Definite integral of $e^{-e^x} dx$ and the zero problem How can I evaluate
$$\int\limits_{0}^{a} {e}^{-e^x} dx$$
After converting $e$ to its equivalent series, I get an integral with a $1/n$ but $n$ my limit starts from $0$ and division by zero is undefined.
Is there any workaround?
 A: $$\int_0^ae^{e^{-x}}\mathrm{d}x=\int_0^a1+\sum_{m=1}^{\infty}\frac{(e^{-x})^m}{m!}\mathrm{d}x=a+\sum_{m=1}^{\infty}\frac1{m!}\int_0^ae^{-mx}\mathrm{d}x=a+\sum_{m=1}^{\infty}\frac1{m!}\frac{1-e^{-am}}{m}=a+\sum_{m=1}^{\infty}\frac1{m\cdot{m!}}-\sum_{m=1}^{\infty}\frac{(e^{-a})^m}{m\cdot{m!}}.$$ Notice that the function $E$ defined by $$E(x):=\int_0^x\frac{e^t-1}{t}\mathrm{d}t$$ has the Maclaurin series expansion $$\sum_{m=1}^{\infty}\frac{x^m}{m\cdot{m!}}.$$ Therefore, we can write that $$\int_0^ae^{e^{-x}}\mathrm{d}x=a+E(1)-E(e^{-a})=a+\int_{e^{-a}}^1\frac{e^t-1}{t}\mathrm{d}t=a+\int_{e^{-a}}^1\frac{e^t}{t}\mathrm{d}t-\int_{e^{-a}}^1\frac1{t}\mathrm{d}t$$ $$=a-\ln\left(\frac1{e^{-a}}\right)+\int_{e^{-a}}^1\frac{e^t}{t}\mathrm{d}t=\int_{e^{-a}}^1\frac{e^t}{t}\mathrm{d}t.$$ However, this latter equality could have been achieved instead by substituting $t=e^{-x}$ in the original integral. So actually, this approach of using Maclaurin series expansions is not a useful one.
To evaluate $$\int_{e^{-a}}^1\frac{e^t}{t}\mathrm{d}t,$$ I suggest you look into the exponential integral function (https://en.wikipedia.org/wiki/Exponential_integral and https://mathworld.wolfram.com/ExponentialIntegral.html). The answer can be expressed as $\operatorname{Ei}(1)-\operatorname{Ei}(e^{-a}).$ It cannot be expressed using elementary functions.
