I need help finding the solution of the next PDE using the Fourier-Cosine transform:
$$ 0<x<\pi,\;y\in\mathbb{R}^+ \qquad \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0$$ $$0<x<\pi \qquad \frac{\partial u}{\partial y}(x,0)=0$$ $$0<y\qquad u(0,y)=0$$ $$0<y\qquad u(\pi,y)=e^{-y} $$
Here is what I've done so far, first I obtain the Transform of the PDE: $$ \mathscr{F}_c\left\lbrace \frac{\partial^2u}{\partial x^2}\right\rbrace=\int_0^{\infty} \frac{\partial^2u}{\partial x^2}(x,y)\cos(\alpha y)dy=\frac{\partial^2}{\partial x^2}\int_0^{\infty} u(x,y)\cos(\alpha y)dy=\frac{\partial^2U}{\partial x^2}(x,\alpha)=\frac{d^2U}{dx^2}$$ $$ \mathscr{F}_c\left\lbrace \frac{\partial^2u}{\partial y^2}\right\rbrace=\int_0^{\infty} \frac{\partial^2u}{\partial y^2}(x,y)\cos(\alpha y)dy=-\frac{\partial u}{\partial y}(x,0)-\alpha^2U=-\alpha^2U$$ So we have the ODE : $$ \frac{d^2U}{dx^2}-\alpha^2U=0\rightarrow U(x,\alpha)=c_1e^{\alpha x}+c_2e^{-\alpha x}$$ Obtaining the Fourier transform of the remaining initial conditions: $$ \mathscr{F}_c\left\lbrace u(0,y) \right\rbrace=\int_0^{\infty}u(0,y)\cos(\alpha y)dy=\int_0^{\infty}(0)\cos(\alpha y)dy=0=U(0,\alpha)$$ $$ \mathscr{F}_c\left\lbrace u(\pi,y) \right\rbrace=\int_0^{\infty}u(\pi,y)\cos(\alpha y)dy=\int_0^{\infty}e^{-y}\cos(\alpha y)dy=\frac{1}{\alpha^2+1}=U(\pi,\alpha)$$ We now obtain constants $c_1$ and $c_2$ with the given conditions: $$U(0,\alpha)=c_1e^{0}+c_2e^{-0}=c1+c_2=0 \longrightarrow c_1=-c_2 $$ $$ U(\pi,\alpha)=ce^{\alpha \pi}-ce^{-\alpha \pi}=2c\frac{e^{\alpha\pi}-e^{-\alpha\pi}}{2}=2c\sinh(\alpha\pi)=\frac{1}{\alpha^2+1}\rightarrow c=\frac{1}{2\sinh{\alpha\pi}(\alpha^2+1)}$$ Substitute $c$ back into the solution: $$ U(x,\alpha)=\frac{1}{2\sinh{\alpha\pi}(\alpha^2+1)}\left(e^{\alpha x}-e^{-\alpha x}\right)=\frac{\sinh(\alpha x)}{\sinh(\alpha \pi)}\cdot\frac{1}{\alpha^2+1}$$ Finally we obtain the inverse Fourier transform: $$u(x,y)=\mathscr{F}_c^{-1}\left\lbrace U(x,\alpha)\right\rbrace=\frac{1}{2\pi}\int_0^{\infty}\frac{\sinh(\alpha x)}{\sinh(\alpha \pi)}\cdot\frac{\cos(\alpha y)}{\alpha^2+1}d\alpha$$ And that's pretty much where I'm stuck, is there a way to solve the integral? Did I do something wrong on my formulation? Or should I just leave the answer in terms of the integral?