Respuesta :
Replace [tex]x\mapsto \tan^{-1}(x)[/tex] :
[tex]\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \int_0^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx[/tex]
Split the integral at x = 1, and consider the latter one over [1, ∞) in which we replace [tex]x\mapsto\frac1x[/tex] :
[tex]\displaystyle \int_1^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx = \int_0^1 \frac{\ln\left(\frac1x\right)}{\sqrt[3]{x} \left(1+\frac1{x^2}\right)} \frac{dx}{x^2} = - \int_0^1 \frac{\ln(x)}{\sqrt[3]{x} (1+x^2)} \, dx[/tex]
Then the original integral is equivalent to
[tex]\displaystyle \int_0^1 \frac{\ln(x)}{1+x^2} \left(\sqrt[3]{x} - \frac1{\sqrt[3]{x}}\right) \, dx[/tex]
Recall that for |x| < 1,
[tex]\displaystyle \sum_{n=0}^\infty x^n = \frac1{1-x}[/tex]
so that we can expand the integrand, then interchange the sum and integral to get
[tex]\displaystyle \sum_{n=0}^\infty (-1)^n \int_0^1 \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \ln(x) \, dx[/tex]
Integrate by parts, with
[tex]u = \ln(x) \implies du = \dfrac{dx}x[/tex]
[tex]du = \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \, dx \implies u = \dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac23}}{2n+\frac23}[/tex]
[tex]\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \int_0^1 \left(\dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac13}}{2n-\frac13}\right) \, dx \\\\ = \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{\left(2n+\frac43\right)^2} - \frac1{\left(2n+\frac23\right)^2}\right) \\\\ = \frac94 \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)[/tex]
Recall the Fourier series we used in an earlier question [27217075]; if [tex]f(x)=\left(x-\frac12\right)^2[/tex] where 0 ≤ x ≤ 1 is a periodic function, then
[tex]\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi n x)}{n^2}[/tex]
[tex]\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(2\pi(3n+1)x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(2\pi(3n+2)x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(2\pi(3n)x)}{(3n)^2}\right)[/tex]
[tex]\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(6\pi n x + 2\pi x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(6\pi n x + 4\pi x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(6\pi n x)}{(3n)^2}\right)[/tex]
Evaluate f and its Fourier expansion at x = 1/2 :
[tex]\displaystyle 0 = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(3n+1)^2} + \sum_{n=0}^\infty \frac{(-1)^n}{(3n+2)^2} + \sum_{n=1}^\infty \frac{(-1)^n}{(3n)^2}\right)[/tex]
[tex]\implies \displaystyle -\frac{\pi^2}{12} - \frac19 \underbrace{\sum_{n=1}^\infty \frac{(-1)^n}{n^2}}_{-\frac{\pi^2}{12}} = - \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)[/tex]
[tex]\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right) = \frac{2\pi^2}{27}[/tex]
So, we conclude that
[tex]\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \frac94 \times \frac{2\pi^2}{27} = \boxed{\frac{\pi^2}6}[/tex]
