Feynman's trick II

Formula involving integrals of logarithm function

We prove the following formula involving the integral of powers of the function $\ln(1-x)$. \[\int_{0}^{1} \frac{\ln^n(1-x)\ln(x)}{x} dx = (-1)^{n+1}n!\left[\left(\frac{n+1}{2}\right) \zeta(n+2) - \frac{1}{2} \sum_{k=1}^{n-1}\zeta(k+1)\zeta(n+1-k)\right]\] Some examples for the first naturals: \begin{align*} &n=1& \int_{0}^{1} \frac{\ln(1-x)\ln(x)}{x} dx &= \zeta(3)\\ &n=2& \int_{0}^{1} \frac{\ln^2(1-x)\ln(x)}{x} dx &= -3\zeta(4)+\zeta(2)\zeta(2) = -\frac{\pi^4}{180} \\ &n=3& \int_{0}^{1} \frac{\ln^3(1-x)\ln(x)}{x} dx & = 12\zeta(5)-6\zeta(2)\zeta(3) = 12\zeta(5)-\pi^2\zeta(3)\\ &n=4& \int_{0}^{1} \frac{\ln^4(1-x)\ln(x)}{x} dx & =-60\zeta(6)+24\zeta(2)\zeta(4)+12\zeta(3)\zeta(3) = 12\zeta^2(3)-\frac{2\pi^6}{105}\\ &n=5& \int_{0}^{1} \frac{\ln^5(1-x)\ln(x)}{x} dx &= 360\zeta(7) -120\zeta(2)\zeta(5)-120\zeta(3)\zeta(4) = -\frac{4\pi^4\zeta(3)}{3} - 20\pi^2\zeta(5) +360\zeta(7) \end{align*} Proof \begin{align*} \int_{0}^{1} \frac{\ln^n(1-x)\ln(x)}{x} dx = & \int_{0}^{1} \frac{\left[\frac{d^n}{dt^n}\Big|_{t=0+} (1-x)^t\right] \left[\frac{d}{ds}\Big|_{s=0+} x^s\right]}{x} dx\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\frac{d}{ds}\Big|_{s=0+} \int_{0}^{1} \frac{ (1-x)^t x^s}{x} dx \\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\frac{d}{ds}\Big|_{s=0+} \int_{0}^{1} (1-x)^t x^{s-1} dx\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\frac{d}{ds}\Big|_{s=0+} B(t+1,s) \\ \end{align*} Now, recall the series expansion of the beta function (proof in Appendix 1): \[B(x,y) = \sum_{j=0}^{\infty} \frac{(1-y)_{j}}{j!(x+j)} \quad y>0\] Therefore \begin{align*} \frac{d^n}{dt^n}\Big|_{t=0+}\frac{d}{ds}\Big|_{s=0+} B(t+1,s) = & \frac{d^n}{dt^n}\Big|_{t=0+}\frac{d}{ds}\Big|_{s=0+} \sum_{j=0}^{\infty} \frac{(1-s)_{j}}{j!(t+1+j)}\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\left[\lim_{s \to 0+}\frac{d}{ds} \sum_{j=0}^{\infty} \frac{(1-s)_{j}}{j!(t+1+j)}\right]\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\left[\lim_{s \to 0+} \sum_{j=0}^{\infty} \frac{(1-s)_{j}(\psi^{(0)}(1-s) -\psi^{(0)}(1-s+j))}{j!(t+1+j)}\right]\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\left[ \sum_{j=0}^{\infty} \frac{(1)_{j}(\psi^{(0)}(1) -\psi^{(0)}(1+j))}{j!(t+1+j)}\right]\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\left[ \sum_{j=0}^{\infty} \frac{\Gamma(j+1)(-\gamma -\psi^{(0)}(1+j))}{j!(t+1+j)}\right]\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\left[ \sum_{j=0}^{\infty} \frac{-\Gamma(j+1)H_{j}}{j!(t+1+j)}\right]\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\left[ \sum_{j=0}^{\infty} \frac{-H_{j}}{(t+1+j)}\right]\\ =& \lim_{t \to 0+} \sum_{j=0}^{\infty}\frac{d^n}{dt^n}\frac{-H_{j}}{(t+1+j)}\\ =& \lim_{t \to 0+} \sum_{j=0}^{\infty}\frac{(-1)^{n+1}n!H_{j}}{(t+1+j)^{n+1}}\\ =& \sum_{j=0}^{\infty}\frac{(-1)^{n+1}n!H_{j}}{(1+j)^{n+1}} \end{align*} Now we know that $\displaystyle H_{j+1} = \frac{1}{j+1} + H_{j} $ therefore \begin{align*} \sum_{j=0}^{\infty}\frac{(-1)^{n+1}n!H_{j}}{(1+j)^{n+1}} = & \sum_{j=0}^{\infty}\frac{(-1)^{n+1}n!H_{j+1}}{(1+j)^{n+1}} - \sum_{j=0}^{\infty}\frac{(-1)^{n+1}n!}{(1+j)^{n+2}} \\ =& (-1)^{n+1}n!\left[\sum_{j=0}^{\infty}\frac{H_{j+1}}{(1+j)^{n+1}} - \sum_{j=0}^{\infty}\frac{1}{(1+j)^{n+2}}\right] \end{align*} If $m=j+1$ then \begin{align*} (-1)^{n+1}n!\left[\sum_{j=0}^{\infty}\frac{H_{j+1}}{(1+j)^{n+1}} - \sum_{j=0}^{\infty}\frac{1}{(1+j)^{n+2}}\right] = & (-1)^{n+1}n!\left[\sum_{m=1}^{\infty}\frac{H_{m}}{m^{n+1}} - \sum_{m=1}^{\infty}\frac{1}{m^{n+2}}\right]\\ =& (-1)^{n+1}n!\left[\sum_{m=1}^{\infty}\frac{H_{m}}{m^{n+1}} - \zeta(n+2) \right] \end{align*} Now, recall this beutiful result proven by Euler (Proof in Appendix 2): \[\sum_{n=1}^{\infty} \frac{H_{n}}{n^r} = \left(1+\frac{r}{2}\right) \zeta(r+1) - \frac{1}{2} \sum_{k=1}^{r-2}\zeta(k+1)\zeta(r-k)\] Then \begin{align*} (-1)^{n+1}n!\left[\sum_{m=1}^{\infty}\frac{H_{m}}{m^{n+1}} - \zeta(n+2) \right] =& (-1)^{n+1}n!\left[\left(1+\frac{n+1}{2}\right) \zeta(n+2) - \frac{1}{2} \sum_{k=1}^{n-1}\zeta(k+1)\zeta(n+1-k) - \zeta(n+2) \right]\\ =& (-1)^{n+1}n!\left[\left(\frac{n+1}{2}\right) \zeta(n+2) - \frac{1}{2} \sum_{k=1}^{n-1}\zeta(k+1)\zeta(n+1-k) \right] \end{align*} Therefore \[\boxed{ \int_{0}^{1} \frac{\ln^n(1-x)\ln(x)}{x} dx = (-1)^{n+1}n!\left[\left(\frac{n+1}{2}\right) \zeta(n+2) - \frac{1}{2} \sum_{k=1}^{n-1}\zeta(k+1)\zeta(n+1-k) \right]}\]

