How to find $\int \sum_{n=0}^\infty \frac {(\ln x)^n}{x^n.n!}dx $? I know how to find the integral of a normal power series like the power series of $ \sin (x) $, but have no idea how to integrate the $( \ln x )^n $ term inside the summation.
 A: In fact this cannot be done in terms of elementary functions. However, by invoking the Incomplete Gamma Function we get that
$$\int \frac{\log^n(x)}{x^n}\mathrm dx=\int \frac{\log^n(x)}{x^{n-1}}\frac{\mathrm dx}x\stackrel{\log(x)=t}=\int \frac{t^n}{e^{(n-1)t}}\mathrm dt =\int t^ne^{-(n-1)t}\mathrm dt$$
The structure of the last integral is clearly recognizable as the one given by the Gamma Function. Thus, we can further obtain
$$\int t^ne^{-(n-1)t}\mathrm dt\stackrel{(n-1)t\mapsto t}=\int \left(\frac t{n-1}\right)^ne^{-t}\frac{\mathrm dt}{n-1}=\frac1{(n-1)^{n+1}}\int t^ne^{-t}\mathrm dt$$
Hence the integrand vanishes while $t$ is approaching towards infinity we may write the latter integral as
$$\int (t')^ne^{-t'}\mathrm dt'=-\int_t^\infty (t')^ne^{-t'}\mathrm dt'=-\Gamma(n+1,t)$$
Putting the aforementioned results together and resubsitute $t=(n-1)\log(x)$ we finally obtain a anti-derivative, which WolframAlpha confirms, namely

$$\therefore~\int \frac{\log^n(x)}{x^n}\mathrm dx~=~-\frac1{(n-1)^{n+1}}\Gamma(n+1,(n-1)\log(x))$$

A: The series sums to $e^{(\ln x)/x}= e^{\ln x^{1/x}} = x^{1/x}.$ So your problem is the same as finding $\int x^{1/x}\,dx$ on $(0,\infty).$
