recategorized by
509 views
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

  1. $\log_e \frac{\beta}{\alpha}$
  2. $\log_e \frac{1+ \beta}{1 + \alpha}$
  3. $\log_e \frac{1+\alpha }{1+ \beta}$
  4. $\infty$
recategorized by

Please log in or register to answer this question.

Related questions

0 votes
0 votes
1 answer
2
Arjun asked Sep 23, 2019
452 views
Given that $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$, the value of $$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+xy+y^2)} dxdy$$ is$\sqrt{\pi/3}$$...
0 votes
0 votes
2 answers
3
Arjun asked Sep 23, 2019
512 views
The value of $$\displaystyle \lim_{n \to \infty} \left[ (n+1) \int_0^1 x^n \ln(1+x) dx \right]$$ is$0$$\ln 2$$\ln 3$$\infty$