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we know that $$ \ln(x)=\frac{1}{x} \int_{0}^{1}dt \frac{1}{1+xt} $$

then given a cuadrature formula inside $(0,1)$ is that true

$$ \ln(x)= \frac{1}{x}\sum_{i}\frac{w_{i}}{1+xt_{i}} $$

wht other rational approximations of logarithm are useful based on rational functions ?? $ R(x)$

Michael Hardy
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Jose Garcia
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  • We know this? I don't. It appears to me that the integral is incorrect. – Ron Gordon May 23 '13 at 19:27
  • @RonGordon, I also thought the integral was incorrect, but if we perform the change of variables $u=1+xt$, then $du=xdt$, this $x$ cancels the other one outside, and $u$ ranges from $1$ to $x$ as in the usual definition. – Gold May 23 '13 at 19:31
  • @user1620696: no, $dt=du/x$ and the $x$ outside does not cancel. You end up with $\log{(1+x)}/x^2$. – Ron Gordon May 23 '13 at 19:33
  • @RonGordon, sorry, I didn't note that, I thought the wrong way arround. It's really incorrect. – Gold May 23 '13 at 19:34
  • so my expression should be $ ln(x)= x\int_{0}^{1} \frac{dt}{1+xt} $ – Jose Garcia May 23 '13 at 19:37
  • @JoseGarcia, this new expression isn't $\ln(x)$, rather it is $\ln(x+1)$. – Gold May 24 '13 at 14:04
  • OK you are right perhapsh i should use this expression plus a power series involving teh difference $ln(x+1)-ln(x) $ – Jose Garcia May 24 '13 at 18:56

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