Question

`\displaystyle \rm \int_{1 - √(\pi x) \large \frac{ {d}^{ {1/2 }} }{ {dx}^(1/2) } \small(1) }^{ \sum \limits_(n = 1)^ \infty \frac{4}{4 {n}^(2) - 1 } } \frac{arctan( (2 - x)/(1 + 2x)) }{ {x}^(2) - 4x - 1} dx`​

Answer 1

There's nothing particularly tricky about the limits of integration. The upper limit is a telescoping series converging to 2,

`\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)`

The lower limit reduces to 0 using the Riemann-Liouville definition of the fractional derivative. For
`q\in\Bbb Q`, let

`\displaystyle (d^q)/(dx^q) f(x) = \frac1{\Gamma(\lceil q\rceil-q)} (d^(\lceil q\rceil))/(dx^(\lceil q\rceil)) \int_a^x (x-t)^(\lceil q\rceil-q-1) f(t) \, dt`

With
`a=0`,
`q=\frac12` and
`\lceil q\rceil=1`, it follows that

`\displaystyle (d^(1/2))/(dx^(1/2)) 1 = \frac1{\Gamma\left(\frac12\right)} \frac d{dx} \int_0^x (x-t)^(-1/2) \, dt = \frac1{√(\pi x)}`

Let

`\displaystyle I = \int_0^2 (\tan^(-1)\left((2-x)/(1+2x)\right))/(x^2-4x-1) \, dx`

Observe that

`f(x) = (2-x)/(1+2x) = f^(-1)(x)`

is its own inverse, so by substituting
`(2-x)/(1+2x)\mapsto x`, we get the equivalent integral

`\displaystyle \int_0^2 (\tan^(-1)(x))/(x^2-4x-1) \, dx`

We have the identity

`\tan^(-1)(x) + \tan^(-1)\left((2-x)/(1+2x)\right) = \tan^(-1)(2)`

so that

`\displaystyle I + I = \int_0^2 (\tan^(-1)\left((2-x)/(1+2x)\right) + \tan^(-1)(x))/(x^2-4x-1) \, dx`

`\implies \displaystyle I = \frac{\tan^(-1)(2)}2 \int_0^2 (dx)/((x-2)^2-5)`

The remaining integral is trivial,

`\displaystyle \int_0^2 (dx)/((x-2)^2-5) = \int_(-2)^0 (dx)/(x^2-5) \\\\ ~~~~~~~~ = \frac1{2\sqrt5} \int_(-2)^0 \left(\frac1{x-\sqrt5} - \frac1{x+\sqrt5}\right) \, dx \\\\ ~~~~~~~~= -(\ln(2+\sqrt5))/(2\sqrt5) \\\\ ~~~~~~~~ = -\frac1{\sqrt5} \tanh^(-1)\left(\frac2{\sqrt5}\right)`

Then the integral we want is

`I = \displaystyle \int_0^2 (\tan^(-1)\left((2-x)/(1+2x)\right))/(x^2-4x-1) \, dx = \boxed{-\frac1{2\sqrt5} \tan^(-1)(2) \, \tanh^(-1)\left(\frac2{\sqrt5}\right)} \approx -0.357395`