calculate-2-1-2x-x-x-1-dx-

Question Number 91619 by abdomathmax last updated on 02/May/20

c a l c u l a t e \int_{2}^{+ \infty} \frac{{(- 1)}^{[2 x]}}{x [x] - 1} d x

Commented by mathmax by abdo last updated on 02/May/20

= \sum_{n = 2}^{\infty} \int_{n}^{n + 1} \frac{{(- 1)}^{[2 x]}}{n x - 1} d x

n < x < n + 1 \Rightarrow 2 n < 2 x < 2 n + 2 \Rightarrow 2 x \in [2 n, 2 n + 1] \cup [2 n + 1, 2 n + 2 [

= \sum_{n = 2}^{\infty} \int_{n}^{n + \frac{1}{2}} \frac{{(- 1)}^{2 n}}{n x - 1} + \sum_{n = 2}^{\infty} \int_{n + \frac{1}{2}}^{n + 1} \frac{{(- 1)}^{2 n + 1}}{n x - 1} d x

= \sum_{n = 2}^{\infty} \frac{1}{n} {[l n (n x - 1)]}_{n}^{n + \frac{1}{2}} - \sum_{n = 2}^{\infty} \frac{1}{n} {[l n (n x - 1)]}_{n + \frac{1}{2}}^{n + 1}

= \sum_{n = 2}^{\infty} \frac{1}{n} {l n (n^{2} + \frac{n}{2} - 1) - l n (n^{2} - 1))}

- \sum_{n = 2}^{\infty} \frac{1}{n} {l n (n^{2} + n - 1) - l n (n^{2} + \frac{n}{2} - 1)}

= \sum_{n = 2}^{\infty} \frac{1}{n} l n (\frac{n^{2} + \frac{n}{2} - 1}{n^{2} - 1}) - \sum_{n = 2}^{\infty} \frac{1}{n} l n (\frac{n^{2} + n - 1}{n^{2} + \frac{n}{2} - 1})}

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