Evaluation of the integral

Question

How to evaluate the following integral? $$\int\frac{e^{-x}dx}{\sqrt x+ae^{-x}}$$ where $a$ is a constant.

Answer 1

If you are ready to use special functions, the only way I can see is $$I=\int\frac{e^{-x}}{\sqrt x+ae^{-x}}\,dx=\int \frac{dx}{a+e^x \sqrt{x}}$$ $$\frac{dx}{a+e^x \sqrt{x}}=\sum_{n=0}^\infty (-1)^n \left(\frac{e^{-x}}{\sqrt{x}}\right)^{n+1}\,a^n$$ $$I=\sum_{n=0}^\infty (-1)^n a^n \int\left(\frac{e^{-x}}{\sqrt{x}}\right)^{n+1}\,dx=\sum_{n=0}^\infty (-1)^n a^n J_n$$
$$J_n=\int \left(\frac{e^{-x}}{\sqrt{x}}\right)^{n+1}\,dx=-x^{\frac{1-n}{2}} E_{\frac{n+1}{2}}((n+1) x)$$ where appear the exponential integral function.

Edit

For illustration, using $a=1$ and integrating between $1$ and $10$, the partial sums (up to $p$) would be $$\left( \begin{array}{cc} p & \text{partial sum} \\ 0 & 0.27879186 \\ 1 & 0.22989135 \\ 2 & 0.24162796 \\ 3 & 0.23842973 \\ 4 & 0.23936339 \\ 5 & 0.23907879 \\ 6 & 0.23916815 \\ 7 & 0.23913948 \\ 8 & 0.23914883 \\ 9 & 0.23914574 \\ 10 & 0.23914677 \end{array} \right)$$ while the "exact" result obtained by numerical integration would be $0.23914652$.