evaluation of this exponential integral? how could i evaluate
$$ \int_{0}^{a(E)}\sqrt{E-16\pi ^{2}e^{4x}} $$
where 'a' is the point so $ E-16\pi^{2}e^{4a}=0 $
this appear s in Quantum mechanic so i think the answer is something like
$$ E^{1/2}log(E) $$ with some constants but it goes like this

For any $K>1$ we have:
$$\begin{eqnarray*}\int_{0}^{\log K}\sqrt{K-e^{x}}\,dx =\int_{1}^{K}\frac{\sqrt{K-u}}{u}\,du&=&\sqrt{K}\int_{1/K}^{1}\sqrt{1-u}\,\frac{du}{u}\\&=&2\sqrt{K}\left(\text{arctanh}\left(\sqrt{1-\frac{1}{K}}\right)-\sqrt{1-\frac{1}{K}}\right)\\&=&\sqrt{K}\log\left(\frac{\sqrt{K}+\sqrt{K-1}}{\sqrt{K}-\sqrt{K-1}}\right)-2\sqrt{K-1}\\[0.2cm]&=&\color{red}{2\sqrt{K}\log(\sqrt{K}+\sqrt{K-1})-2\sqrt{K-1}}\end{eqnarray*}$$
hence you are right.
