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The following series and integrals relate log(2) to its third and fourth convergents, $\frac{2}{3}$ and $\frac{7}{10}$.

$$\begin{align} \log\left(2\right)-\frac{2}{3} &= \sum_{k=1}^\infty \frac{1}{(2k+1)(2k+2)(2k+3)}\\ &= \int_0^1\frac{x^2(1-x)^2}{1+x^2}dx\\ &= \frac{1}{2}\int_0^1 \frac{x^2(1-x)}{1+x}dx \\ \\ \frac{7}{10}-\log\left(2\right) &= \sum_{k=2}^\infty \frac{1}{(2k+2)(2k+3)(2k+4)} \\ &= \int_0^1\frac{x^4(1-x)^2}{1+x^2}dx\\ &= \frac{1}{2}\int_0^1 \frac{x^5(1-x)}{1+x} dx \\ &= \frac{24}{5}\sum_{k=0}^{\infty} \frac{1}{(4k+2)(4k+3)(4k+4)(4k+5)(4k+6)} \\ &= \frac{1}{5}\int_{0}^{1} \frac{x(1-x)^4}{1-x^4}dx = \frac{1}{5}\int_{0}^{1} \frac{x(1-x)^3}{(1+x)(1+x^2)}dx \\ \end{align} $$

Similarly, for $\log(3)$ and $\log(5)$, we have

$$\begin{align} \log(3)-1 &= \sum_{k=0}^\infty \frac{2}{(3 k+2) (3 k+3) (3 k+4)} \\ &= \int_{0}^{1} \frac{x(1-x)^2}{1-x^3}=\int_{0}^{1} \frac{x (1-x)}{1+x+x^2} dx \\ \\ \frac{11}{10}-\log(3) &= \sum_{k=1}^{\infty} \frac{8}{(3k+1)(3k+2)(3k+3)(3k+4)(3k+5)} \\ &= \frac{1}{3}\int_{0}^{1} \frac{x^3(1-x)^4}{1-x^3}=\int_{0}^{1} \frac{x^3 (1-x)^3}{1+x+x^2} dx \\ \log(5)-\frac{3}{2} &= \sum_{k=0}^\infty \left(\frac{8}{(5k+3)(5k+5)(5k+7)}+\frac{2}{(5k+4)(5k+5)(5k+6)} \right) \\ \end{align}$$

Q Is there a similar integral for $\log(5)-\frac{3}{2}$?

Generalization

Extending the method used by Olivier Oloa on the series for $\log(2n+1)-H_n$, one gets the following Dalzell-type integrals

$$\log(7)-\frac{11}{6}=\int_0^1 \frac{x^3 (1-x) (1+3 x+6 x^2+3 x^3+ x^4)}{1+x+x^2+x^3+x^4+x^5+x^6} dx$$ check

$$\log(9)-\frac{25}{12}=\int_0^1 \frac{x^4 (1-x) (1+3 x+6 x^2+10 x^3+6 x^4+3 x^5+x^6)}{1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8} dx$$

check

The terms in the larger polynomial in the numerator are $\frac{(n+1)(n+2)}{2}x^n$ in the first half and then mirror.

In general, $$\log(2n+1)-H_n=\int_{0}^{1} \frac{x^n(1-x)}{\sum_{k=0}^{2n}x^k} \left( \frac{n(n+1)}{2}x^{n-1}+\sum_{k=0}^{n-2}\frac{(k+1)(k+2)}{2}\left(x^k+x^{2(n-1)-k}\right)\right)dx$$

This integral proves $$\log(2n+1)>H_n$$ which agrees with $$\lim_{n->\infty} (log(2n+1)-H_n)=\log(2)-\gamma$$

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By using a partial fraction decomposition of the summand, you get $$ \begin{align} \log(5)-\frac{3}{2}&=\sum_{k=0}^\infty \left(\frac{8}{(5k+3)(5k+5)(5k+7)}+\frac{2}{(5k+4)(5k+5)(5k+6)} \right)\\\\ &=\sum_{k=0}^\infty \int_0^1 \left(x^{5k+6}+x^{5k+5}+x^{5k+3}+x^{5k+2}-4x^{5k+4}\right)dx\\\\&=\int_0^1 \sum_{k=0}^\infty \left(x^{5k+6}+x^{5k+5}+x^{5k+3}+x^{5k+2}-4x^{5k+4}\right)dx \end{align} $$ that is

$$ \log(5)-\frac{3}{2}=\int_0^1 \frac{x^2(1-x)(1+3x+x^2)}{1+x+x^2+x^3+x^4}\:dx. $$

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    $\begingroup$ Olivier: are you sure of your result? For me, your integral isn't convergent wolframalpha.com/input/?i=integrate+from+x%3D0+to+1+x^2(5%2B5x-4x^2%2B5x^3%2B5x^4)%2F(5*(1-x^5))+dx $\endgroup$ – FDP Feb 16 '16 at 9:43
  • $\begingroup$ For me, it's $\displaystyle \int_0^1 \frac{x^2(1+x-4x^2+x^3+x^4)}{1-x^5}dx$ (found with PARI GP LLL Stuff, confirmed by Wolfram Alpha) $\endgroup$ – FDP Feb 16 '16 at 11:12
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    $\begingroup$ or $\displaystyle \int_0^1 \dfrac{x^2(x-1)^2(x^2+3x+1)}{1-x^5}dx$ showing the positivity of $\log(5)-\dfrac{3}{2}$ $\endgroup$ – FDP Feb 16 '16 at 11:53
  • $\begingroup$ @FDP Good catch! There was a wrong factor 1/5, thank you! $\endgroup$ – Olivier Oloa Feb 16 '16 at 20:38
  • $\begingroup$ @FDP Nice! Now the denominators for the integral of $log(n+1)$ are $\sum_{k=0}^{n} x^k$ and coefficients $1,1,1,1,-4$ agree with those in math.stackexchange.com/questions/883348/series-for-logarithms. But this is misleading because the exponents are not in order, $-4$ is the central one in this regrouping. In the numerator, beautiful symmetry in polynomial $(1+3x+x^2)$ that multiplies the same numerator as in the second integral for $\log(2)$, namely $x^2(1-x)$. $\endgroup$ – Jaume Oliver Lafont Feb 19 '16 at 5:49

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