Integral of $\int x^{-x} dx$

Question

Question: $\int x^{-x} dx =$ ?

Hint: $$ e^{x\ln \frac{1}{x}} = \sum_{n=0}^\infty \frac{x^n}{n!} \left(\ln\left(\frac{1}{x}\right)\right)^n$$

I figure since $\int x^{-x} dx = \int e^{x\ln \frac{1}{x}} dx$, maybe I should find the correlation between $\int e^{x\ln \frac{1}{x}} dx$ and $\int e^{-x\ln \frac{1}{x}} dx$. But I still can't think of the connection between the two

I used WolframAlpha to solve it but it didn't show the process

Answer 1

It seems to me that we should use the hint and then swap integral and summation (if we can?) so the main issue would be to evaluate:

$$\int x^n [\ln(\frac{1}{x})]^n dx \ \ (\text{removed the} \frac{1}{n!})$$

Not sure if I'm right but it seems helpful to repeated integration by parts to arrive at what I think should be:

$$ [\ln(\frac{1}{x})]^n \frac{x^{n+1}}{n+1} + [n \ln(\frac{1}{x})]^{n-1} \frac{x^{n+1}}{(n+1)^2} + [n (n-1) \ln(\frac{1}{x})]^{n-2} \frac{x^{n+1}}{(n+1)^3} + \dots + [n! \frac{x^{n+1}}{(n+1)^n}]$$

Edit: Or evaluate:

$$\int x^n [\ln(x)]^n dx \ \ (\text{removed the} \frac{(-1)^n}{n!})$$

I think that's

$$\frac{(\ln x)^n x^{n+1}}{n+1} - \frac{(\ln x)^{n-1} (n) x^{n+2}}{(n+1)(n+2)} + ... \text{something}$$