Evaluate $int_0^1 log left( frac{x^2+sqrt{3}x+1}{x^2-sqrt{3}x+1} right) frac{dx}{x} $












5












$begingroup$


I'm trying to evaluate this integral but i'm having a lot of problems with the standards method.



$$int_0^1
log
left(
frac{x^2+sqrt{3}x+1}{x^2-sqrt{3}x+1}
right)
frac{dx}{x}
$$



I've tried integration by parts, taylor expansion of the $log$ function and some substitution.



Wolfram says that the results is $frac{pi^2}{3}$. I think that it divide the integral in four parts factorizing the argument of the logarithm with complex numbers, anyway it seems to take too long as approach.
Do you have any suggestions?



I've also noticed that the function is even so we could also write the integral as
$$
frac{1}{2}
int_{-1}^1
log
left(
frac{x^2+sqrt{3}x+1}{x^2-sqrt{3}x+1}
right)
frac{dx}{x}
$$






definite-integrals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Both the numerator and denominator are invariant under the transformation $p(x) mapsto x^2 p(frac{1}{x})$, which suggests you might get some mileage out of applying the substitution $u = frac{1}{x}$ and rearranging.
    $endgroup$
    – Travis
    Jan 7 at 13:13
















5












$begingroup$


I'm trying to evaluate this integral but i'm having a lot of problems with the standards method.



$$int_0^1
log
left(
frac{x^2+sqrt{3}x+1}{x^2-sqrt{3}x+1}
right)
frac{dx}{x}
$$



I've tried integration by parts, taylor expansion of the $log$ function and some substitution.



Wolfram says that the results is $frac{pi^2}{3}$. I think that it divide the integral in four parts factorizing the argument of the logarithm with complex numbers, anyway it seems to take too long as approach.
Do you have any suggestions?



I've also noticed that the function is even so we could also write the integral as
$$
frac{1}{2}
int_{-1}^1
log
left(
frac{x^2+sqrt{3}x+1}{x^2-sqrt{3}x+1}
right)
frac{dx}{x}
$$






definite-integrals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Both the numerator and denominator are invariant under the transformation $p(x) mapsto x^2 p(frac{1}{x})$, which suggests you might get some mileage out of applying the substitution $u = frac{1}{x}$ and rearranging.
    $endgroup$
    – Travis
    Jan 7 at 13:13














5












5








5





$begingroup$


I'm trying to evaluate this integral but i'm having a lot of problems with the standards method.



$$int_0^1
log
left(
frac{x^2+sqrt{3}x+1}{x^2-sqrt{3}x+1}
right)
frac{dx}{x}
$$



I've tried integration by parts, taylor expansion of the $log$ function and some substitution.



Wolfram says that the results is $frac{pi^2}{3}$. I think that it divide the integral in four parts factorizing the argument of the logarithm with complex numbers, anyway it seems to take too long as approach.
Do you have any suggestions?



I've also noticed that the function is even so we could also write the integral as
$$
frac{1}{2}
int_{-1}^1
log
left(
frac{x^2+sqrt{3}x+1}{x^2-sqrt{3}x+1}
right)
frac{dx}{x}
$$






definite-integrals







share|cite|improve this question









$endgroup$




I'm trying to evaluate this integral but i'm having a lot of problems with the standards method.



$$int_0^1
log
left(
frac{x^2+sqrt{3}x+1}{x^2-sqrt{3}x+1}
right)
frac{dx}{x}
$$



I've tried integration by parts, taylor expansion of the $log$ function and some substitution.



Wolfram says that the results is $frac{pi^2}{3}$. I think that it divide the integral in four parts factorizing the argument of the logarithm with complex numbers, anyway it seems to take too long as approach.
Do you have any suggestions?



