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

1079




1079








  • 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
    Jan 10 at 20:02










  • $begingroup$
    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
    Jan 10 at 20:08



















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
    Jan 10 at 20:02










  • $begingroup$
    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
    Jan 10 at 20:08
















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
    Jan 10 at 20:02










  • $begingroup$
    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
    Jan 10 at 20:08














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

2,9701739




2,9701739












  • $begingroup$
    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
    Jan 10 at 20:02










  • $begingroup$
    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
    Jan 10 at 20:08


















  • $begingroup$
    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
    Jan 10 at 20:02










  • $begingroup$
    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
    Jan 10 at 20:08
















$begingroup$
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
Jan 10 at 20:02




$begingroup$
@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
Jan 10 at 20:02












$begingroup$
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
Jan 10 at 20:08




$begingroup$
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
Jan 10 at 20:08











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

288k33280660




288k33280660












  • $begingroup$
    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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1300-talet

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Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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1300-talet

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Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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Display a custom attribute below product name in the front-end Magento 1.9.3.8

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0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
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