Apendix 1: Series expansion for the beta function

From the definition of the beta function: \[B(x,y) = \int_{0}^{1} t^{x-1}(1-t)^{y-1}dt\] Recall the series expansion of $(1-t)^{y-1}$ \[(1-t)^{y-1} = \sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j}t^{j}\] \begin{align*} \int_{0}^{1} t^{x-1}(1-x)^{y-1}dt =& \int_{0}^{1} t^{x-1}\sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j}t^{j}dt\\ =& \sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j} \int_{0}^{\infty} t^{x+j-1}dt\\ =& \sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j} \frac{1}{(x+j)} \end{align*} Then we have \begin{align*} \sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j}\frac{1}{(x+j)} =& \sum_{j=0}^{\infty}\binom{j-y}{j}\frac{1}{(x+j)}\\ =& \sum_{j=0}^{\infty}\frac{(j-y)!}{j!(-y)!}\frac{1}{(x+j)}\\ =& \sum_{j=0}^{\infty}\frac{\Gamma(j-y+1)}{j!\Gamma(1-y)} \frac{1}{(x+j)}\\ =& \sum_{j=0}^{\infty}\frac{(1-y)_{j}}{j!(x+j)} \end{align*} where the last equality is the definnition of the Pochhammer polynomials. Therefore \[\boxed{B(x,y) = \sum_{j=0}^{\infty}\frac{(1-y)_{j}}{j!(x+j)}}\]

Appendix 2: Proof of Euler's theorem for harmonic series

I found a beautiful proof of this result which relies on the residue theorem. I really liked this proof because it shows that the famous result due to Euler is just a residue of certain kernel function. I present here the original proof in the spirit of Flajolet et al. [1] but I complete the omitted parts by the authors in the original paper.

We need the following Lemma due to Cauchy and Lindelöf.

