I'm working on Exercise 20 from Chapter 8 of Stein and Shakarchi, which asks you to prove the following:
(a) The function $$\int_0^x \frac{d\zeta}{\sqrt{\zeta)(\zeta-1)(\zeta-\lambda)}}$$ with $\lambda \in \mathbb R, \lambda \neq 1$, maps the upper half-plane conformally to a rectangle, one of whose vertices is the image of the point at infinity.
(b) In the case of $\lambda = -1$, the image of $$\int_0^x \frac{d\zeta}{\sqrt{\zeta(\zeta^2-1)}}$$ is a square whose side lengths are $\frac{\Gamma^2(1/4)}{2\sqrt{2\pi}}$.
I believe I've solved (a) by applying Prop. 4.1 from the chapter, which tells you about the vertices and angles of the polygon that's mapped to by the real axis. But I'm really not sure how to proceed for part (b). (I don't have a good grasp on how to tackle problems about conformal maps and how to determine things about the image)
Note: this is a homework problem, so hints but not full solutions would be appreciated!
Hints: Let $$ F(z)=\int_0^z \frac{d\zeta}{\sqrt{\zeta(\zeta^2-1)}}.$$ The image of $F(z)$ is a rectangle and has four vertices $F(0)=0, F(1),F(\infty)$ and $F(1).$
If $$|F(1)-F(0)|=|F(\infty)-F(1)|$$ holds, then the image is a square.
Note that \begin{align} F(1)-F(0)&=\int_0^1 \frac{d\zeta}{\sqrt{\zeta(\zeta^2-1)}}=\frac{1}{i}\int_0^1 \frac{d\zeta}{\sqrt{\zeta(1-\zeta^2)}},\\ F(\infty)-F(1)&=\int_1^\infty \frac{d\zeta}{\sqrt{\zeta(\zeta^2-1)}}=\int_0^1 \frac{dt}{\sqrt{t(1-t^2)}}\quad (t=1/\zeta). \end{align} Next $$ \int_0^1 \frac{dt}{\sqrt{t(1-t^2)}}=\int_0^{\frac{\pi}{2}} \frac{d\theta }{\sqrt{\sin \theta }} =\frac{1}{2}\text{B}\left(\frac{1}{2},\frac{1}{4}\right),$$ where $\text{B}(x,y)$ is the beta function.