Integral of $(\ln(x)/x)^n$

Question

How would you go about integrating the function $\left(\frac{\ln(x)}{x}\right)^n$ for some integer $n≥2$?

I've managed to recursively calculate it using integration by parts to get a nasty series but WolframAlpha says it's equal to $$\frac{-\Gamma(n+1,(n-1)\ln(x))}{(n-1)^{n+1}}$$

where $\Gamma(a,x)$ is the incomplete gamma function. Why is this the case?

Answer 1

As Gary suggested, let $x=e^{\frac{t}{n-1} }$, then $t=(n-1)\ln x$ and $$ \begin{aligned} I & =\int \frac{\left(\frac{t}{n-1}\right)^n}{e^{\frac{n t}{n-1}}} \cdot \frac{1}{n-1} e^{\frac{t}{n-1}} d t \\ & =\frac{1}{(n-1)^{n+1}} \int t^n e^{-t} d t \\ & =-\frac{1}{(n-1)^{n+1}} \Gamma(n+1, t)\\&=-\frac{1}{(n-1)^{n+1}} \Gamma(n+1,(n-1)\ln x) \end{aligned} $$

Answer 2

Since Lai give an nice answer for your question using the Gamma incomplete function. I will try avoid (in some sense) the special function. Now, we can start slightly different using the substitution $t:=1/x$, then $dt=-1/x^2 dx$ and we have for $n\in {\bf N}$, $$\int \frac{\ln^{n}(x)}{x^{n}}dx=\int \frac{\ln^{n}\left(\frac{1}{t}\right)}{\frac{1}{t^{n}}}\left(\frac{-1}{t^{2}}\right)dt=(-1)^{n+1}\int t^{n-2}\ln^{n}(t) dt$$ The last primitive is part of article in Wikipedia about the Sophomore dream and as indicated in the article can be find using integration by parts and using the falling factorial notation $(n)_{k}$, that is, for $(m,n)\in {\bf N}\times {\bf N}$

$$\int t^{m}\ln^{n}(t)dt=\frac{t^{m+1}}{m+1}\sum_{0\leqslant k\leqslant n}(-1)^{k}\frac{(n)_{k}}{(m+1)^{k}}\ln^{n-k}(t).$$

Then, the result it is follows, setting $m:=n-2$.

Note: We can show that the falling factorial it is related with the Gamma function, since $(t)_{k}=\frac{\Gamma(t+1)}{\Gamma(t-k+1)}$.