find-the-value-of-0-1-x-2x-1-2-dx-

Question Number 27309 by abdo imad last updated on 04/Jan/18

f i n d t h e v a l u e o f \int_{0}^{\propto} \frac{{(- 1)}^{[x]}}{{(2 x + 1)}^{2}} d x

Answered by prakash jain last updated on 05/Jan/18

\int_{0}^{\infty} \frac{{(- 1)}^{⌊ x ⌋}}{{(2 x + 1)}^{2}} d x

= \underset{i = 0}{\sum^{\infty}} \int_{i}^{i + 1} \frac{{(- 1)}^{i}}{{(2 x + 1)}^{2}} d x

= \underset{i = 0}{\sum^{\infty}} {(- 1)}^{i} {[- \frac{1}{2 x + 1}]}_{i}^{i + 1}

= \underset{i = 0}{\sum^{\infty}} {(- 1)}^{i + 1} [\frac{1}{2 i + 1} - \frac{1}{2 i + 3}]

= \underset{i = 0}{\sum^{\infty}} {(- 1)}^{i + 1} \frac{1}{2 i + 1} - \underset{i = 0}{\sum^{\infty}} {(- 1)}^{i + 1} \frac{1}{2 i + 3}

= \underset{i = 0}{\sum^{\infty}} {(- 1)}^{i + 1} \frac{1}{2 i + 1} - \underset{i = 1}{\sum^{\infty}} {(- 1)}^{i} \frac{1}{2 i + 1}

= \underset{i = 0}{\sum^{\infty}} {(- 1)}^{i + 1} \frac{1}{2 i + 1} - \underset{i = 0}{\sum^{\infty}} {(- 1)}^{i} \frac{1}{2 i + 1} + 1

= - \frac{π}{4} - \frac{π}{4} + 1 = 1 - \frac{π}{2}