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The Gaussian approximation to a nonresonant thermonuclear rate already gives the saddle and curvature. In its notation $E_G=4E_B$, so the same result readsThe full width ${\Delta }_{1/e}$ joins the two points at $e^{-1}$ of the peak of the approximating Gaussian function; it is neither its standard deviation ${\sigma }_G$ nor its full width at half maximum $2\sqrt{2\log 2}{\sigma }_G$. These are local Gaussian approximation widths, not exact equal-height endpoints of the asymmetric kernel.

Why keep the astrophysical S factor inside the integral? If $S(E)$ is positive and smooth, stationarity of $\log S(E)-\Phi (E)$ gives a leading peak displacement $\delta E\simeq {\sigma }_G^2(d\log S/dE{)}_{E_0}$. The slow-variation test is therefore $|d\log S/dE|{\sigma }_G\ll 1$, together with small curvature corrections. More generally any energy-dependent prefactor shifts the saddle; a temperature-dependent prefactor also changes the local temperature exponent of a thermonuclear reaction. A sharp nuclear feature can invalidate this expansion altogether.

Choose $T=15\mathrm{MK}$ and $E_G=493\mathrm{keV}$ for an illustrative proton pair. Then $k_{B}T=1.293\mathrm{keV}$, $E_0=5.905\mathrm{keV}$ and ${\Delta }_{1/e}=6.380\mathrm{keV}$. The relative Maxwell-Boltzmann velocity distribution has energy-density maximum $k_{B}T/2=0.646\mathrm{keV}$, far below the reaction kernel's maximum. Now choose the deliberately artificial enhancementNumerical integration raises the rate by about $54.2\%$ at the same temperature, even though the enhancement is away from $E_0$. This is a sensitivity experiment, not a claimed proton–proton nuclear level or a physical nuclear line-shape model. It illustrates why extrapolating $S(E_0)$ without checking nearby structure can fail.