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

$$findthevalueof{\int }_0^{\propto }\frac{{(-1)}^{\left[x\right]}}{{(2x+1)}^2}dx$$

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

$${\int }_0^{\mathrm{\infty }}\frac{{(-1)}^{\lfloor x\rfloor }}{{(2x+1)}^2}dx$$$$=\underset{i=0}{\sum ^{\mathrm{\infty }}}{\int }_i^{i+1}\frac{{(-1)}^i}{{(2x+1)}^2}dx$$$$=\underset{i=0}{\sum ^{\mathrm{\infty }}}{(-1)}^i{[-\frac{1}{2x+1}]}_i^{i+1}$$$$=\underset{i=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{i+1}[\frac{1}{2i+1}-\frac{1}{2i+3}]$$$$=\underset{i=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{i+1}\frac{1}{2i+1}-\underset{i=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{i+1}\frac{1}{2i+3}$$$$=\underset{i=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{i+1}\frac{1}{2i+1}-\underset{i=1}{\sum ^{\mathrm{\infty }}}{(-1)}^i\frac{1}{2i+1}$$$$=\underset{i=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{i+1}\frac{1}{2i+1}-\underset{i=0}{\sum ^{\mathrm{\infty }}}{(-1)}^i\frac{1}{2i+1}+1$$$$=-\frac{\pi }{4}-\frac{\pi }{4}+1=1-\frac{\pi }{2}$$

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