Does $\int_0^1 \frac{\ln^k(1-x)}{x}\, dx = (-1)^k\Gamma(k+1)\zeta(k+1)$ for positive integers $k$?

Question

Some time ago I came across the following integral on the internet.

\begin{equation} \int_0^1 \frac{\ln^2(1-x)}{x}\, dx =2 \zeta(3) \end{equation}

I think I can generalise this integral to the following...

\begin{equation} \boxed{\int_0^1 \frac{\ln^k(1-x)}{x}\, dx = (-1)^k\Gamma(k+1)\zeta(k+1)\,, \quad k \in \mathbb{Z}^+} \end{equation}

I've attempted a couple of integrals like these and they often involve interchanging an infinite sum and integral.

I'm from a physics background and I don't know how to do this rigourously using arguments from analysis. My question is about how to justify this step in the calculation below and whether the value of the integral is correct.

Here's my attempt. Let the left hand side of the above equation be equal to $I$.

Firstly we can perform the simple shift $(x \to 1-x)$ and get

\begin{equation} I = \int_0^1 \frac{\ln^k(x)}{1-x}\, dx \end{equation}

From here use the geometric series $\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$ noting that $|x|<1$. Interchanging the sum and the integral (justification needed) we have...

\begin{equation} I = \sum_{n=0}^\infty \int_0^1 x^n \ln^{k}(x)\, dx \end{equation}

Now use the change of variable $u=-\ln(x)$ and after some tidying up we have...

\begin{equation} I = (-1)^k \sum_{n=0}^\infty \int_0^\infty u^k e^{-(n+1)u}\, du \end{equation} Finally letting $t=(n+1)u$ ... \begin{align} I &= (-1)^k \sum_{n=0}^\infty \int_0^{\infty} e^{-t} \big(\frac{t}{n+1}\big)^k\,\frac{dt}{n+1} \\ &= (-1)^k \sum_{n=0}^\infty \big(\frac{1}{n+1}\big)^{k+1} \int_0^{\infty} t^{k} e^{-t}\,dt \\ \implies I&= (-1)^k \Gamma(k+1)\zeta(k+1) \end{align} As above. Here I've used the standard definitions of the Riemann zeta function $\zeta$ and the gamma function $\Gamma$.

Is this correct? How can I justify the swapping of the sum and the integral?

Thanks

Answer 1

This looks good and is probably the easiest way to do it.

To justify the sum-integral exchange, there are a few standard results. Had you: $$\int_0^a\frac{\ln^kx}{1-x}\,\mathrm{d}x$$With $a<1$ I think you could use uniform convergence of the geometric series to conclude. But with $a=1$, there is a potential problem.

By far the most common way and most (usually) easy way to justify things is the dominated convergence theorem.

Here, I notice that: $$\left|\sum_{j=0}^nx^j\ln^k x\right|<\frac{\ln^k(1/x)}{1-x}$$For all $n,k\in\Bbb N$ and all $0<x<1$. The right hand side is integrable on $(0,1)$ thus the dominated convergence theorem applies to let you exchange the series with the integral. In fact, since the power series for $(1-x)^{-1}$ has all positive coefficients and $\ln^k$ had constant sign on $(0,1)$, the monotone convergence theorem applies (which is a slightly more elementary result and can be used in different situations) to justify the exchange.

Answer 2

See article by Kölbig in Math. Comp. 39 (160) (1981) 647–654 entitled "Closed Expressions for $\int_0^1 t^{-1}\log^{n-1}t \log^p (1-t) dt$":

Define $$ s_{n,p}\equiv \frac{(-)^{n+p-1}}{(n-1)!p!}\int_0^1 t^{-1}\log^{n-1}t \log^p(1-t)dt $$ then $$ s_{n,p}=s_{p,n}=\sum_{k=1}^p \frac{(-)^{k+1}}{k!} \sum_{m_i}\frac{H_p(m_1,\ldots m_k)}{m_1\cdots m_k}\zeta(m_1)\cdots \zeta(m_k), $$ where $$ H_p(m_1,\ldots,m_k)=\sum_{p_i}\binom{m_1}{p_1} \cdots\binom{m_k}{p_k}, $$ the sum over $m_i$ over all sets of integers which satisfy $m_i\ge 2$, $\sum_{i=1}^k=n+p$, and the sum over $p_i$ over all sets of integers which satisfy $1\le p_i\le m_i-1$, $\sum_{i=1}^kp_i=p$. Examples with $s_{n,p}=\sum_{k=1}^p (-)^{k+1}\alpha_k(n,p)/k!$ are $\alpha_1(n,p)=(n+p-1)!\zeta(n+p)/(n!p!)$ or $\alpha_2(n,2)=\sum_{\nu=2}^n \zeta(\nu) \zeta(n-\nu+2)$. The reference provides an explicit table for $n,p\le 4$.