0 0 votes Let $0 < \alpha < \beta < 1$. Then $$ \Sigma_{k=1}^{\infty} \int_{1/(k+\beta)}^{1/(k+\alpha)} \frac{1}{1+x} dx$$ is equal to $\log_e \frac{\beta}{\alpha}$ $\log_e \frac{1+ \beta}{1 + \alpha}$ $\log_e \frac{1+\alpha }{1+ \beta}$ $\infty$ Calculus isi2015-mma calculus definite-integral summation non-gatecse + – Arjun 1.0k views answer comment Share Follow Print See all 4 Comments 4 4 Comments reply `JEET commented Oct 4, 2019 reply Follow flag $\ln |x + c| $ ? 0 0 replyShare slow_but_detemined commented Jan 27, 2020 reply Follow flag B ? by applying the telescoping technique. 1 1 replyShare neeraj_bhatt commented Sep 9, 2020 reply Follow flag B is the answer 0 0 replyShare KAUNIL commented Mar 10, 2023 reply Follow flag https://math.stackexchange.com/a/4655871/952652 0 0 replyShare Please log in or register to add a comment.