INTEGRAL REPRESENTATION STILTJIES CONSTANT

In today´s post we want to prove this beautiful integral representation of the Stieltjes constants

$\gamma_n=\int_1^\infty \frac{\left\{ x\right\}}{x^2} \left(n-\ln(x) \right)\ln^{n-1}(x)\,dx$

The Stieltjes constants are given by the limit

$\gamma_{n}=\lim _{N \rightarrow \infty}\left\{\sum_{k=1}^{N} \frac{(\ln k)^{n}}{k}-\int_{1}^{N} \frac{(\ln x)^{n}}{x} d x\right\}=\lim _{N \rightarrow \infty}\left\{\sum_{k=1}^{N} \frac{(\ln k)^{n}}{k}-\frac{(\ln N)^{n+1}}{n+1}\right\}$

We start by proving the following result:

$\int\frac{\ln^n(x)}{x^2}\,dx=-\sum_{r=0}^n\frac{n!}{r!}\frac{\ln^{r}(x)}{x}$ (1)

Proof:

$\begin{aligned} I_n&=\int\frac{\ln^n(x)}{x^2}\,dx\\ &=-\frac{\ln^n(x)}{x}+n\int\frac{\ln^{n-1}(x)}{x^2}\,dx\\ &=-\frac{\ln^n(x)}{x}+nI_{n-1}\\ &=-\frac{\ln^n(x)}{x}-n\frac{\ln^{n-1}(x)}{x}+n(n-1)\int\frac{\ln^{n-2}(x)}{x^2}\,dx\\ &=-\frac{\ln^n(x)}{x}-n\frac{\ln^{n-1}(x)}{x}-n(n-1)(n-2)\frac{\ln^{n-3}(x)}{x}+n(n-1)(n-2)\int\frac{\ln^{n-2}(x)}{x^2}\,dx\\ &=-\sum_{r=0}^n\frac{n!}{r!}\frac{\ln^{r}(x)}{x} \end{aligned}$

With the aid of (1) we now prove the following result

$\begin{aligned} \int_1^\infty\frac{\left\{x\right\}}{x^2}\ln^n(x)\,dx=n!-\gamma_n-\sum_{r=0}^{n-1}\frac{n!}{r!}\gamma_r \end{aligned}$ (2)

Proof:

$\begin{aligned} &\int_1^\infty \frac{\left\{x\right\}}{x^2}\ln^n(x)\,dx\\&=\int_1^\infty \frac{\ln^n(x)}{x}\,dx-\lim_{N \to \infty}\sum_{k=1}^{N-1}k \int_k^{k+1}\frac{\ln^n(x)}{x^2}\,dx\\ &=\int_1^\infty \frac{\ln^n(x)}{x}\,dx-\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( -\sum_{r=0}^n\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=\int_1^\infty \frac{\ln^n(x)}{x}\,dx+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^n\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=\int_1^\infty \frac{\ln^n(x)}{x}\,dx+\lim_{N \to \infty}\left(\sum_{k=1}^{N-1}k \frac{\ln^n(k+1)}{k+1}-\sum_{k=1}^{N-1}k \frac{\ln^n(k)}{k} \right)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-1}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=\int_1^\infty \frac{\ln^n(x)}{x}\,dx+\lim_{N \to \infty}\left(\sum_{k=2}^{N}(k-1) \frac{\ln^n(k)}{k}-\sum_{k=1}^{N-1}k \frac{\ln^n(k)}{k} \right)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-1}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=\int_1^\infty \frac{\ln^n(x)}{x}\,dx+\lim_{N \to \infty}\left(\ln^n(N)-\sum_{k=1}^{N} \frac{\ln^n(k)}{k} \right)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-1}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=-\lim_{N \to \infty}\left(\sum_{k=1}^{N} \frac{\ln^n(k)}{k}-\int_1^N \frac{\ln^n(x)}{x}\,dx\right)+\lim_{N \to \infty}\ln^n(N)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-1}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=-\gamma_n+\lim_{N \to \infty}\ln^n(N)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-1}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=-\gamma_n+\lim_{N \to \infty}\ln^n(N)+n\lim_{N \to \infty}\sum_{k=1}^{N-1}k\left(\frac{\ln^{n-1}(k+1)}{k+1}-\frac{\ln^{n-1}(k)}{k} \right)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-2}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=-\gamma_n+\lim_{N \to \infty}\ln^n(N)+n\lim_{N \to \infty}\left(\sum_{k=2}^{N}\frac{(k-1)\ln^{n-1}(k)}{k}-\sum_{k=1}^{N-1}k\frac{\ln^{n-1}(k)}{k} \right)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-2}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=-\gamma_n+\lim_{N \to \infty}\ln^n(N)+n\lim_{N \to \infty}\left(\ln^{n-1}(N)-\sum_{k=1}^{N}\frac{\ln^{n-1}(k)}{k} \right)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-2}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=-\gamma_n+n\lim_{N \to \infty}\ln^{n-1}(N)-n\lim_{N \to \infty}\left(\sum_{k=1}^{N}\frac{\ln^{n-1}(k)}{k}-\frac{\ln^{n}(N)}{n} \right)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-2}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=-\gamma_n-n\gamma_{n-1}+n\lim_{N \to \infty}\ln^{n-1}(N)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-2}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ \end{aligned}$