Lemma. Let $\xi$ be a kernel function and let $r(s)$ be a rational function with $\mathscr{O}(s^{-2})$ at infinity. Then \[ \underbrace{\sum_{\alpha } \left\{\operatorname{Res}\left(r(s)\xi(s), \alpha\right) \Big| \alpha \textrm{ is a pole of } \xi \textrm{ that is not a pole of } r(s) \right\}}_{\textrm{infinite series}} = -\underbrace{\sum_{\beta } \left\{\operatorname{Res}\left(r(s)\xi(s), \beta \right) \Big| \beta \textrm{ is a pole of } r(s) \right\}}_{\textrm{finite series}}\]

We will apply the lemma to the kernel $\displaystyle \xi(s) = (\psi(-s)+\gamma)^2$

Propositon. If $\displaystyle \xi(s) = (\psi(-s)+\gamma)^2$ then
\[\sum_{\alpha } \left\{\operatorname{Res}\left(r(s)\xi(s), \alpha\right) \Big| \alpha \textrm{ is a pole of } \xi \textrm{ that is not a pole of } r(s) \right\} = 2\sum_{n=1}^{\infty}H_{n}r(n)+ \sum_{n=1}^{\infty}r'(n)\]
Proof

The kernel $\xi(s)$ has a countable number of pole at the positive integers.
Let $ \displaystyle n\in \mathbb{N}$, then the function $\displaystyle (\psi(-s)+\gamma)$ has the following expansion at $s \to n$.

\[\psi(-s)+\gamma = \frac{1}{(s-n)}+ H_{n} + \sum_{k=1}^{\infty}\left[(-1)^K H_{n}^{k+1} - \zeta(k+1)\right](s-n)^{k}\] Therefore
\begin{align*} (\psi(-s)+\gamma)^2 =& \left[\frac{1}{(s-n)}+ H_{n} + \sum_{k=1}^{\infty}\underbrace{\left[(-1)^K H_{n}^{k+1} - \zeta(k+1)\right]}_{A_{k}}(s-n)^{k}\right]^2\\ =& \left(\frac{1}{(s-n)}+ H_{n}\right)^2 + 2\left(\frac{1}{(s-n)}+ H_{n}\right)\sum_{k=1}^{\infty}A_{k}(s-n)^{k} + \left(\sum_{k=1}^{\infty}A_{k}(s-n)^{k}\right)^2\\ =& \frac{1}{(s-n)^2}+\frac{2H_{n}}{(s-n)} + H^2_{n} + 2\left(\frac{1}{(s-n)}+ H_{n}\right)\sum_{k=1}^{\infty}A_{k}(s-n)^{k} + \left(\sum_{k=1}^{\infty}A_{k}(s-n)^{k}\right)^2\\ =& \frac{1}{(s-n)^2}+\frac{2H_{n}}{(s-n)} + H^2_{n} + 2A_{1} + \mathscr{O}(s-n) \end{align*} Now, suppose $r(s)$ is a rational function that can be expanded with its power series (in reality, we have assumed something weaker: that $r(s)$ is $\mathcal{O}(s^{-2}$)). Then for $s \to n$: \[r(s) = r(n)+ r'(n)(s-n)+\frac{r''(n)}{2!}(s-n)^2+\frac{r'''(n)}{3!}(s-n)^3+... \] Therefore \begin{align*} r(s)(\psi(-s)+\gamma)^2 =& r(n)(\psi(-s)+\gamma)^2+ r'(n)(s-n)(\psi(-s)+\gamma)^2+\frac{r''(n)}{2!}(s-n)^2(\psi(-s)+\gamma)^2+...\\ =& \frac{r(n)}{(s-n)^2}+\frac{2H_{n}r(n)}{(s-n)} + H^2_{n}r(n) + 2A_{1}r(n) + \mathscr{O}(s-n)r(n) + \frac{r'(n)}{(s-n)}+2H_{n}r'(n) + H^2_{n}r'(n)(s-n) + 2A_{1}r'(n)(s-n) + \mathscr{O}(s-n)r'(n)(s-n) + ... \\ =& H^2_{n}r(n)+ 2A_{1}r(n)+2H_{n}r'(n)+\frac{r''(n)}{2!}+\frac{1}{s-n}\underbrace{\left[2H_{n}r(n)+r'(n)\right]}_{\textrm{residue at } n} +\frac{r(n)}{(s-n)^2} + \mathscr{O}(s-n) \end{align*} \[\therefore \operatorname{Res}\left(r(s)(\psi(-s)+\gamma)^2, n\right) = 2H_{n}r(n)+r'(n)\] Summing each $n \in \mathbb{N}\setminus\left\{0\right\}$ \[\boxed{\sum_{n=1}^{\infty} \operatorname{Res}\left(r(s)(\psi(-s)+\gamma)^2, n\right) = 2\sum_{n=1}^{\infty}H_{n}r(n)+ \sum_{n=1}^{\infty}r'(n)}\] or, equivalently \[\boxed{\sum_{\alpha } \left\{\operatorname{Res}\left(r(s)\xi(s), \alpha\right) \Big| \alpha \textrm{ is a pole of } \xi \textrm{ that is not a pole of } r(s) \right\} = 2\sum_{n=1}^{\infty}H_{n}r(n)+ \sum_{n=1}^{\infty}r'(n)}\]