I've also noticed that the function is even so we could also write the integral as
$$
frac{1}{2}
int_{-1}^1
log
left(
frac{x^2+sqrt{3}x+1}{x^2-sqrt{3}x+1}
right)
frac{dx}{x}
$$






definite-integrals




definite-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 13:05









FabioFabio

1029




1029








  • 1




    $begingroup$
    Both the numerator and denominator are invariant under the transformation $p(x) mapsto x^2 p(frac{1}{x})$, which suggests you might get some mileage out of applying the substitution $u = frac{1}{x}$ and rearranging.
    $endgroup$
    – Travis
    Jan 7 at 13:13














  • 1




    $begingroup$
    Both the numerator and denominator are invariant under the transformation $p(x) mapsto x^2 p(frac{1}{x})$, which suggests you might get some mileage out of applying the substitution $u = frac{1}{x}$ and rearranging.
    $endgroup$
    – Travis
    Jan 7 at 13:13








1




1




$begingroup$
Both the numerator and denominator are invariant under the transformation $p(x) mapsto x^2 p(frac{1}{x})$, which suggests you might get some mileage out of applying the substitution $u = frac{1}{x}$ and rearranging.
$endgroup$
– Travis
Jan 7 at 13:13




$begingroup$
Both the numerator and denominator are invariant under the transformation $p(x) mapsto x^2 p(frac{1}{x})$, which suggests you might get some mileage out of applying the substitution $u = frac{1}{x}$ and rearranging.
$endgroup$
– Travis
Jan 7 at 13:13










2 Answers
2






active

oldest

votes


















13












$begingroup$

Let $$I(theta)=int_0^1frac{ln(x^2+2xcostheta+1)}xdx,$$
differentiating it gives $$I'(theta)=int_0^1-frac{2sintheta}{1+2xcostheta+x^2}dx\
=-theta$$
(since $thetain(-pi,pi)$)

The integral need to find is $$Ileft(frac16piright)-Ileft(frac56piright)=int_{5/6pi}^{1/6pi}-theta dtheta=frac13pi^2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Brilliant, i'll add this trick to my collection thank you
    $endgroup$
    – Fabio
    Jan 7 at 13:24






  • 1




    $begingroup$
    @Fabio one can actually generalize this method to see that $$int_0^1logbigg(frac{x^2+ax+1}{x^2-ax+1}bigg)frac{mathrm{d}x}{x}=frac{pi^2}2-piarccosfrac{a}2$$
    $endgroup$
    – clathratus
    2 days ago










  • $begingroup$
    I didn't noticed, besides allow to give a more elegant answer I guess this will be quite useful in the future, thanks for the observation
    $endgroup$
    – Fabio
    2 days ago



















8












$begingroup$

Let us consider that for any $alphain S^1$



$$ int_{0}^{1}frac{log(1-alpha x)}{x},dx = -sum_{ngeq 1}int_{0}^{1}frac{alpha^n x^{n}}{nx},dx=-sum_{ngeq 1}frac{alpha^n}{n^2}=-text{Li}_2(alpha) $$
and by setting $zeta=expleft(frac{2pi i}{12}right)$ we have



$$ int_{0}^{1}logleft(frac{1-sqrt{3}x+x^2}{1+sqrt{3}x+x^2}right),frac{dx}{x}=int_{0}^{1}left[log(1-zeta x)+log(1-zeta^{11} x)-log(1-zeta^5 x)-log(1-zeta^7 x)right]frac{dx}{x} $$
hence your integral equals
$$ text{Li}_2(zeta)+text{Li}_2(zeta^{11})-text{Li}_2(zeta^5)-text{Li}_2(zeta^7)=2text{Re},text{Li}_2(zeta)-2text{Re},text{Li}_2(zeta^5). $$
Last trick: $text{Re},text{Li}_2(e^{itheta})$ is an elementary function, since it is the formal primitive of the sawtooth wave $sum_{ngeq 1}frac{sin(nx)}{n}$ (piecewise-linear), hence a periodic and piecewise-parabolic function. Putting everything together, the outcome is simply $2zeta(2)=frac{pi^2}{3}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's impressive! But i don't understand some things sorry for the ignorance: what's the set $S^1$?
    $endgroup$
    – Fabio
    Jan 7 at 13:31






  • 1




    $begingroup$
    @Fabio: the unit circle centered at the origin of the complex plane.
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 13:52











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









13












$begingroup$

Let $$I(theta)=int_0^1frac{ln(x^2+2xcostheta+1)}xdx,$$
differentiating it gives $$I'(theta)=int_0^1-frac{2sintheta}{1+2xcostheta+x^2}dx\
=-theta$$
(since $thetain(-pi,pi)$)

The integral need to find is $$Ileft(frac16piright)-Ileft(frac56piright)=int_{5/6pi}^{1/6pi}-theta dtheta=frac13pi^2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Brilliant, i'll add this trick to my collection thank you
    $endgroup$
    – Fabio
    Jan 7 at 13:24






