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.
On
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 $$2 \sinh ^{-1}\left(\frac{1}{\sqrt{a}}\right) \simeq \log (4)-\log (a)+\frac{a}{2}-\frac{3 a^2}{16}$$
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.