Question Number 27309 by abdo imad last updated on 04/Jan/18
![find the value of ∫_0 ^∝ (((−1)^([x]) )/((2x+1)^2 ))dx](https://www.tinkutara.com/question/Q27309.png)
$${find}\:{the}\:{value}\:\:{of}\:\:\int_{\mathrm{0}} ^{\propto} \:\:\:\frac{\left(−\mathrm{1}\right)^{\left[{x}\right]} }{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\: \\ $$
Answered by prakash jain last updated on 05/Jan/18
![∫_0 ^∞ (((−1)^(⌊x⌋) )/((2x+1)^2 ))dx =Σ_(i=0) ^∞ ∫_i ^(i+1) (((−1)^i )/((2x+1)^2 ))dx =Σ_(i=0) ^∞ (−1)^i [−(1/(2x+1))]_i ^(i+1) =Σ_(i=0) ^∞ (−1)^(i+1) [(1/(2i+1))−(1/(2i+3))] =Σ_(i=0) ^∞ (−1)^(i+1) (1/(2i+1))−Σ_(i=0) ^∞ (−1)^(i+1) (1/(2i+3)) =Σ_(i=0) ^∞ (−1)^(i+1) (1/(2i+1))−Σ_(i=1) ^∞ (−1)^i (1/(2i+1)) =Σ_(i=0) ^∞ (−1)^(i+1) (1/(2i+1))−Σ_(i=0) ^∞ (−1)^i (1/(2i+1))+1 =−(π/4)−(π/4)+1=1−(π/2)](https://www.tinkutara.com/question/Q27346.png)
$$\int_{\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{\lfloor{x}\rfloor} }{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{{i}} ^{{i}+\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{i}} }{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}} \left[−\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}\right]_{{i}} ^{{i}+\mathrm{1}} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{2}{i}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{i}+\mathrm{3}}\right] \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{i}+\mathrm{1}}−\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{i}+\mathrm{3}} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{i}+\mathrm{1}}−\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}} \frac{\mathrm{1}}{\mathrm{2}{i}+\mathrm{1}} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{i}+\mathrm{1}}−\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}} \frac{\mathrm{1}}{\mathrm{2}{i}+\mathrm{1}}+\mathrm{1} \\ $$$$=−\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{4}}+\mathrm{1}=\mathrm{1}−\frac{\pi}{\mathrm{2}} \\ $$