  • 1




    $begingroup$
    @Fabio one can actually generalize this method to see that $$int_0^1logbigg(frac{x^2+ax+1}{x^2-ax+1}bigg)frac{mathrm{d}x}{x}=frac{pi^2}2-piarccosfrac{a}2$$
    $endgroup$
    – clathratus
    2 days ago










  • $begingroup$
    I didn't noticed, besides allow to give a more elegant answer I guess this will be quite useful in the future, thanks for the observation
    $endgroup$
    – Fabio
    2 days ago
















13












$begingroup$

Let $$I(theta)=int_0^1frac{ln(x^2+2xcostheta+1)}xdx,$$
differentiating it gives $$I'(theta)=int_0^1-frac{2sintheta}{1+2xcostheta+x^2}dx\
=-theta$$
(since $thetain(-pi,pi)$)

The integral need to find is $$Ileft(frac16piright)-Ileft(frac56piright)=int_{5/6pi}^{1/6pi}-theta dtheta=frac13pi^2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Brilliant, i'll add this trick to my collection thank you
    $endgroup$
    – Fabio
    Jan 7 at 13:24






  • 1




    $begingroup$
    @Fabio one can actually generalize this method to see that $$int_0^1logbigg(frac{x^2+ax+1}{x^2-ax+1}bigg)frac{mathrm{d}x}{x}=frac{pi^2}2-piarccosfrac{a}2$$
    $endgroup$
    – clathratus
    2 days ago










  • $begingroup$
    I didn't noticed, besides allow to give a more elegant answer I guess this will be quite useful in the future, thanks for the observation
    $endgroup$
    – Fabio
    2 days ago














13












13








13





$begingroup$

Let $$I(theta)=int_0^1frac{ln(x^2+2xcostheta+1)}xdx,$$
differentiating it gives $$I'(theta)=int_0^1-frac{2sintheta}{1+2xcostheta+x^2}dx\
=-theta$$
(since $thetain(-pi,pi)$)

The integral need to find is $$Ileft(frac16piright)-Ileft(frac56piright)=int_{5/6pi}^{1/6pi}-theta dtheta=frac13pi^2$$






share|cite|improve this answer









$endgroup$



Let $$I(theta)=int_0^1frac{ln(x^2+2xcostheta+1)}xdx,$$
differentiating it gives $$I'(theta)=int_0^1-frac{2sintheta}{1+2xcostheta+x^2}dx\
=-theta$$
(since $thetain(-pi,pi)$)

The integral need to find is $$Ileft(frac16piright)-Ileft(frac56piright)=int_{5/6pi}^{1/6pi}-theta dtheta=frac13pi^2$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 7 at 13:21









Kemono ChenKemono Chen

2,8531738




2,8531738












  • $begingroup$
    Brilliant, i'll add this trick to my collection thank you
    $endgroup$
    – Fabio
    Jan 7 at 13:24






  • 1




    $begingroup$
    @Fabio one can actually generalize this method to see that $$int_0^1logbigg(frac{x^2+ax+1}{x^2-ax+1}bigg)frac{mathrm{d}x}{x}=frac{pi^2}2-piarccosfrac{a}2$$
    $endgroup$
    – clathratus
    2 days ago










  • $begingroup$
    I didn't noticed, besides allow to give a more elegant answer I guess this will be quite useful in the future, thanks for the observation
    $endgroup$
    – Fabio
    2 days ago


















  • $begingroup$
    Brilliant, i'll add this trick to my collection thank you
    $endgroup$
    – Fabio
    Jan 7 at 13:24






  • 1




    $begingroup$
    @Fabio one can actually generalize this method to see that $$int_0^1logbigg(frac{x^2+ax+1}{x^2-ax+1}bigg)frac{mathrm{d}x}{x}=frac{pi^2}2-piarccosfrac{a}2$$
    $endgroup$
    – clathratus
    2 days ago










  • $begingroup$
    I didn't noticed, besides allow to give a more elegant answer I guess this will be quite useful in the future, thanks for the observation
    $endgroup$
    – Fabio
    2 days ago
















$begingroup$
Brilliant, i'll add this trick to my collection thank you
$endgroup$
– Fabio
Jan 7 at 13:24




$begingroup$
Brilliant, i'll add this trick to my collection thank you
$endgroup$
– Fabio
Jan 7 at 13:24




