3

prove that $$\int_{0}^{1}\dfrac{x^m\ln{x}}{x-1}dx=\sum_{r=m+1}^{\infty}\dfrac{1}{r^2}$$

This true?

my idea

let $$I(m)=\int_{0}^{1}\dfrac{x^m}{x-1}dx\Longrightarrow I'(m)=\int_{0}^{1}\dfrac{x^m\ln{x}}{x-1}dx$$

and we must find $$I(m)=\int_{0}^{1}\dfrac{x^m}{x-1}dx=-\int_{0}^{1}\dfrac{(1-t)^m}{t}dt$$ then we can't prove it. Thank you everyone can help, and This problem is from:How prove $\frac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{n}\frac{1}{t^2}<e$

Community
  • 1
math110
  • 93,304

3 Answers3

4

Here is how,

$$ \int_{0}^{1}\dfrac{x^m\ln{x}}{x-1}dx=-\sum_{k=0}^{\infty}\int_{0}^{1}x^{m+k}\ln(x)dx$$

$$= \sum_{k=0}^{\infty}\frac{1}{ (k+m)^2+2(k+m)+1 }$$

$$ = \sum_{k=m}^{\infty}\frac{1}{ k^2+2k+1 }$$

$$ = \sum_{k=m}^{\infty}\frac{1}{ (k+1)^2 }= \sum_{r=m+1}^{\infty}\frac{1}{ r^2 }=\psi'(m+1),$$

where $\psi(x)$ is the digamma function.

Note:

$$ \sum_{m+1}^{\infty}\frac{1}{r^2}=\psi'(m+1). $$

Mhenni Benghorbal
  • 47,431
  • 1
    oh,Thank you, it's nice+1 – math110 Jul 12 '13 at 10:10
  • @math110: You are welcome. – Mhenni Benghorbal Jul 12 '13 at 10:16
  • 2
    IMHO, it will be useful to include some justification why one is legal to exchange the order of integration and summation in the first equality. – achille hui Jul 12 '13 at 10:24
  • 2
    You can exchange the order of integration if you consider the series expansion for $\log(x)/(1-x) = \sum \log(x) x^n$ converges uniformly for $|x|<1-\epsilon$ and that the leftover part of the integral must be small since $\log(x)~(1-x)$ for $x$ near $1$. – abnry Aug 09 '14 at 21:54
2

Hint: $\frac{1}{x-1} = -(1 + x + x^2 + \ldots)$.

Tunococ
  • 10,303
1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}{x^{m}\ln\pars{x} \over x - 1}\,\dd x =\sum_{r = m + 1}^{\infty}{1 \over r^{2}}}$

\begin{align}&\color{#66f}{\large\int_{0}^{1}{x^{m}\ln\pars{x} \over x - 1}\,\dd x} =\lim_{\mu \to m}\partiald{}{\mu}\int_{0}^{1}{1 - x^{\mu} \over 1 - x}\,\dd x =\lim_{\mu \to m}\partiald{\Psi\pars{\mu + 1}}{\mu}=\Psi\,'\pars{m + 1} \\[3mm]&=\partiald{}{z}\sum_{r = 1}^{\infty}\pars{{1 \over r} - {1 \over r + z}} _{z\ =\ m } =\sum_{r = 1}^{\infty}{1 \over \pars{r + m}^{2}} =\color{#66f}{\large\sum_{r\ =\ m + 1}^{\infty}{1 \over r^{2}}} \end{align}

See ${\bf 6.3.22}$ and ${\bf 6.3.16}$.

Felix Marin
  • 89,464
  • This poses more problems than it brings solutions: why are the interchanges of limits and integrals, of integrals and derivatives, of derivatives and series, and so forth, valid? (Note also that the détour by the digamma function is quite unnecessary.) – Did Aug 09 '14 at 23:41