We are ready to prove Euler's theorem

Theorem (Euler). For integer $\geq 2$, \[\sum_{n=1}^{\infty} \frac{H_{n}}{n^q} = \left(1+\frac{q}{2}\right)\zeta(q+1) -\frac{1}{2}\sum_{k=1}^{q-2} \zeta(k+1)\zeta(q-k)\] Proof If we apply the last result to the function $\displaystyle r(s)= \frac{1}{s^q}$ we have \[\sum_{n=1}^{\infty} \operatorname{Res}\left(\frac{(\psi(-s)+\gamma)^2}{s^q}, n\right) = 2\sum_{n=1}^{\infty}\frac{H_{n}}{n^q} -q\zeta(q+1)\] By the first lemma, \[\displaystyle -\sum_{\beta} \left\{\operatorname{Res}\left(\frac{(\psi(-s)+\gamma)^2}{s^q}, \beta \right) \Big| \beta \textrm{ is a pole of } \frac{1}{s^q} \right\} = 2\sum_{n=1}^{\infty}\frac{H_{n}}{n^q} -q\zeta(q+1)\] Given that $\displaystyle r(s)= \frac{1}{s^q}$ has a single pole at $s=0$ of order $q$ we have: \[ -\operatorname{Res}\left(\frac{(\psi(-s)+\gamma)^2}{s^q}, 0\right) =2\sum_{n=1}^{\infty}\frac{H_{n}}{n^q} -q\zeta(q+1) \]

To find the residue we can expand $\displaystyle \frac{(\psi(-s)+\gamma)^2}{s^q}$:
First, recall the series expansion of $\psi(-s)+\gamma$ at $s=0$:
\[\psi(-s)+\gamma = \frac{1}{s} - \sum_{k=1}^{\infty} \zeta(k+1)s^{k}\] Therefore \begin{align*} (\psi(-s)+\gamma)^2 =& \left[\frac{1}{s} -\sum_{k=1}^{\infty} \zeta(k+1)s^{k}\right]^2 \\ =& \frac{1}{s^2} -2\sum_{k=1}^{\infty} \zeta(k+1)s^{k-1} + \left(\sum_{k=1}^{\infty} \zeta(k+1)s^{k}\right)^2\\ =& \frac{1}{s^2} -2\sum_{k=1}^{\infty} \zeta(k+1)s^{k-1} + \sum_{k=1}^{\infty}\sum_{j=1}^{k} \zeta(j+1)\zeta(k-j+2)s^{k+1}\\ \end{align*} Then \begin{align*} \frac{(\psi(-s)+\gamma)^2}{s^q} = \frac{1}{s^{2+q}} -2\underbrace{\sum_{k=1}^{\infty} \zeta(k+1)s^{k-1-q}}_{A} + \underbrace{\sum_{k=1}^{\infty}\sum_{j=1}^{k} \zeta(j+1)\zeta(k-j+2)s^{k+1-q}}_{B} \end{align*} From here is easy to show that the residue in A is the coefficient of $\displaystyle s^{k-1-q}=s^{-1}$ \[ \Longrightarrow k-1-q=-1 \Longrightarrow k=q\] and the residue in B is coefficient of $\displaystyle s^{k+1-q}=s^{-1}$ where \[\Longrightarrow k+1-q=-1 \Longrightarrow k=q-2\] Therefore \[\operatorname{Res}\left(\frac{(\psi(-s)+\gamma)^2}{s^q},0\right) = -2\zeta(q+1) + \sum_{j=1}^{q-2} \zeta(j+1)\zeta(q-j) \] Then \[ 2\zeta(q+1) - \sum_{j=1}^{q-2} \zeta(j+1)\zeta(q-j) = 2\sum_{n=1}^{\infty}\frac{H_{n}}{n^q} -q\zeta(q+1)\] \[\boxed{\therefore\sum_{n=1}^{\infty} \frac{H_{n}}{n^q} = \left(1+\frac{q}{2}\right)\zeta(q+1) -\frac{1}{2}\sum_{k=1}^{q-2} \zeta(k+1)\zeta(q-k)}\]

[1] Philippe Flajolet, Bruno Salvy. Euler Sums and Contour Integral Representations. [Research Report] RR-2917, INRIA. 1996. ffinria-00073780f