \end{align*}$$

where we use $\sin\dfrac\pi{10}=\dfrac{\sqrt5-1}4$.

The simplification in $(*)$ follows from toying with the binomial series. Into

$$(1+t)^a = \sum_{n\ge0} \binom an t^n = \sum_{n\ge0} \frac{\Gamma(n-a)}{\Gamma(-a)\Gamma(n+1)} (-t)^n,$$

insert $t=-1$ on both sides:

$$\sum_{n\ge\color{red}1} \frac{\Gamma(n-a)}{\Gamma(-a)\Gamma(n+1)} + 1 = 0 \implies \sum_{n\ge1} \frac{\Gamma(n-a)}{\Gamma(n+1)} = -\Gamma(-a)$$