Does the repeated integral of $\ln x$ have a pattern?
Andrew Henderson 
I was reminiscing with a friend and we remembered a question asked to us in our high school calculus class.
Find an expression for the $n^{\text{th}}$ derivative and the $n^{\text{th}}$ integral of $\ln x$ (ignoring integration constants$^1$).
The first part is easy:
$$f(x) = \ln x$$ $$f'(x) = \frac{1}{x}$$ $$f''(x) = -\frac{1}{x^2}$$ $$f'''(x) = \frac{2}{x^3}$$ $$\vdots$$ $$f^{(n)}(x) = \frac{(-1)^{n-1}\cdot(n-1)!}{x^n}$$
The second part seems to have no pattern:
$$f(x) = \ln x$$ $$\int f(x)\ dx = x\ln x - x$$ $$\iint f(x) \ dx^2 = \frac12x^2\ln x - \frac34x^2$$ $$\iiint f(x) \ dx^3 = \frac16x^3\ln x - \frac{11}{36}x^3$$ $$\vdots$$
It looks like the answer is something like:
$$f^{(-n)}(x) = \frac{1}{n!}x^n\ln x - \ ?x^n$$
From the looks of it, I think the question boils down to finding a general term for:
$$\int x^n \ln x \ dx$$
And then also trying to figure out what pattern the coefficients $?$ take.
$^1$As user Semiclassical points out, this is just equivalent to computing successive antiderivatives of:
$$\int_0^x f(t) \ dt$$
$\endgroup$ 52 Answers
$\begingroup$I guess @GrumpyParsnip is right, I should refrain from answering questions in the comments.
Here is what i wrote in the comment.
Yes, it has a pattern. I brightly remember reading that on some Wikipedia page, but for the life of me, I can't find the source (hence i write this as a comment). So here's the answer: Define $$f_n(x)=\frac{x^n}{n!}(\ln x- H_n).$$ Then by differentiation we find that $$\frac{d}{dx} f_n(x)=f_{n-1}(x)$$, and since $f_0(x)=\ln x$, this answers your question.
$\endgroup$ 1 $\begingroup$As you have established that $$\log^{(-n)}(x)=\frac{x^n}{n!}(\log(x)-c_n),$$ you can find a recurrence for $c_n$:
$$\left(\log^{(-n)}(x)\right)'=\frac{x^{n-1}}{(n-1)!}\log(x)+\frac{x^{n-1}}{n!}(1-nc_n)=\frac{x^{n-1}}{(n-1)!}(\log(x)-c_{n-1}).$$
Hence by identification,
$$c_n=c_{n-1}+\frac1n.$$
$\endgroup$ 1
