# Integrate the following using INFINITE SERIES (no binomial please)?

## $\setminus \int \frac{\setminus \ln \left(1 + {x}^{2}\right)}{x} ^ w \mathrm{dx}$

May 12, 2018

$\int \ln \frac{1 + {x}^{2}}{x} ^ w = C + {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(n + 1\right) {x}^{2 n + 3 - w} / \left(2 n + 3 - w\right)$

#### Explanation:

Using the MacLaurin expansion of:

$\ln \left(1 + t\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {t}^{n + 1} / \left(n + 1\right)$

let: $t = {x}^{2}$: to get

$\ln \left(1 + {x}^{2}\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 2} / \left(n + 1\right)$

then divide by ${x}^{w}$ and integrate term by term:

$\ln \frac{1 + {x}^{2}}{x} ^ w = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 2 - w} / \left(n + 1\right)$

$\int \ln \frac{1 + {x}^{2}}{x} ^ w = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(n + 1\right) \int {x}^{2 n + 2 - w} \mathrm{dx}$

$\int \ln \frac{1 + {x}^{2}}{x} ^ w = C + {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(n + 1\right) {x}^{2 n + 3 - w} / \left(2 n + 3 - w\right)$

The series has radius of convergence at least equal to the original series $R = 1$, but has sense only if for every $n$:

$2 n + 3 - w \ne 0$

$2 n + 3 \ne w$

thus $w$ cannot be an odd integer number except for $w = 1$.