0 votes 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-gate + – Arjun asked Sep 23, 2019 • recategorized Nov 17, 2019 by Lakshman Bhaiya Arjun 510 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply `JEET commented Oct 4, 2019 reply Follow Share $\ln |x + c| $ ? 0 votes 0 votes slow_but_detemined commented Jan 27, 2020 reply Follow Share B ? by applying the telescoping technique. 1 votes 1 votes neeraj_bhatt commented Sep 9, 2020 reply Follow Share B is the answer 0 votes 0 votes KAUNIL commented Mar 10, 2023 reply Follow Share https://math.stackexchange.com/a/4655871/952652 0 votes 0 votes Please log in or register to add a comment.