Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to verify if the integral $$\int_1^{\infty}\ln((\exp(1/x)+(n-1))/n)dx, \;\;n>0$$

is convergent or divergent? I made some handy calculations but couldn't find any way for it. Please give me help.

share|improve this question
2  
Divergent. Hint: find a simple equivalent of the integrand at infinity. –  Did Nov 1 '12 at 8:14
add comment

2 Answers

up vote 4 down vote accepted

Note that $e^{1/x}+(n-1)/n\geq e^{1/x}$ and so $\ln (e^{1/x}+(n-1)/n)\geq 1/x$. Hence the given integral is greater than or equal to $$\int_1^\infty\frac{1}{x}dx=\lim_{x\to\infty}\ln x=\infty$$

share|improve this answer
    
Your first step is a nice trick. Thank you very much. –  Basil R Nov 1 '12 at 8:31
    
The idea is nice but the logarithm is not the one in the question hence its lower bound does not apply, as soon as $n\gt1$. (So much for the upvotes and for the acceptance 14 minutes after the answer is posted.) –  Did Nov 1 '12 at 8:38
add comment

HINT: Besides to @pritman's answer; consider the integrand as $f(x)=\ln(1+\frac{e^{x^{-1}}-1}{n})$. I agree with @did that $f(x)=\ln(1+\frac{e^{x^{-1}}-1}{n})\sim\frac{1}{nx}$ when $n$ tends to infinity. Can you do the rest?

share|improve this answer
    
Thanks for the answer. –  Basil R Nov 1 '12 at 8:31
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.