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I've been trying to solve the following integral:

$$I_n=\int_0^1\frac1x\ln^n(x)\ln^{8-n}(1-x)~\mathrm dx$$

for $n\in[2,6]$. It can be computed as a limit of derivatives of the Beta function:

$$I_n=\lim_{x\to0^+}\frac{\partial^8}{\partial x^n\partial y^{8-n}}B(x,y)\bigg|_{y=1}$$

the derivatives of which can be given by WolframAlpha. However, I'm having difficulty completing the last step and taking the limit myself. I've tried tackling this manually by taking the constant term in the Laurent expansion of each part, which isn't very difficult since WolframAlpha factors expressions nicely. Unfortunately though there is a lot of steps and I am prone to error, and I don't know if I can get WolframAlpha to give me the answer.

Is there a way to get WolframAlpha to give me the answer to these problems? If not I would appreciate it if someone could provide me the solution to these integrals, preferably in terms of the Riemann zeta function or the polygamma function.

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  • $\begingroup$ Sorry just to be clear, you are looking for a closed form of $$\lim_{x\to0^+}\frac{\partial^8}{\partial x^n\partial y^{8-n}}B(x,y)\bigg|_{y=1}$$ right? $\endgroup$ – Tom Himler Jul 10 at 2:33
  • $\begingroup$ I know it can be written in terms of $\psi^{(n)}(1)$, and hence in terms of $\zeta(n)$. The issue is moreso that this is messy to compute. (And I'm only asking for those specific values.) $\endgroup$ – Simply Beautiful Art Jul 10 at 4:55
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Here is an approach that reduces the problem to a whole bunch of Euler sums of weight 9. Not sure in practice if it is any more useful than using the derivative of the Beta function since beyond the first few lowest orders (corresponding to your $n = 6$ and $5$) the Euler sums are not very "pretty."

Starting with the generating function for the (unsigned) Stirling numbers of the first kind, namely $$\frac{\ln^k (1 - x)}{k!} = \sum_{n = k}^\infty (-1)^k \begin{bmatrix} n \\ k \end{bmatrix} \frac{x^n}{n!}, \qquad k \in \mathbb{N}$$ reindexing $k \mapsto 8 - k$ where we restrict $k$ to $2,3,4,5$, and $6$, one has $$\ln^k (1 - x) = (8 - k)! \sum_{n = 8-k}^\infty (-1)^k \begin{bmatrix} n \\ 8 - k \end{bmatrix} \frac{x^n}{n!}.$$

Now for your integral, which I will call $I_k$, we have \begin{align} I_k &= \int_0^1 \frac{1}{x} \ln^k \ln^{8 - k} (1 - x) \, dx\\ &= (8 - k)! (-1)^k \sum_{n = 8 - k}^\infty \frac{1}{n!} \begin{bmatrix} n \\ 8 -k \end{bmatrix} \int_0^1 x^{n - 1} \ln^k x \, dx\\ &= (8 - k)! (-1)^k \sum_{n = 8 - k}^\infty \frac{1}{n!} \begin{bmatrix} n \\ 8 -k \end{bmatrix} \frac{d^k}{ds^k} \left [\int_0^1 x^{n + s - 1} \right ]_{s = 0}\\ &= (8 - k)! (-1)^k \sum_{n = 8 - k}^\infty \frac{1}{n!} \begin{bmatrix} n \\ 8 -k \end{bmatrix} \frac{d^k}{ds^k} \left [\frac{1}{n + s} \right ]_{s = 0}\\ &= (8 - k)! (-1)^k \sum_{n = 8 - k}^\infty \frac{1}{n!} \begin{bmatrix} n \\ 8 -k \end{bmatrix} \cdot \frac{(-1)^k k!}{n^{k + 1}}\\ &= (8 - k)! \, k! \sum_{n = 8 - k}^\infty \frac{1}{n^{k + 1} n!} \begin{bmatrix} n \\ 8 -k \end{bmatrix}\tag1 \end{align}

Now the (unsigned) Stirling numbers of the first kind can be related to the generalised harmonic numbers $H^{(p)}_n$. The first few are: \begin{align} \begin{bmatrix} n \\ 2 \end{bmatrix} &= (n - 1)! H_{n - 1}\\[1ex] \begin{bmatrix} n \\ 3 \end{bmatrix} &= \frac{1}{2!}(n - 1)! \left [H^2_{n - 1} - H^{(2)}_{n - 1} \right ]\\[1ex] \begin{bmatrix} n \\ 4 \end{bmatrix} &= \frac{1}{3!} (n - 1)! \left [H^3_{n - 1} - 3H_{n - 1} H^{(2)}_{n - 1} + 2 H^{(3)}_{n - 1} \right ] \end{align}

As an example, if I take the simplest case of $k = 6$, from (1) we see that \begin{align} I_6 &= 1440 \sum_{n = 2}^\infty \begin{bmatrix} n \\ 2 \end{bmatrix} \frac{1}{n^7 n!}\\ &= 1400 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^8}\\ &= 1440 \sum_{n =1}^\infty \frac{H_n}{n^8} - 1440 \sum_{n = 1}^\infty \frac{1}{n^9}\\ &= 1440 \left (4 \zeta (9) - \zeta (7) \zeta (2) - \zeta (5) \zeta (4) - \zeta (3) \zeta (6) \right ), \end{align} which is clearly doable.

As you can see, taking $k = 5$ would be a real struggle, and as for $k = 2,3$ and $4$, well the resulting Euler sums of weight 9 can, I suppose in principle, all be found, but how one actually goes about finding them is beyond me.

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    $\begingroup$ cries in CAS thanks for the alternative approach. $\endgroup$ – Simply Beautiful Art Jul 10 at 12:44

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