= Solution
Interpret the malformed printed domain inequality as the first quadrant $x>0,y>0$, as indicated by the two axis boundary conditions. There is also a genuine missing hypothesis: decay of the boundary data does not require the <harmonic function> itself to be bounded. For zero data, both $q=0$ and $q=xy$ satisfy the <Laplace equation> and both edge values, but the <Wirtinger derivative> of the latter is
$$
q_z=\tfrac12(q_x-iq_y)=\tfrac12(y-ix)=-\tfrac i2z\ne0.
$$
Thus \b[the data do not determine an arbitrary solution's derivative without a growth restriction]. We derive the <integral representation> for the bounded solution, which is also the standard decaying solution when the continuous edge data decay. Any other admissible unbounded solution can differ by a zero-boundary <harmonic function>.
The <conformal map> $w=z^2$ sends the quadrant bijectively to the <complex upper half-plane>: the positive real axis maps to $s>0$, and the positive imaginary axis maps to $s<0$. Define its real-axis boundary function by
$$
f(s)=\begin{cases}g_2(\sqrt s),&s>0,\\g_1(\sqrt{-s}),&s<0,\end{cases}
\qquad f(0)=g_1(0)=g_2(0).
$$
The compatibility at the corner makes $f$ continuous there. The <Dirichlet Poisson integral in a quadrant> follows from the upper-half-plane <Poisson kernel>:
$$
q(x,y)=\frac1\pi\int_{\mathbb R}
\frac{\operatorname{Im}(z^2)f(s)}{|s-z^2|^2}\,ds.
$$
It is bounded by the supremum of the bounded edge data, and the kernel's <approximate identity> property supplies both <Dirichlet boundary conditions>. More explicitly, for $z^2=a+ic$, where $a=x^2-y^2$ and $c=2xy>0$, this integral is
$$
q(x,y)=\frac{2c}{\pi}\int_0^\infty r\left[
\frac{g_2(r)}{(r^2-a)^2+c^2}
+\frac{g_1(r)}{(r^2+a)^2+c^2}\right]dr.
$$
It displays directly which part of the real-line boundary corresponds to each original axis.
To obtain its derivative, first take real data and use the <Schwarz integral formula>. A <holomorphic function> whose <real part> is the Poisson solution in the $w$-plane is
$$
F(w)=\frac1{\pi i}\int_{\mathbb R}f(s)
\left[\frac1{s-w}-\frac{s}{1+s^2}\right]ds.
$$
The regularizing subtraction is independent of $w$ and affects only the imaginary normalization; it permits bounded data without an artificial integrability assumption. Then $q=\operatorname{Re}F(z^2)$. By the <chain rule> and the <Wirtinger derivative>,
$$
q_z=zF'(z^2)=\frac z{\pi i}\int_{\mathbb R}\frac{f(s)}{(s-z^2)^2}\,ds.
$$
The factor is $z$, not $2z$: taking the <real part> contributes the factor $1/2$ in the complex derivative. <Differentiation under the integral sign> is justified uniformly on compact interior subsets because the denominator stays away from the real boundary and decays quadratically at infinity. Splitting the real integral at zero and substituting $s=r^2$ or $s=-r^2$ gives the <holomorphic derivative of a quadrant Dirichlet solution>:
$$
\boxed{q_z(z)=\frac{2z}{\pi i}\int_0^\infty r\left[
\frac{g_2(r)}{(r^2-z^2)^2}
+\frac{g_1(r)}{(r^2+z^2)^2}\right]dr,\qquad x,y>0.}
$$
This converges for bounded data, and in particular for the stated sufficiently decaying data. No interior pole is encountered since $\operatorname{Im}(z^2)=2xy>0$. For complex-valued data the same formula holds by applying the real-data construction separately to their real and imaginary parts and using linearity.
An equivalent form, often convenient when the data's derivatives decay, comes from <integration by parts>. Since
$$
\frac{2r}{(r^2-z^2)^2}=-\frac d{dr}\frac1{r^2-z^2},\qquad
\frac{2r}{(r^2+z^2)^2}=-\frac d{dr}\frac1{r^2+z^2},
$$
the two endpoint terms are $-g_2(0)/z^2$ and $+g_1(0)/z^2$. They cancel by the specified compatibility, leaving
$$
\boxed{q_z(z)=\frac z{\pi i}\left[
\int_0^\infty\frac{g_2'(r)}{r^2-z^2}\,dr
+\int_0^\infty\frac{g_1'(r)}{r^2+z^2}\,dr\right].}
$$
This is the <harmonic quadrant solution from tangential boundary derivatives> applied to the actual derivatives $g_1',g_2'$; the supplied $g_1,g_2$ in this question are boundary values, not derivatives. Omitting those primes would solve a different boundary problem.
For completeness, <bounded Dirichlet uniqueness in a quadrant> selects the solution just constructed. The difference of two bounded solutions is zero on both axes. <Odd reflection> across each axis gives a bounded harmonic extension to the punctured plane; the <removable singularity for a bounded harmonic function> fills the corner. The <harmonic Liouville theorem> makes the extension constant, and oddness makes it zero. Decaying data give a decaying Poisson extension: split the real-axis integral into a fixed compact interval and a tail with arbitrarily small data; the compact contribution tends to zero as $|z^2|\to\infty$, while the tail is bounded by that small supremum.
As literally printed without boundedness or a growth condition, the most one can assert is
$$
\boxed{q_z=(q_{\rm bounded})_z+v_z,\qquad
\Delta v=0,\quad v|_{x=0}=v|_{y=0}=0.}
$$
The <zero-boundary harmonic growth in a quadrant> includes $v=cxy$ and more generally $v=\operatorname{Im}(z^{2m})$. This is an actual nonuniqueness of the printed problem, not a defect in the integral for the normalized bounded solution.
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