Respuesta :
There's nothing particularly tricky about the limits of integration. The upper limit is a telescoping series converging to 2,
[tex]\displaystyle \sum_{n=1}^\infty \frac4{4n^2-1} = 2 \sum_{n=1}^\infty \left(\frac1{2n-1} - \frac1{2n+1}\right) \\\\ ~~~~~~~~ = 2 \left(\left(1-\frac13\right)+\left(\frac13-\frac15\right) + \left(\frac15 - \frac17\right) + \cdots\right)[/tex]
The lower limit reduces to 0 using the Riemann-Liouville definition of the fractional derivative. For [tex]q\in\Bbb Q[/tex], let
[tex]\displaystyle \frac{d^q}{dx^q} f(x) = \frac1{\Gamma(\lceil q\rceil-q)} \frac{d^{\lceil q\rceil}}{dx^{\lceil q\rceil}} \int_a^x (x-t)^{\lceil q\rceil-q-1} f(t) \, dt[/tex]
With [tex]a=0[/tex], [tex]q=\frac12[/tex] and [tex]\lceil q\rceil=1[/tex], it follows that
[tex]\displaystyle \frac{d^{1/2}}{dx^{1/2}} 1 = \frac1{\Gamma\left(\frac12\right)} \frac d{dx} \int_0^x (x-t)^{-1/2} \, dt = \frac1{\sqrt{\pi x}}[/tex]
Let
[tex]\displaystyle I = \int_0^2 \frac{\tan^{-1}\left(\frac{2-x}{1+2x}\right)}{x^2-4x-1} \, dx[/tex]
Observe that
[tex]f(x) = \dfrac{2-x}{1+2x} = f^{-1}(x)[/tex]
is its own inverse, so by substituting [tex]\frac{2-x}{1+2x}\mapsto x[/tex], we get the equivalent integral
[tex]\displaystyle \int_0^2 \frac{\tan^{-1}(x)}{x^2-4x-1} \, dx[/tex]
We have the identity
[tex]\tan^{-1}(x) + \tan^{-1}\left(\dfrac{2-x}{1+2x}\right) = \tan^{-1}(2)[/tex]
so that
[tex]\displaystyle I + I = \int_0^2 \frac{\tan^{-1}\left(\frac{2-x}{1+2x}\right) + \tan^{-1}(x)}{x^2-4x-1} \, dx[/tex]
[tex]\implies \displaystyle I = \frac{\tan^{-1}(2)}2 \int_0^2 \frac{dx}{(x-2)^2-5}[/tex]
The remaining integral is trivial,
[tex]\displaystyle \int_0^2 \frac{dx}{(x-2)^2-5} = \int_{-2}^0 \frac{dx}{x^2-5} \\\\ ~~~~~~~~ = \frac1{2\sqrt5} \int_{-2}^0 \left(\frac1{x-\sqrt5} - \frac1{x+\sqrt5}\right) \, dx \\\\ ~~~~~~~~= -\frac{\ln(2+\sqrt5)}{2\sqrt5} \\\\ ~~~~~~~~ = -\frac1{\sqrt5} \tanh^{-1}\left(\dfrac2{\sqrt5}\right)[/tex]
Then the integral we want is
[tex]I = \displaystyle \int_0^2 \frac{\tan^{-1}\left(\frac{2-x}{1+2x}\right)}{x^2-4x-1} \, dx = \boxed{-\frac1{2\sqrt5} \tan^{-1}(2) \, \tanh^{-1}\left(\dfrac2{\sqrt5}\right)} \approx -0.357395[/tex]
