How to find a conformal map from the upper half-plane to this L-shaped domain? I want to find an integral representation of the conformal map that sends the upper half plane $\text {Im}z>0$ onto the infinite L-shaped region $$
\Omega = \{z = x+iy; \ x > 0, \ y > 0, \ \min(a,\frac yb) < 1 \}
$$ 

which is a slight modification of 
this question, where we change the vertex of the turn from the point $1+i$ to the more general point $1+bi$ in the first quadrant. (This is an exercise in Gamelin's text)
By the Schwarz-Christoffel formula, the solution is of the form $$
F(w) = K\int_{z_1}^w (\zeta-z_0)^{-1}(\zeta-z_1)^{-1/2}(\zeta-z_2)^{-1} \ d\zeta
$$
where $z_0, z_1, z_2$ are the numbers that get mapped by the conformal mapping to the vertices at $i \infty, 0, +\infty$ respectively. I do not know what are these preimages of the vertices.
In the linked question they are given as $+1, -1, 0, \infty$ for the particular case.
Ahlfors says in his book that in general there is no formula for determining the prevertices, and only $3$ of them can be arbitrarily chosen. Since we have $4$ points to determine in this case, we may assume that $3$ of the prevertices are $z_0=-1, z_1=0, z_2=\infty$. 

I was told that the right decision for the remaining vertex is $b^2$. How can I get this? Is there an intuitive explanation for why this is the right choice of the prevertex $z_2$, perhaps using the symmetry of the domain along the line $y=bx$?

 A: The L-shaped domain has four vertices $0,\infty, 1+bi$ and $i\infty.$ As Ahlfors says, we can choose $3$ of prevertices aibitrarily, so we may assume$$
z=-1\longleftrightarrow w=i\infty,\quad z=0\longleftrightarrow w=0,\quad z=\infty \longleftrightarrow w=1+bi.$$ Let the remaining prevertex be $z=a$ (it's value is unknown at the present time).   Then the mapping function has the form
$$
F(z) = K\int_{0}^z (\zeta+1)^{-1}\zeta^{-1/2}(\zeta-a)^{-1} \ d\zeta.\tag{1}
$$
We have to determine the value of $a$ so that $F(\infty)=1+bi.$ 
\begin{align}
F(\infty)&=F(i\infty)= K\int_0^\infty \frac{idt}{(it+1)\sqrt{it}(it-a)}     \\
 &= -\frac{K}{\sqrt{2}}\int_0^\infty \frac{(1+i)(t^2+a+it(1-a))}{(t^2+1)\sqrt{t}(t^2+a^2)}dt    \\&=I+J+i(I-J),
\end{align}
where $$
I=-\frac{K}{\sqrt{2}}\int_0^\infty \frac{t^2+a}{(t^2+1)\sqrt{t}(t^2+a^2)}dt,\quad J=-\frac{K}{\sqrt{2}}\int_0^\infty \frac{(a-1)\sqrt{t}}{(t^2+1)(t^2+a^2)}dt.$$
Since $F(i\infty)$ should be $1+bi,$ $K,$ $a$ should be determined so that $$
I+J=1,\quad I-J=b$$ holds. I could not get $a=b^2$, however $(1)$ is an integral expression for $w=F(z)$.
EDIT:  $$a=\frac{1}{b^2}$$ will be true, not $a=b^2.$
In the case $b=1$, we can find $F(z)$ explicitly. When $b=1$, we have $a=1.$ Thus $$
-\frac{K}{\sqrt{2}}\int_0^\infty \frac{dt}{(t^2+1)\sqrt{t}}=-\frac{\pi K}{2}=1
$$
gives $K=-\frac{2}{\pi}$ and we get explicitly
$$F(z)=-\frac{2}{\pi}\int_0^z \frac{d\zeta }{(\zeta^2 -1)\sqrt{\zeta }}=\frac{1}{\pi}\left(\log \frac{1+\sqrt{z}}{1-\sqrt{z}}+\tan^{-1}\sqrt{z}\right).$$
