I need to integrate $\int_0^{\Omega}\sqrt{\log(1+1/\theta)}d\theta$ whenever $\Omega$ is very small.
2026-04-11 23:44:58.1775951098
How to evaluate the integral defined below analytically,
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You could use the double inequality (see here) $$1-\frac 1t \leq \log(t)\leq t-1$$ which makes $$\frac 1{\sqrt{x+1}}\leq \sqrt{\log \left(1+\frac{1}{x}\right)}\leq \frac 1{\sqrt{x}}$$ So, integrating $$\int_0^y\frac {dx}{\sqrt{x+1}}\leq \int_0^y\sqrt{\log \left(1+\frac{1}{x}\right)}\,dx\leq \int_0^y\frac {dx}{\sqrt{x}}$$ $$2 \sqrt{y+1}-2\leq \int_0^y\sqrt{\log \left(1+\frac{1}{x}\right)}\,dx\leq 2 \sqrt{y}$$ I do not see what more could be done (at least by myself).