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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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Jaume Oliver Lafont
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  • I was just searching my books for a proof that the value of pi we learn in school is slightly greater than pi. So, I'm glad I found this question ! – Saikat Feb 06 '16 at 06:38
  • @user230452 You may find that proof here: https://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80 – Jaume Oliver Lafont Feb 06 '16 at 06:56

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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

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

Jaume Oliver Lafont
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    You answered your own question ? – Saikat Feb 06 '16 at 06:40
  • Do you mind explaining the first four steps ? – Saikat Feb 06 '16 at 06:40
  • @user230452 Yes, I answered it, I read somewhere that this is OK. For the first one, you may change variable $t=(1+x^2)$ so $2xdx=dt$ and then use $\int\frac{dt}{t}=log(t)$. For the others, the process is parallel to https://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80#Details_of_evaluation_of_the_integral – Jaume Oliver Lafont Feb 06 '16 at 06:49
  • @user230452 You also have the third one in detail here: https://en.wikipedia.org/wiki/Talk:Proof_that_22/7_exceeds_%CF%80#Proof_that_natural_log_.282.29_exceeds_2.2F3 – Jaume Oliver Lafont Feb 06 '16 at 07:10
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    You posted an answer at nearly the exact same time as you posted the question? – fosho Feb 06 '16 at 07:40
  • @Daniel http://blog.stackoverflow.com/2011/07/its-ok-to-ask-and-answer-your-own-questions/ – qwr Feb 06 '16 at 07:43
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    @qwr That's the "somewhere" I didn't manage to find when answering to user230452... thank you! – Jaume Oliver Lafont Feb 06 '16 at 07:47
  • How do you get the hunch when faced with an inequality in between log2 and two rational numbers to start the approach from integral calculus ? – Saikat Feb 06 '16 at 07:54
  • @user230452 hint: write the answer first and then the question. – Jaume Oliver Lafont Feb 06 '16 at 07:57
  • Write the answer first and then the question ? – Saikat Feb 06 '16 at 08:02
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    @user230452 The exponents in http://www.wolframalpha.com/input/?i=int_0%5E1+x%5E4(1-x)%5E4dx%2F(1%2Bx%5E2) may be changed to find rational approximations to $log(2)$ and $\pi$. – Jaume Oliver Lafont Feb 06 '16 at 09:50
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    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. – Jyrki Lahtonen Feb 06 '16 at 17:46
  • @user230452 The seventh integral is evaluated here to prove that $\frac{7}{12}<\log(2)<\frac{5}{6}$ http://math.stackexchange.com/a/1648329/134791 – Jaume Oliver Lafont Feb 09 '16 at 23:51