0

Is there a closed form expression for $$\int_{0}^{1} \frac{(x^2 - 3x)^n \log(x)}{x}\,dx$$

The motivation behind the question comes from this post. The OP wanted to evaluate the integral without using $Li_3$. He gave a closed form in terms of zeta functions and logarithms, which makes me hopeful that this has a closed form.

Applying the binomial theorem, and pulling out from the integral,

$$ \begin{align} \int_{0}^{1} \frac{(x^2 - 3x)^n \log(x)}{x} dx &= \sum_{k=0}^{n} \binom{n}{k} (-3)^{n-k} \int_{0}^{1} x^{2k+(n-k)} \frac{\log(x)}{x} dx = \sum_{k=0}^{n} \binom{n}{k} (-3)^{n-k} \int_{0}^{1} x^{k+n-1} {\log(x)} dx \end{align} $$

The inner integral can be evaluated.

$$ \begin{align} \int_{0}^{1} \frac{(x^2 - 3x)^n \log(x)}{x} \,dx &= \sum_{k=0}^{n} \binom{n}{k} (-3)^{n-k} \int_{0}^{1} x^{2k+(n-k)} \frac{\log(x)}{x} dx = -\sum_{k=0}^{n} \binom{n}{k} (-3)^{n-k} \frac{1}{(n+k)^2} \end{align} $$

Is there a better form for this sum?

Thanks.

Maxime Jaccon
  • 2,770
  • @teadawg1337 The integral for $n=1$ should be $\int_{0}^{1} \frac{ (x^2 - 3x) \log(x)}{x} dx $. See: https://www.wolframalpha.com/input?i2d=true&i=Integrate%5Blog%5C%2840%29x%5C%2841%29Divide%5B%5C%2840%29Power%5Bx%2C2%5D-3x%5C%2841%29%2Cx%5D%2C%7Bx%2C0%2C1%7D%5D – Maxime Jaccon Jan 01 '25 at 04:26
  • But $\frac{x^2-3x}{x}=x-3$... Even if the $x$ in the denominator mattered here, we have an improper integral anyway due to the natural logarithm. $\lim_{x\to0^+}\frac{x^2-3x}{x}=\lim_{x\to0^+}(x-3)=-3$. – teadawg1337 Jan 01 '25 at 04:31
  • @teadawg1337 The claim that $\int_0^1 (x-3)\log(x)~dx=26/9$ appears spurious according to WolframAlpha. – Semiclassical Jan 01 '25 at 04:33
  • 1
    @teadawg1337 $\log x$ is integrable on $[0,1]$... – Semiclassical Jan 01 '25 at 04:49
  • 1
    @Semiclassical True, I meant to communicate that the natural log term makes it an improper integral since $(x-3)\log{(x)}$ is undefined at $x=0$. – teadawg1337 Jan 01 '25 at 04:58
  • 3
    Regarding the sum, I can only manage to squeeze a generalized hypergeometric function out of it. I can try to put together a proof if you'd like, but the form you have now is much better computationally. – teadawg1337 Jan 01 '25 at 05:07
  • 2
    by @teadawg1337 comment : $$-\frac{3^n e^{i n \pi } , _3F_2\left(-n,n,n;1+n,1+n;\frac{1}{3}\right)}{n^2}$$ – Mariusz Iwaniuk Jan 01 '25 at 11:10

1 Answers1

0

I do not want this question to be browsing the list of 'unanswered' questions: the comments made under this post provide a satisfactory response to the original question. The form I have derived is probably the best computationally (alternatives are in terms of hypergeometric functions).

$$ \int_{0}^{1} \frac{(x^2 - 3x)^n \log(x)}{x} \,dx = -\sum_{k=0}^{n} \binom{n}{k} (-3)^{n-k} \frac{1}{(n+k)^2} $$

Maxime Jaccon
  • 2,770