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 need help with this integral:

$\Large {\int_0^\infty \frac{dx}{x\sqrt{1+x}}} $

What I did: Substitute $\sqrt {1+x} = t$. Then the integral turns into $ \int_1^\infty 2dt/(t^2-1) $. Now I replaced $\infty$ by a variable $m$ and took the limit as $m \rightarrow \infty$. However I don't get a proper answer.

share|improve this question
3  
$\Large {\int_0^\infty \frac{dx}{x\sqrt{1+x}}} $ is not converghent at lower bound. –  JJacquelin Aug 14 at 6:31

2 Answers 2

The improper integral diverges. For if $x\le 1$ then $\frac{1}{x\sqrt{x+1}}\gt \frac{1}{2x}$. And it is easy to show that $\int_0^1 \frac{1}{2x}\,dx$ diverges.

share|improve this answer

Your work is correct as long as we consider the antiderivative $$\int\frac{dx}{x\sqrt{1+x}}=2\int\frac{dt}{t^2-1}=\int \Big(\frac{1}{t-1}-\frac{1}{t+1} \Big){dt}=\log \Big(\frac{1-t}{1+t}\Big)$$ Now, the problem starts with the integral because of the lower bound and not with the upper bound. What you could show is that, if $a \gt 0$, $${\int_a^\infty \frac{dx}{x\sqrt{1+x}}}=2 \sinh ^{-1}\left(\frac{1}{\sqrt{a}}\right)$$ For small positive values of $a$, a Taylor expansion is $$2 \sinh ^{-1}\left(\frac{1}{\sqrt{a}}\right) \simeq \log (4)-\log (a)+\frac{a}{2}-\frac{3 a^2}{16}$$

share|improve this answer

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.