Dalzell integral

The equation $$ \int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi $$

proves that $\frac{22}{7}-\pi>0$ because the integrand is positive.

Some Dalzell-type integrals for $\log(2n+1)$

This type of proof may be used for other inequalities relating constants to their rational approximations.

For example, the answer by Olivier Oloa to this question

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

extends to other differences between logarithms of odd integers $\log(2n+1)$ and the corresponding harmonic numbers $H_n$, such as

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


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


As a consequence of the series $$ log(2n+1)=H_n+\sum_{k=1}^{\infty}\left(\sum_{i=-n}^{n}\frac{1}{(2n+1)k+i}-\frac{1}{k}\right) $$ (see https://math.stackexchange.com/a/1602945/134791)

the integral expressions above generalize to $$\log(2n+1)-H_n=\int_{0}^{1} \frac{x^n(1-x)}{\sum_{k=0}^{2n}x^k} \left( \frac{n(n+1)}{2}x^{n-1}+\sum_{k=0}^{n-2}\frac{(k+1)(k+2)}{2}\left(x^k+x^{2(n-1)-k}\right)\right)dx$$

The fact that this integral has non-negative integrand proves that $\log(2n+1) \ge H_n$, following the pattern of the integral proof that $\frac{22}{7}>\pi$ shown above.


Q Can this formula be reduced to $$\log(2n+1)-H_n=\int_{0}^{1} \frac{x^n(1-x)\sum_{k=0}^{2n-2}a(k)x^k}{\sum_{k=0}^{2n}x^k} dx$$ with a simple expression for $a(k)$, such as a ratio of factorials?

  • $\begingroup$ Yes, we can take $a(k)=\binom{\min(k,2n-2-k)}{2}$. $\endgroup$ – Chen Wang May 10 '16 at 12:07

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