$\begin{aligned} &=-\gamma_n-n\gamma_{n-1}+n\lim_{N \to \infty}\ln^{n-1}(N)+n(n-1)\lim_{N \to \infty}\left(\ln^{n-2}(N)-\sum_{k=1}^N\frac{\ln^{n-2}(k)}{k}\right)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-3}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=-\gamma_n-n\gamma_{n-1}-n(n-1)\lim_{N \to \infty}\left(\sum_{k=1}^N\frac{\ln^{n-2}(k)}{k}-\frac{\ln^{n-1}(N)}{n-1}\right)+n(n-1)\lim_{N \to \infty}\ln^{n-2}(N)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-3}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &=-\gamma_n-n\gamma_{n-1}-n(n-1)\gamma_{n-2}+n(n-1)\lim_{N \to \infty}\ln^{n-2}(N)+\lim_{N \to \infty}\sum_{k=1}^{N-1}k \left( \sum_{r=0}^{n-3}\frac{n!}{r!}\frac{\ln^{r}(x)}{x}\right)\Bigg|_k^{k+1}\\ &= \cdots \\ &=-\gamma_n-\sum_{r=2}^{n-1}\frac{n!}{r!}\gamma_r-\frac{n!}{2!}\lim_{N \to \infty} \ln^2(N)+n!\lim_{N \to \infty}\left(\ln(N)-\sum_{k=2}^{N}\frac{\ln(k)}{k} \right)+n!\lim_{N \to \infty}\left(\sum_{k=2}^{N}\frac{(k-1)}{k}-\sum_{k=1}^{N-1}1\right)\\ &=-\gamma_n-\sum_{r=2}^{n-1}\frac{n!}{r!}\gamma_r-n!\lim_{N \to \infty}\left(\sum_{k=1}^{N}\frac{\ln(k)}{k} -\frac{\ln^2(N)}{2}\right)-n!\lim_{N \to \infty}\left(\sum_{k=2}^{N}\frac{(k-1)}{k}-\ln(N) \right)\\ &=-\gamma_n-\sum_{r=1}^{n-1}\frac{n!}{r!}\gamma_r-n!\lim_{N \to \infty}\left(-1+\sum_{k=1}^{N}\frac{(k-1)}{k}-\ln(N) \right)\\ &=n!-\gamma_n-\sum_{r=1}^{n-1}\frac{n!}{r!}\gamma_r-n!\gamma_0\\ &=n!-\gamma_n-\sum_{r=0}^{n-1}\frac{n!}{r!}\gamma_r \qquad \blacksquare \end{aligned}$

With the result (2), we can now prove our goal integral:

$\begin{aligned} \int_1^\infty \frac{\left\{ x\right\}}{x^2} \left(n-\ln(x) \right)\ln^{n-1}(x)\,dx&=n\int_1^\infty \frac{\left\{ x\right\}}{x^2} \ln^{n-1}(x)\,dx-\int_1^\infty \frac{\left\{ x\right\}}{x^2} \ln^{n}(x)\,dx\\ &=n\left((n-1)!-\gamma_{n-1}-\sum_{r=0}^{n-2}\frac{(n-1)!}{r!}\gamma_r \right)-\left( n!-\gamma_n-\sum_{r=0}^{n-1}\frac{n!}{r!}\gamma_r\right)\\ &=n(n-1)!-n\gamma_{n-1}-n\sum_{r=0}^{n-2}\frac{(n-1)!}{r!}\gamma_r -n!+\gamma_n+\frac{n!}{(n-1)!}\gamma_{n-1}+\sum_{r=0}^{n-2}\frac{n!}{r!}\gamma_r\\ &=\gamma_n+n!-n!+n\gamma_{n-1}-n\gamma_{n-1}+\sum_{r=0}^{n-2}\frac{n!}{r!}\gamma_r-\sum_{r=0}^{n-2}\frac{n!}{r!}\gamma_r\\ &=\gamma_n \qquad \blacksquare \end{aligned}$

Search This Blog

MY MATEX NOTES