1




1




$begingroup$
@Fabio one can actually generalize this method to see that $$int_0^1logbigg(frac{x^2+ax+1}{x^2-ax+1}bigg)frac{mathrm{d}x}{x}=frac{pi^2}2-piarccosfrac{a}2$$
$endgroup$
– clathratus
2 days ago




$begingroup$
@Fabio one can actually generalize this method to see that $$int_0^1logbigg(frac{x^2+ax+1}{x^2-ax+1}bigg)frac{mathrm{d}x}{x}=frac{pi^2}2-piarccosfrac{a}2$$
$endgroup$
– clathratus
2 days ago












$begingroup$
I didn't noticed, besides allow to give a more elegant answer I guess this will be quite useful in the future, thanks for the observation
$endgroup$
– Fabio
2 days ago




$begingroup$
I didn't noticed, besides allow to give a more elegant answer I guess this will be quite useful in the future, thanks for the observation
$endgroup$
– Fabio
2 days ago











8












$begingroup$

Let us consider that for any $alphain S^1$



$$ int_{0}^{1}frac{log(1-alpha x)}{x},dx = -sum_{ngeq 1}int_{0}^{1}frac{alpha^n x^{n}}{nx},dx=-sum_{ngeq 1}frac{alpha^n}{n^2}=-text{Li}_2(alpha) $$
and by setting $zeta=expleft(frac{2pi i}{12}right)$ we have



$$ int_{0}^{1}logleft(frac{1-sqrt{3}x+x^2}{1+sqrt{3}x+x^2}right),frac{dx}{x}=int_{0}^{1}left[log(1-zeta x)+log(1-zeta^{11} x)-log(1-zeta^5 x)-log(1-zeta^7 x)right]frac{dx}{x} $$
hence your integral equals
$$ text{Li}_2(zeta)+text{Li}_2(zeta^{11})-text{Li}_2(zeta^5)-text{Li}_2(zeta^7)=2text{Re},text{Li}_2(zeta)-2text{Re},text{Li}_2(zeta^5). $$
Last trick: $text{Re},text{Li}_2(e^{itheta})$ is an elementary function, since it is the formal primitive of the sawtooth wave $sum_{ngeq 1}frac{sin(nx)}{n}$ (piecewise-linear), hence a periodic and piecewise-parabolic function. Putting everything together, the outcome is simply $2zeta(2)=frac{pi^2}{3}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's impressive! But i don't understand some things sorry for the ignorance: what's the set $S^1$?
    $endgroup$
    – Fabio
    Jan 7 at 13:31






  • 1




    $begingroup$
    @Fabio: the unit circle centered at the origin of the complex plane.
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 13:52
















8












$begingroup$

Let us consider that for any $alphain S^1$



$$ int_{0}^{1}frac{log(1-alpha x)}{x},dx = -sum_{ngeq 1}int_{0}^{1}frac{alpha^n x^{n}}{nx},dx=-sum_{ngeq 1}frac{alpha^n}{n^2}=-text{Li}_2(alpha) $$
and by setting $zeta=expleft(frac{2pi i}{12}right)$ we have



$$ int_{0}^{1}logleft(frac{1-sqrt{3}x+x^2}{1+sqrt{3}x+x^2}right),frac{dx}{x}=int_{0}^{1}left[log(1-zeta x)+log(1-zeta^{11} x)-log(1-zeta^5 x)-log(1-zeta^7 x)right]frac{dx}{x} $$
hence your integral equals
$$ text{Li}_2(zeta)+text{Li}_2(zeta^{11})-text{Li}_2(zeta^5)-text{Li}_2(zeta^7)=2text{Re},text{Li}_2(zeta)-2text{Re},text{Li}_2(zeta^5). $$
Last trick: $text{Re},text{Li}_2(e^{itheta})$ is an elementary function, since it is the formal primitive of the sawtooth wave $sum_{ngeq 1}frac{sin(nx)}{n}$ (piecewise-linear), hence a periodic and piecewise-parabolic function. Putting everything together, the outcome is simply $2zeta(2)=frac{pi^2}{3}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's impressive! But i don't understand some things sorry for the ignorance: what's the set $S^1$?
    $endgroup$
    – Fabio
    Jan 7 at 13:31






  • 1




    $begingroup$
    @Fabio: the unit circle centered at the origin of the complex plane.
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 13:52














