= Solution
{c}
The core contributes no change in $\varphi$, since $\varphi'=0$ there. Integrating the radial <gravitational field> across the shell, with $u=r/r_0$, gives
$$
\varphi(2r_0)-\varphi(0)
=-4\pi G\rho_0r_0^2\int_1^2\frac{2/e-(u+1)e^{-u}}{u^2}\,du.
$$
Using the supplied <antiderivative>,
$$
\begin{aligned}
\int_1^2\frac{2/e-(u+1)e^{-u}}{u^2}\,du
&=\left[-\frac2{eu}+\frac{e^{-u}}u\right]_1^2\\
&=\left(-\frac1e+\frac1{2e^2}\right)-\left(-\frac2e+\frac1e\right)
=\frac1{2e^2}.
\end{aligned}
$$
Hence
$$
\boxed{\varphi(2r_0)-\varphi(0)=-\frac{2\pi}{e^2}G\rho_0r_0^2,\qquad\alpha=-\frac{2\pi}{e^2}.}
$$
The negative sign agrees with $\varphi'(r)<0$ in the material shell under the paper's <gravitational potential sign conventions>.
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