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How can we evaluate the definite integral $$\displaystyle \int_{0}^{1}{\frac{\ln{x}}{x^2-x-1}dx}$$

I tried many times but still have no idea.

pxchg1200
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    What level of calculus are you in, just wondering? The solution on Wolfram Alpha doesn't look too pretty... :) – apnorton Mar 08 '13 at 03:09
  • well,I tried $x^2-x-1=(x-x_{0})(x-x_{1})$ where $ x_{0}=\frac{\sqrt{5}+1}{2},x_{1}=\frac{1-\sqrt{5}}{2}$,then I can do nothing with $\frac{1}{\sqrt{5}}\int_{0}^{1}{\ln{x}(\frac{1}{x-x_{0}}-\frac{1}{x-x_{1}})dx} $... – pxchg1200 Mar 08 '13 at 03:13
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    Maple gets $\dfrac{1}{\sqrt{5}} \left( \text{dilog}\left(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\right) - \text{dilog}\left(\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\right)\right)$. Numerically this appears to be $\dfrac{\pi^2}{5 \sqrt{5}}$. – Robert Israel Mar 08 '13 at 03:19
  • This problem is a special case of the following problem. – Mhenni Benghorbal Mar 08 '13 at 04:24

1 Answers1

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For what it's worth, here are some computations to help you get further from where you stopped.

For $|x_0|>1$, which is the case of your $x_0=\frac{1+\sqrt{5}}{2}$, we have: $$ \int_0^1\frac{\ln x}{x-x_0}dx=-\frac{1}{x_0}\int_0^1\frac{\ln x}{1-\frac{x}{x_0}}dx=-\frac{1}{x_0}\int_0^1\ln x\sum_{k\geq 0}\left(\frac{x}{x_0}\right)^kdx $$ $$ =-\frac{1}{x_0}\sum_{k\geq 0}\frac{1}{x_0^k}\int_0^1x^k\ln x dx=\sum_{k\geq 0}\frac{x_0^{-(k+1)}}{(k+1)^2}. $$ Now a power series computation starting from $(1-y)^{-1}=\sum_{k\geq 0}y^k$ yields $$ \sum_{k\geq 0}\frac{y^{k+1}}{(k+1)^2}=-\int_0^y\frac{\ln(1-t)}{t}dt=\mathrm{Li}_2(y) $$ the dilogarithm. So $$ \int_0^1\frac{\ln x}{x-x_0}dx=\mathrm{Li}_2\left(\frac{1}{x_0}\right)\quad\forall\;|x_0|>1. $$

I think I'll let you handle the case $-1\leq x_1\leq 1$.

Julien
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