Integrating $\int_0^ex^{1/x}\ \mathrm dx$

Compute $$\int_0^ex^{1/x}\;\mathrm dx.$$

There is an analytical anti-derivative found in this answer. How does one compute this?

Using the anti-derivative approach we have

$$\int x^{1/x}\;\mathrm d x=x + \frac{\log^2x}{2}-\sum^\infty_{n=2}\sum^n_{k=0}\frac{\log^{n-k}x\;}{x^{n-1}(n-k)!(n-1)^{k+1}}+C$$

Now there is a problem for this anti-derivative as it gets near $0$ (apparent from the $\log$ s). I do not know how to prove this but if $f(x)=x^{1/x}$ and $\lim_{x\to0}f(x)=0$ can be defined then I want to say that $\lim_{x\to0}F(x)=0$. Assuming my non-rigorous logic and some logic from this question we have $$\int_0^ex^{1/x}\;\mathrm dx=\sum _{n=2}^{\infty } \frac{ \Gamma (n+1,\;n-1)}{(n-1)^{n+1}\;n!}+\frac{1}{2}+e$$

Update: Using Mathematica I computed some numerical integrals to find that $F(0)\approx1.53328$. Therefore we have $$\int_0^ex^{1/x}\;\mathrm dx=\sum _{n=2}^{\infty } \frac{ \Gamma (n+1,\;n-1)}{(n-1)^{n+1}\;n!}+\frac{1}{2}+e-\lim_{x\to0}\int x^{1/x}\;\mathrm dx$$

Solution 1:

Here is a way a physicist is approaching the problem.

First, the variable change $x=\frac{1}{y}$ gives:

$$\int_0^ex^{1/x}\;\mathrm dx=\int_{\frac{1}{e}}^{\infty}\frac{y^{-y}}{y^2}dy$$

Now, as a first approximation, we use the following inequality:

$$\int_{\frac{1}{e}}^{\infty}\frac{y^{-y}}{y^2}dy<\int_{\frac{1}{e}}^{\infty}\frac{y^{-\frac{1}{e}}}{y^2}dy=\int_{\frac{1}{e}}^{\infty}y^{-\frac{1}{e}-2}dy=\frac{e^{2+\frac{1}{e}}}{e+1}$$

Finally:

$$0<\int_0^ex^{1/x}\;\mathrm dx<\frac{e^{2+\frac{1}{e}}}{e+1};\;e>0$$

This result can be probably improved.