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Find out $$\int_{0}^{1} \frac{\ln(-x^2+x+1)}{x(1-x)} \rm{d}x$$.

They say when it comes to logarithms in integrals first thing to check if $[x → \frac{1}{x}]$ works or not .

My try :

$$I = \int_{0}^{1} \frac{\ln(-x^2+x+1)}{x(1-x)} \rm{d}x = \int_{1}^{\infty} \frac{\ln(\frac{-1}{x^2} + \frac{1}{x} + 1)}{\frac{1}{x}(1-\frac{1}{x})}.\frac{-1}{x^2} \rm{d}x [x → \frac{1}{x}] = - \int_{1}^{\infty} \frac{\ln(\frac{x^2+x-1}{x^2})}{x-1} \rm{d}x = \int_{1}^{\infty} \frac{2\ln(x)}{x-1} \rm{d}x - \int_{1}^{\infty} \frac{\ln(x^2+x-1)}{x-1} \rm{d}x$$

No idea how to proceed further , all the approaches are welcome (except Feynman ones) .

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    $\begingroup$ Is it possible that you have forgotten to change the limits of integration? $\endgroup$ Commented May 21 at 16:25
  • $\begingroup$ @SeverinSchraven no I don't think so $\endgroup$
    – Ash_Blanc
    Commented May 21 at 16:27
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    $\begingroup$ Ok, then how did you replace $x$ by $1/x$ without needing to adapt the limits of integration? $\endgroup$ Commented May 21 at 16:32
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    $\begingroup$ Now you can see that you cannot split the integral as you did in the end as the first one will be $\infty$ while the second one is $-\infty$. $\endgroup$ Commented May 21 at 16:41
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    $\begingroup$ The very last step is not correct. It is like saying $\int_1^\infty 0 = \int_1^\infty (-x) dx + \int_1^\infty x dx$. The integrals are not finite. $\endgroup$ Commented May 21 at 16:44

2 Answers 2

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\begin{align} &\int_{0}^{1} \frac{\ln(-x^2+x+1)}{x(1-x)} \ dx \\ = &\int_{0}^{1} \frac{\ln(-x^2+x+1)}{x} \ dx +\int_{0}^{1} \frac{\ln(-x^2+x+1)}{1-x} \overset{x\to 1-x}{dx}\\ = &\ 2\int_{0}^{1} \frac{\ln(-x^2+x+1)}{x} \ dx\\ = &\ 2\int_{0}^{1} \frac{\ln(1-\frac1{\phi}x)}{x} \ dx + 2\int_{0}^{1} \frac{\ln(1+{\phi}x)}{x} \ dx\\ = &\ - 2 \text{Li}_2(\frac1{\phi})- 2 \text{Li}_2({-\phi})=4\ln^2\phi \end{align} where $\quad\text{Li}_2(\frac1{\phi})=\frac{\pi^2}{10}-\ln^2\phi\quad$ and $\quad\text{Li}_2({-\phi})=-\frac{\pi^2}{10}-\ln^2\phi$.

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  • $\begingroup$ Whats $\phi$ here ? $\endgroup$
    – Ash_Blanc
    Commented May 21 at 16:56
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    $\begingroup$ @Ash_Blanc Golden ratio $\phi=\frac{1+\sqrt5}2$ $\endgroup$
    – Quanto
    Commented May 21 at 17:04
  • $\begingroup$ It's lil insane though the actual ans is in terms of $\pi$ not $\phi$ $\endgroup$
    – Ash_Blanc
    Commented May 21 at 17:07
  • $\begingroup$ Though I got one idea this integral converts into $2\int_{0}^{1} \frac{\ln(-x^2+x+1)}{x} \rm{d}x$ $\endgroup$
    – Ash_Blanc
    Commented May 21 at 17:23
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$$I=\int \frac{\log(-x^2+x+1)}{x(1-x)} \,dx=\int \frac{\log(\phi-x)+\log \left(x-\frac{1}{\phi }\right)}{x(1-x)} \,dx$$ $$\frac 1 {x(1-x)}=\frac{1}{x}-\frac{1}{x-1}$$ $$I=\int \frac{\log (\phi -x)}{x}\,dx-\int \frac{\log (\phi -x)}{x-1} \,dx+\int \frac{\log \left(x-\frac{1}{\phi }\right)}{x}\,dx-\int \frac{\log \left(x-\frac{1}{\phi }\right)}{x-1}\,dx$$

Four simple integrals

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  • $\begingroup$ whats after this step ? $\endgroup$
    – Ash_Blanc
    Commented May 22 at 6:10
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    $\begingroup$ @Ash_Blanc nothing left $\endgroup$
    – Martin.s
    Commented May 22 at 7:49
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    $\begingroup$ @Martin.s how to find out the value of last four integrals ?? $\endgroup$
    – Ash_Blanc
    Commented May 22 at 7:58

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