Question

`\rm\sum_(n = 1)^ \infty ( - 1 {)}^(n + 1) \int^( \infty )_(0) \frac{ \sqrt{ {x}^(2n - 3) } \: {tan}^( - 1)( {x}^(2n - 1) )}{1 + {x}^(2n - 1) } \: dx\\`​

Answer 1

Substitute
`y = x^{\frac{2n-1}2}` and
`dy = \frac{2n-1}2 x^{\frac{2n-3}2} \, dx`. Then the integral is

`\displaystyle \int_0^\infty \frac{x^{\frac{2n-3}2} \tan^(-1)\left(x^(2n-1)\right)}{1 + x^(2n-1)} \, dx = \frac2{2n-1} \underbrace{\int_0^\infty (\tan^(-1)\left(y^2\right))/(1 + y^2) \, dy}_I`

which is to say, the integral is now independent of n. Then

`\displaystyle \sum_(n=1)^\infty (-1)^(n+1) \int_0^\infty \frac{x^{\frac{2n-3}2} \tan^(-1)\left(x^(2n-1)\right)}{1+x^(2n-1)} \, dx = I \sum_(n=1)^\infty (2(-1)^(n+1))/(2n-1)`

Let's evaluate the sum. Recall that if |x| < 1, we have

`\displaystyle \frac1{1-x} = \sum_(n=0)^\infty x^n`

which means

`\displaystyle \frac1{1+x^2} = \frac1{1-(-x^2)} = \sum_(n=0)^\infty \left(-x^2\right)^n = \sum_(n=0)^\infty (-1)^n x^(2n)`

By the fundamental theorem of calculus, integrating both sides gives

`\displaystyle \tan^(-1)(x) = \sum_(n=0)^\infty (-1)^n (x^(2n+1))/(2n+1) = \sum_(n=1)^\infty (-1)^(n+1) (x^(2n-1))/(2n-1)`

As x approaches 1 from below, we have

`\displaystyle \lim_(x\to1^-) \tan^(-1)(x) = \lim_(x\to1^-) \sum_(n=1)^\infty (-1)^(n+1) (x^(2n-1))/(2n-1)`

`\displaystyle \frac\pi4 = \sum_(n=1)^\infty ((-1)^(n+1))/(2n-1) \left(\lim\limits_(x\to1^-) x^(2n-1)\right) = \sum_(n=1)^\infty ((-1)^(n+1))/(2n-1)`

and so

`\displaystyle \sum_(n=1)^\infty (-1)^(n+1) \int_0^\infty \frac{x^{\frac{2n-3}2} \tan^(-1)\left(x^(2n-1)\right)}{1+x^(2n-1)} \, dx = \frac\pi2 I`

Now compute the remaining integral. First split it at y = 1 :

`\displaystyle I = \int_0^1 (\tan^(-1)\left(y^2\right))/(1 + y^2) \, dy + \int_1^\infty (\tan^(-1)\left(y^2\right))/(1 + y^2) \, dy`

In the second integral, notice that replacing
`z=\frac1y` and
`dy = -(dz)/(z^2)` yields

`\displaystyle \int_1^\infty (\tan^(-1)\left(y^2\right))/(1 + y^2) \, dy = - \int_1^0 \frac{\tan^(-1)\left(\frac1{z^2}\right)}{1 + \frac1{z^2}} \, (dz)/(z^2)  = \int_0^1 \frac{\tan^(-1)\left(\frac1{z^2}\right)}{1+z^2} \, dz`

The inverse tangent function has the property

`\tan^(-1)(x) = \frac\pi2 - \tan^(-1)\left(\frac1x\right)`

so it follows that

`\displaystyle \int_0^1 \frac{\tan^(-1)\left(\frac1{z^2}\right)}{1+z^2} \, dz = \frac\pi2 \int_0^1 (dz)/(1+z^2) - \int_0^1 (\tan^(-1)(z^2))/(1+z^2) \, dz`

and hence

`\displaystyle I = \int_0^1 (\tan^(-1)\left(y^2\right))/(1 + y^2) \, dy + \frac\pi2 \int_0^1 (dz)/(1+z^2) - \int_0^1 (\tan^(-1)(z^2))/(1+z^2) \, dz`

`\displaystyle I = \frac\pi2 \int_0^1 (dz)/(1+z^2)`

`I = \frac\pi2 * \frac\pi4 = \frac{\pi^2}8`

Then the original expression has an exact value of

`\displaystyle \sum_(n=1)^\infty (-1)^(n+1) \int_0^\infty \frac{x^{\frac{2n-3}2} \tan^(-1)\left(x^(2n-1)\right)}{1+x^(2n-1)} \, dx = \frac\pi2 * \frac{\pi^2}8 = \boxed{(\pi^3)/(16)}`