Making Fourier transform by $x\to k$ we have
\begin{equation*}
\hat{u}_{yy}-k^2 \hat{u}=0, \qquad \hat{u}(0,k)=\hat{h}(k)
\end{equation*}
with
\begin{multline*}
\hat{h}(k)=\frac{1}{2\pi} \int_{-\pi}^\pi \sin (x) e^{-ikx}\,dx=\frac{1}{i\pi} \int_0^\pi \sin (x) \sin(kx)\,dx\\=
\frac{1}{2i\pi} \int_0^\pi \bigl(\cos ((k-1)x)-\cos ((k+1)x)\bigr)\,dx\\=
\frac{1}{2i\pi}\bigr( \frac{\sin((k-1)\pi )}{k-1}-\frac{\sin ((k+1)\pi )}{k+1}\bigr)=
\frac{1}{i\pi} \frac{\sin (k\pi)}{k^2-1}.\end{multline*}
Then $\hat{u}(k,y)=A(k)e^{-|k|y}+B(k)e^{|k|y}$ and we need to take $B(k)=0$, $A(k)=\hat{h}(k)$. So
\begin{equation*}
\hat{u}(k,y)=\frac{1}{i\pi} \frac{\sin (k\pi)}{k^2-1} e^{-|k|y}
\end{equation*}
and
\begin{equation*}
u(x,y)=\frac{1}{i\pi}\int_{-\infty}^\infty \frac{\sin (k\pi)}{k^2-1} e^{-|k|y+ik x}\,dk=
\frac{4}{\pi}\int_0^\infty \frac{\sin (k\pi)}{k^2-1} e^{-ky}\sin (kx)\,dk.
\end{equation*}
