It is an integral from a note.
We concerend conformal map on $\mathbb{C}_+=\left\{ z:\text{Im}z>0 \right\} $ which satisfies $$ \text{Im}f\left( z \right) >0,\text{Im}f'\left( z \right) \ne 0\qquad on~\mathbb{C}_+ $$ A such example is given that $$ f_{\alpha}\left( z \right) =\int_0^{\infty}{\left( z+t \right) ^{\frac{1}{2}}\left( 1+t \right) ^{-\alpha -2}}dt $$ for $\alpha>-1/2$.
And I want to show this integral is meaningful. So we should prove two things.One is that it is convergent,and the other is holormorphic.The latter one is easy by Fubini theorem and Morera theorem.However,I don't know how to show this integral is convergent,for it is hard for me to find an estimation.And I want to split $\mathbb{C}_+$ but failed. Besides,I think that $\sqrt{z+t}$ has no bounds,and the integral seems not to be convergent?
Assuming that $z$ doesn't belong to the real interval $(-\infty,0)$, the map $g_{\alpha,z}(t) = \left( z+t \right) ^{\frac{1}{2}}\left( 1+t \right) ^{-\alpha -2}$ is continuous on $[0, \infty)$ and
$$\vert g_{\alpha,z}(t) \vert \le \left(\vert z \vert+t \right) ^{\frac{1}{2}}\left( 1+t \right) ^{-\alpha -2} = \left(\frac{\vert z \vert+t}{(1+t)}\right)^{1/2}\left( 1+t \right) ^{-\alpha -3/2}$$
As $\lim\limits_{t \to \infty} \left(\frac{\vert z \vert+t}{(1+t)}\right)^{1/2} = 1$ and $\alpha +3/2 \gt 1$, the integral is convergent.