= Solution
Insert the position and momentum resolutions of the identity to obtain the unitary <Fourier transform>
$$
\widetilde\psi(p)=\frac1{\sqrt{2\pi\hbar}}\int_{\mathbb R}e^{-ipx/\hbar}\psi(x)\,dx,\qquad
\psi(x)=\frac1{\sqrt{2\pi\hbar}}\int_{\mathbb R}e^{ipx/\hbar}\widetilde\psi(p)\,dp.
$$
The kinetic term is multiplication by $p^2/(2m)$, and the potential has momentum kernel $\langle p|V|p'\rangle=(2\pi\hbar)^{-1}\int V(x)e^{-i(p-p')x/\hbar}\,dx$. Thus the <Schrödinger equation> becomes
$$
\left(\frac{p^2}{2m}-E\right)\widetilde\psi(p)
=-\frac1{2\pi\hbar}\int_{\mathbb R^2}V(x)e^{-i(p-p')x/\hbar}\widetilde\psi(p')\,dx\,dp'.
$$
For the attractive <delta potential>, $V(x)=-(\hbar^2\lambda/m)\delta(x)$, the kernel is the constant $-\hbar\lambda/(2\pi m)$, giving the stated positive right-hand side. For $\widetilde\psi=N/(p^2+\alpha^2)$ with $\alpha>0$, constancy of the left side requires $E=-\alpha^2/(2m)$. Since $\int dp/(p^2+\alpha^2)=\pi/\alpha$, its coefficient equation gives
$$
\boxed{\alpha=\hbar\lambda,\qquad E=-\frac{\hbar^2\lambda^2}{2m}.}
$$
Finally, transforming $\sqrt\lambda e^{-\lambda|x|}$ gives $\widetilde\psi=2\lambda^{3/2}\hbar^2/[\sqrt{2\pi\hbar}(p^2+\hbar^2\lambda^2)]$, verifying the same momentum-space form and energy.
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