Proof $\int_{1}^{\infty}x^n(r)^\left(x-1\right) = {r^{-1}n!\over\ln^{n+1}(r^{-1})}$

Question

I was playing around with some series today and came across an interesting fact.

I've solved for $\sum_{x=1}^{\infty}x^2r^\left(x-1\right) = {1+r\over(1-r)^3}$ where $r < 1$

Then I tried degree n = 3 for: $\sum_{x=1}^{\infty}x^3r^\left(x-1\right) = {1-4r+r^2\over(1-r)^4}$

When I tried to find a general solution for n, it became to complicated but I noticed $$\sum_{x=1}^{\infty}x^nr^\left(x-1\right) \approx \int_{1}^{\infty}x^nr^\left(x-1\right)$$ So I tried solving this integral but again it proved to be too tedious/complicated.

Using an integral calculator I noticed $\int_{1}^{\infty}x^3(\frac13)^\left(x-1\right) = {18\over \ln^4 (3) }$ or in general $$\int_{1}^{\infty}x^n(r)^\left(x-1\right) = {r^{-1}n!\over\ln^{n+1}(r^{-1})}$$

Could someone proof that is true for when $r < 1$?

Answer 1

It can be show that $$\int_{0}^{\infty}x^{n}e^{-x}dx=n!$$ by repeated integration by parts (i.e. induction).

Here is the proof: let $I_{n}=\int_{0}^{\infty}x^{n}e^{-x}dx$ Then $$I_{n}=\left[-x^{n}e^{-x}\right]_{0}^{\infty}-\int_{0}^{\infty}-nx^{n-1}e^{-x}dx=nI_{n-1}$$
Combined with $I_{0}=\int_{0}^{\infty}e^{-x}dx=1$

This is more commonly stated as the identity $\Gamma(n+1)=n!$, where $\Gamma$ is the Gamma function.

Then your integral $$\int_{0}^{\infty}x^{n}r^{x-1}dx=\frac{1}{r}\int_{0}^{\infty}x^{n}e^{\log (r) x}dx$$ Then let $u=-\log(r)x$, to get $$\frac{1}{r}\frac{(-1)^{n+1}}{\log(r)^{n+1}}\int_{0}^{\infty}u^{n}e^{-u}du=\frac{n!r^{-1}}{\log(r^{-1})^{n+1}}$$

Answer 2

Since $$\int_0^{\infty} x^3 \frac{1}{3}^{(x-1)} dx=\frac{18}{\ln^4(3)}$$ holds I suppose you mean the integral from $0$ to $\infty$ (and not starting from 1).

To prove this, as suggested in my comment above, we can multiply both sides by $r$ to simplify. Furthermore, we can write $\ln(r^{-1})=a$ and thus obtain $$\int_0^{\infty} \frac{x^n}{e^{ax}} dx=\frac{n!}{a^{n+1}}$$ Now, by the integral definition of the Gamma Function we find that $LHS=\frac{\Gamma (n+1)}{a^{n+1}}$ which combined with the famous fact that $\Gamma (n)=(n-1)!$ (also easy to prove with integration by parts) gives the desired identity.