8












8








8





$begingroup$

Let us consider that for any $alphain S^1$



$$ int_{0}^{1}frac{log(1-alpha x)}{x},dx = -sum_{ngeq 1}int_{0}^{1}frac{alpha^n x^{n}}{nx},dx=-sum_{ngeq 1}frac{alpha^n}{n^2}=-text{Li}_2(alpha) $$
and by setting $zeta=expleft(frac{2pi i}{12}right)$ we have



$$ int_{0}^{1}logleft(frac{1-sqrt{3}x+x^2}{1+sqrt{3}x+x^2}right),frac{dx}{x}=int_{0}^{1}left[log(1-zeta x)+log(1-zeta^{11} x)-log(1-zeta^5 x)-log(1-zeta^7 x)right]frac{dx}{x} $$
hence your integral equals
$$ text{Li}_2(zeta)+text{Li}_2(zeta^{11})-text{Li}_2(zeta^5)-text{Li}_2(zeta^7)=2text{Re},text{Li}_2(zeta)-2text{Re},text{Li}_2(zeta^5). $$
Last trick: $text{Re},text{Li}_2(e^{itheta})$ is an elementary function, since it is the formal primitive of the sawtooth wave $sum_{ngeq 1}frac{sin(nx)}{n}$ (piecewise-linear), hence a periodic and piecewise-parabolic function. Putting everything together, the outcome is simply $2zeta(2)=frac{pi^2}{3}$.






share|cite|improve this answer











$endgroup$



Let us consider that for any $alphain S^1$



$$ int_{0}^{1}frac{log(1-alpha x)}{x},dx = -sum_{ngeq 1}int_{0}^{1}frac{alpha^n x^{n}}{nx},dx=-sum_{ngeq 1}frac{alpha^n}{n^2}=-text{Li}_2(alpha) $$
and by setting $zeta=expleft(frac{2pi i}{12}right)$ we have



$$ int_{0}^{1}logleft(frac{1-sqrt{3}x+x^2}{1+sqrt{3}x+x^2}right),frac{dx}{x}=int_{0}^{1}left[log(1-zeta x)+log(1-zeta^{11} x)-log(1-zeta^5 x)-log(1-zeta^7 x)right]frac{dx}{x} $$
hence your integral equals
$$ text{Li}_2(zeta)+text{Li}_2(zeta^{11})-text{Li}_2(zeta^5)-text{Li}_2(zeta^7)=2text{Re},text{Li}_2(zeta)-2text{Re},text{Li}_2(zeta^5). $$
Last trick: $text{Re},text{Li}_2(e^{itheta})$ is an elementary function, since it is the formal primitive of the sawtooth wave $sum_{ngeq 1}frac{sin(nx)}{n}$ (piecewise-linear), hence a periodic and piecewise-parabolic function. Putting everything together, the outcome is simply $2zeta(2)=frac{pi^2}{3}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 7 at 18:21

























answered Jan 7 at 13:25









Jack D'AurizioJack D'Aurizio

288k33280659




288k33280659












  • $begingroup$
    That's impressive! But i don't understand some things sorry for the ignorance: what's the set $S^1$?
    $endgroup$
    – Fabio
    Jan 7 at 13:31






  • 1




    $begingroup$
    @Fabio: the unit circle centered at the origin of the complex plane.
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 13:52


















  • $begingroup$
    That's impressive! But i don't understand some things sorry for the ignorance: what's the set $S^1$?
    $endgroup$
    – Fabio
    Jan 7 at 13:31






  • 1




    $begingroup$
    @Fabio: the unit circle centered at the origin of the complex plane.
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 13:52
















$begingroup$
That's impressive! But i don't understand some things sorry for the ignorance: what's the set $S^1$?
$endgroup$
– Fabio
Jan 7 at 13:31




$begingroup$
That's impressive! But i don't understand some things sorry for the ignorance: what's the set $S^1$?
$endgroup$
– Fabio
Jan 7 at 13:31




1




1




$begingroup$
@Fabio: the unit circle centered at the origin of the complex plane.
$endgroup$
– Jack D'Aurizio
Jan 7 at 13:52




$begingroup$
@Fabio: the unit circle centered at the origin of the complex plane.
$endgroup$
– Jack D'Aurizio
Jan 7 at 13:52


















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Mario Kart Wii

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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
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What does “Dominus providebit” mean?

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5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
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Antonio Litta Visconti Arese

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Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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