Integrating Powers of Logarithms

Question

Throughout this post, let the constant of integration $C$ be $0$ for simplicity's sake.

I've calculated the indefinite integrals of powers of logarithms and they seem be following a pattern. Below are the first few.

\begin{align} \int\log(x)dx &= x\left(\log(x)-1\right)\\ \int\log^2(x)dx &=x\log^2(x)-2\int\log(x)dx\\ &=x\left(\log^2(x)-2\log(x)+2\right)\\ \int\log^3(x)dx &=x\log^3(x)-3\int\log^2(x)dx\\ &=x\left(\log^3(x)-3\log^2(x)+6\log(x)-6\right)\\ \int\log^4(x)dx &=x\log^4(x)-4\int\log^3(x)dx\\ &=x\left(\log^4(x)-4\log^3(x)+12\log^2(x)-24\log(x)+24\right) \end{align}

Clearly, for positive integers $n$ this generalises to

\begin{align} \int\log^n(x)dx =x\sum_{k=0}^{n}a_k\log^k(x) \end{align}

for some integer coefficients $a_k$.

From observing the pattern of the coefficients $a_k$, I conjecture that the general closed form for any integer $n$ is given by

$$\int\log^n(x)dx = n!x\sum_{k=0}^{n}\frac{(-1)^{n-k}}{k!}\log^k(x) \tag1$$

which holds for all the four cases that I've written down above.

However, I'm having trouble proving this to be true.

My attempt at a proof

$(1)$ is equivalent to

$$\frac{\log^n(x)}{n!} = \frac{d}{dx}\left[x\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}\log^k(x)\right]$$

Computing the RHS derivative, we have

\begin{align} \frac{d}{dx}\left[x\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}\log^k(x)\right]&=\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}\log^k(x)+x\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}\frac{d}{dx}\left(\log^k(x)\right)\\ &=\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}\left(\log^k(x)+x\frac{k\log^{k-1}(x)}{x}\right)\\ &=\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}\left(\log^k(x)+k\log^{k-1}(x)\right)\\ &=\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}\log^{k-1}(x)\log(xe^k) \end{align}

However, at this point, I'm stuck. Maybe my $\log$-rules are a little rusty, but I'm not seeing any possible simplification at this point, much less how to show that this expression is actually equal to $\frac{\log^n(x)}{n!}$.

How would one go about proving $(1)$ (assuming that $(1)$ is, in fact, true)? Is there an easier way than evaluating the sum that I am struggling with or is there something that I'm not seeing in this sum that makes an evaluation possible?

Answer 1

Split the sum and re-index,

\begin{align} \frac{d}{dx}\Biggl(x\sum_{k = 0}^{n} (-1)^{n-k}\frac{\log k^ x}{k!}\Biggr) &= \sum_{k = 0}^{n} (-1)^{n-k}\frac{\log^k x}{k!} + x\sum_{k = 1}^{n} (-1)^{n-k} \frac{k\log^{k-1} x}{k!}\frac{1}{x} \\ &= \sum_{k = 0}^{n} (-1)^{n-k}\frac{\log^k x}{k!} + \sum_{k = 1}^{n} (-1)^{n-k} \frac{\log^{k-1} x}{(k-1)!} \\ &= \frac{\log^n x}{n!} + \sum_{m = 0}^{n-1} \bigl((-1)^{n-m} + (-1)^{n-(m+1)}\bigr)\frac{\log^m x}{m!} \\ &= \frac{\log^n x}{n!}. \end{align}