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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.

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Divergent. Hint: find a simple equivalent of the integrand at infinity. –  Did Nov 1 '12 at 8:14
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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 $$\int_1^\infty\frac{1}{x}dx=\lim_{x\to\infty}\ln x=\infty$$

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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
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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?

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Thanks for the answer. –  Basil R Nov 1 '12 at 8:31
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