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Positive integrals $$\int_{0}^{1}\frac{2x(1-x)^2}{1+x^2}dx=\pi-3$$ and $$\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi $$ (https://math.stackexchange.com/a/1618454/134791)

prove that $$3<\pi<\frac{22}{7}$$

Is there a similar argument for the following $\log (2)$ inequality? $$\frac{2}{3}<\log(2)<\frac{7}{10}$$

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There are positive integrals that relate $\log(2)$ to its first four convergents: $0,1,\frac{2}{3},\frac{7}{10}$. $$ \begin{align} \int_0^1\frac{2x}{1+x^2}dx &= \log\left(2\right) \\ \int_0^1\frac{(1-x)^2}{1+x^2}dx &= 1-\log\left(2\right) \\ \int_0^1\frac{x^2(1-x)^2}{1+x^2}dx &= \log\left(2\right)-\frac{2}{3} \\ \int_0^1\frac{x^4(1-x)^2}{1+x^2}dx &=\frac{7}{10}-\log\left(2\right) \\ \end{align} $$

Therefore, $$-\int_0^1\frac{x^2(1-x)^2}{1+x^2}dx<0<\int_0^1\frac{x^4(1-x)^2}{1+x^2}dx$$

$$\frac{2}{3}-\log(2)<0<\frac{7}{10}-\log\left(2\right)$$

$$\frac{2}{3}<\log(2)<\frac{7}{10}$$

A similar set is available with denominators $(1+x)$:

$$\begin{align} \int_0^1 \frac{1}{1+x}dx &= \log(2) \\ \int_0^1 \frac{x}{1+x}dx &= 1-\log(2)\\ \frac{1}{2}\int_0^1 \frac{x^2(1-x)}{1+x} dx &= \log(2)-\frac{2}{3} \\ \frac{1}{2}\int_0^1 \frac{x^5(1-x)}{1+x} dx &= \frac{7}{10}-\log(2) \end{align}$$

and series versions are given by

$$\begin{align} \log(2)-\frac{2}{3} &= \sum_{k=1}^\infty \frac{1}{(2k+1)(2k+2)(2k+3)} \\ \frac{7}{10}-\log(2) &= \sum_{k=2}^\infty \frac{1}{(2k+2)(2k+3)(2k+4)} \\ \end{align} $$

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    $\begingroup$ You answered your own question ? $\endgroup$
    – Saikat
    Feb 6, 2016 at 6:40
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    $\begingroup$ You posted an answer at nearly the exact same time as you posted the question? $\endgroup$
    – fosho
    Feb 6, 2016 at 7:40
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    $\begingroup$ @qwr That's the "somewhere" I didn't manage to find when answering to user230452... thank you! $\endgroup$ Feb 6, 2016 at 7:47
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    $\begingroup$ @user230452 The exponents in wolframalpha.com/input/… may be changed to find rational approximations to $log(2)$ and $\pi$. $\endgroup$ Feb 6, 2016 at 9:50
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    $\begingroup$ That link to blog.stackoverflow does not grant a blanket permission to ask questions with precomposed answers. At least that's how it is applied in Math.SE. This was discussed - heatedly I may add - when one user was systematically doing this and used that same blog post as an excuse. So my advice is to tread carefully. Also read this meta thread and the previous discussions linked to there. If anyone wants to discuss this further, don't do it here, but take it to relevant meta threads instead. $\endgroup$ Feb 6, 2016 at 17:46

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