I would appreciate any comment/correction about what I did for the following problem, I would be very thankful if you let me know the parts of it which may not be very precise:
Let $g(z)$ be defined as $$g(z)=\int_{z_1}^{z} \frac{d\tau}{\sqrt{\tau(\tau-z_1)}}$$ Take the branch of $\displaystyle \sqrt{z(z-z_1)}$ with the branch cut $[0,z_1]$ and $\displaystyle \sqrt{z(z-z_1)}>0 $ for $z>z_1$. I want to check how this function behaves as $z \to \infty$ and most importantly as $z \to 0$.
It behaves like $\ln(z)$ as $z \to \infty$ because $$ g(z)=\int_{z_1}^{z} \frac{d\tau}{\sqrt{\tau^2-z_1\tau}} = \int_{z_1}^{z} \frac{d\tau}{\sqrt{\left( \tau-\frac{z_1}{2}\right)^2-\left( z_1/2\right)^2}} = \int_{\kappa(z_1)}^{\kappa(z)} \frac{d\kappa}{\sqrt{\left( \kappa\right)^2-\left( z_1/2\right)^2}} = \int_{\theta(\kappa(z_1))}^{\theta(\kappa(z))} \sec(\theta) d\theta $$ where $$ \kappa = \tau-\frac{z_1}{2} \ \ \ \ \ \ \text{,} \ \ \ \ \kappa= \frac{z_1}{2} \sec(\theta) \ \ \ \ \ \Rightarrow \ \ \ \ d\kappa =\frac{z_1}{2} \sec(\theta)\tan(\theta) d\theta $$ so $$ g(z) = \int_{\theta(\kappa(z_1))}^{\theta(\kappa(z))} d(ln(\sec\theta+\tan\theta)) = \ln(\sec\theta + \tan \theta) = \ln\left( \frac{2 \kappa}{z_1}+\sqrt{\left( \frac{2 \kappa}{z_1}\right) ^2-1}\right) = $$ $$ \ln\left( \frac{2\tau}{z_1}-1 + \sqrt{\left( \frac{2\tau}{z_1}-1 \right)^2-1 } \right)\Biggl|^{z}_{z_1} = \ln\left( \frac{2\tau}{z_1}-1 + \sqrt{\left( \frac{2\tau}{z_1} \right)^2 - \frac{4\tau}{z_1} } \right)\Biggl|^{z}_{z_1} = $$ $$ \ln\left( (\frac{2\tau}{z_1}) \left( 1+ \sqrt{1-\frac{z_1}{\tau} } -\frac{z_1}{2\tau} \right) \right) \Biggl|^{z}_{z_1} = \ln\left( (\frac{2z}{z_1}) \left( 1+ \sqrt{1-\frac{z_1}{z} } -\frac{z_1}{2z} \right) \right) $$ For large $z$ we have $\displaystyle \sqrt{1-\frac{z_1}{z} } = 1-\frac{z_1}{2z}-\frac{z^2_1}{8z^2}-\cdots$. So for large $z$ we can write $$ g(z)= \ln\left( (\frac{2z}{z_1}) \left( 2 -\frac{z_1}{z}-\frac{z^2_1}{8z^2}-\frac{z_1^3}{16z^3} + O(z^{-4}) \right) \right) = \ln(\frac{4}{z_1})+\ln(z)+\underbrace{\ln\left( 1 -\frac{z_1}{2z}-\frac{z^2_1}{16z^2}-\frac{z_1^3}{32z^3} + O(z^{-4} ) \right)}_{O(z^{-1})} $$ So we are done with the case $z\to \infty$;
Now for the case when $z \to 0$; I argue as follows (what I am not so sure about)
$$g(z)=\ln\left( \frac{2\tau}{z_1}-1 + \sqrt{\left( \frac{2\tau}{z_1}-1 \right)^2-1 } \right)\Biggl|^{z}_{z_1} = \ln\left( \frac{2z}{z_1}-1 + \sqrt{\left( \frac{2z}{z_1}-1 \right)^2-1 } \right)$$ Now, note that $$\sqrt{\left( \frac{2z}{z_1}-1 \right)^2-1 }=\sqrt{ \frac{4z^2}{z^2_1} -\frac{4z}{z_1} }=O(\sqrt{z}) \ \ \ \ \ \ z \to 0 $$ hence $$\frac{2z}{z_1} + \sqrt{\left( \frac{2z}{z_1}-1 \right)^2-1 } = O(\sqrt{z}) \ \ \ \ \ \ z \to 0$$ and hence $$g(z)=\ln(-1)+\ln\left(1- \frac{2z}{z_1} - \sqrt{\left( \frac{2z}{z_1}-1 \right)^2-1 } \right)= \ln(-1) + \ln(1-O(\sqrt{z})) = $$ $$ i\pi + O(\sqrt{z}) \ \ \ \ \ \ z \to 0 $$
Thank you so much in advance !