isn't this wikipedia Definite integrals involving rational or irrational expression wrong!?












9












$begingroup$


photo



isn't this wikipedia Definite integrals involving rational or irrational expression wrong!?



According to my calculations:
$sin[frac{(m+1)pi}{n}]$ while in wiki it is
$sin[frac{(m+1)}{n}]$



seems $pi$ missed.



here is result of my work:



$int_0^infty frac{x^m , dx}{({x^n+a^n)}^r}=frac{(-1)^{r-1}pi a^{m+1-nr}Gamma [(m+1)/n]}{nsin[pi (m+1)/n](r-1)!Gamma[(m+1)/n-r+1]} , 0<m+1<nr$



ref:List of definite integrals



Remark:
@icurays1 in your link there is another wrong $Gamma [frac {(m+1)}{(n-p+1)}]$ while it could be $Gamma [(frac {m+1}{n})-p+1]$
photo






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    @icurays1 i check several times it seems to me $pi$ is missed.
    $endgroup$
    – Neo
    Jan 30 '13 at 15:37






  • 4




    $begingroup$
    What would Jesus Wolfram Alpha say?
    $endgroup$
    – nbubis
    Jan 30 '13 at 15:39






  • 1




    $begingroup$
    I've deleted my previous comment - @neo you might be correct, I've found this which agrees with you. Someone should verify with a published table though, for instance Gradshteyn-Ryzhik
    $endgroup$
    – icurays1
    Jan 30 '13 at 15:56








  • 1




    $begingroup$
    Well, something is wrong on that Wikipedia page. The second expression is a special case of the one you listed above, but the right hand sides do not agree.
    $endgroup$
    – Willie Wong
    Jan 30 '13 at 16:10






  • 3




    $begingroup$
    @nbubis Most of the time Jesus Wolfram Alpha says "calculation too complicated...give me your money and I will do it!" ;-)
    $endgroup$
    – Matemáticos Chibchas
    Jan 30 '13 at 16:25
















9












$begingroup$


photo



isn't this wikipedia Definite integrals involving rational or irrational expression wrong!?



According to my calculations:
$sin[frac{(m+1)pi}{n}]$ while in wiki it is
$sin[frac{(m+1)}{n}]$



seems $pi$ missed.



here is result of my work:



$int_0^infty frac{x^m , dx}{({x^n+a^n)}^r}=frac{(-1)^{r-1}pi a^{m+1-nr}Gamma [(m+1)/n]}{nsin[pi (m+1)/n](r-1)!Gamma[(m+1)/n-r+1]} , 0<m+1<nr$



ref:List of definite integrals



Remark:
@icurays1 in your link there is another wrong $Gamma [frac {(m+1)}{(n-p+1)}]$ while it could be $Gamma [(frac {m+1}{n})-p+1]$
photo






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    @icurays1 i check several times it seems to me $pi$ is missed.
    $endgroup$
    – Neo
    Jan 30 '13 at 15:37






  • 4




    $begingroup$
    What would Jesus Wolfram Alpha say?
    $endgroup$
    – nbubis
    Jan 30 '13 at 15:39






  • 1




    $begingroup$
    I've deleted my previous comment - @neo you might be correct, I've found this which agrees with you. Someone should verify with a published table though, for instance Gradshteyn-Ryzhik
    $endgroup$
    – icurays1
    Jan 30 '13 at 15:56








  • 1




    $begingroup$
    Well, something is wrong on that Wikipedia page. The second expression is a special case of the one you listed above, but the right hand sides do not agree.
    $endgroup$
    – Willie Wong
    Jan 30 '13 at 16:10






  • 3




    $begingroup$
    @nbubis Most of the time Jesus Wolfram Alpha says "calculation too complicated...give me your money and I will do it!" ;-)
    $endgroup$
    – Matemáticos Chibchas
    Jan 30 '13 at 16:25














9












9








9


2



$begingroup$


photo



isn't this wikipedia Definite integrals involving rational or irrational expression wrong!?



According to my calculations:
$sin[frac{(m+1)pi}{n}]$ while in wiki it is
$sin[frac{(m+1)}{n}]$



seems $pi$ missed.



here is result of my work:



$int_0^infty frac{x^m , dx}{({x^n+a^n)}^r}=frac{(-1)^{r-1}pi a^{m+1-nr}Gamma [(m+1)/n]}{nsin[pi (m+1)/n](r-1)!Gamma[(m+1)/n-r+1]} , 0<m+1<nr$



ref:List of definite integrals



Remark:
@icurays1 in your link there is another wrong $Gamma [frac {(m+1)}{(n-p+1)}]$ while it could be $Gamma [(frac {m+1}{n})-p+1]$
photo






integration







share|cite|improve this question











$endgroup$




photo



isn't this wikipedia Definite integrals involving rational or irrational expression wrong!?



According to my calculations:
$sin[frac{(m+1)pi}{n}]$ while in wiki it is
$sin[frac{(m+1)}{n}]$



seems $pi$ missed.



here is result of my work:



$int_0^infty frac{x^m , dx}{({x^n+a^n)}^r}=frac{(-1)^{r-1}pi a^{m+1-nr}Gamma [(m+1)/n]}{nsin[pi (m+1)/n](r-1)!Gamma[(m+1)/n-r+1]} , 0<m+1<nr$



ref:List of definite integrals



Remark:
@icurays1 in your link there is another wrong $Gamma [frac {(m+1)}{(n-p+1)}]$ while it could be $Gamma [(frac {m+1}{n})-p+1]$
photo






integration




integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 20:30









Glorfindel

3,41981830




3,41981830










asked Jan 30 '13 at 15:30









NeoNeo

3341213




3341213












  • $begingroup$
    @icurays1 i check several times it seems to me $pi$ is missed.
    $endgroup$
    – Neo
    Jan 30 '13 at 15:37






  • 4




    $begingroup$
    What would Jesus Wolfram Alpha say?
    $endgroup$
    – nbubis
    Jan 30 '13 at 15:39






  • 1




    $begingroup$
    I've deleted my previous comment - @neo you might be correct, I've found this which agrees with you. Someone should verify with a published table though, for instance Gradshteyn-Ryzhik
    $endgroup$
    – icurays1
    Jan 30 '13 at 15:56








  • 1




    $begingroup$
    Well, something is wrong on that Wikipedia page. The second expression is a special case of the one you listed above, but the right hand sides do not agree.
    $endgroup$
    – Willie Wong
    Jan 30 '13 at 16:10






  • 3




    $begingroup$
    @nbubis Most of the time Jesus Wolfram Alpha says "calculation too complicated...give me your money and I will do it!" ;-)
    $endgroup$
    – Matemáticos Chibchas
    Jan 30 '13 at 16:25


















  • $begingroup$
    @icurays1 i check several times it seems to me $pi$ is missed.
    $endgroup$
    – Neo
    Jan 30 '13 at 15:37






  • 4




    $begingroup$
    What would Jesus Wolfram Alpha say?
    $endgroup$
    – nbubis
    Jan 30 '13 at 15:39






  • 1




    $begingroup$
    I've deleted my previous comment - @neo you might be correct, I've found this which agrees with you. Someone should verify with a published table though, for instance Gradshteyn-Ryzhik
    $endgroup$
    – icurays1
    Jan 30 '13 at 15:56








  • 1




    $begingroup$
    Well, something is wrong on that Wikipedia page. The second expression is a special case of the one you listed above, but the right hand sides do not agree.
    $endgroup$
    – Willie Wong
    Jan 30 '13 at 16:10






  • 3




    $begingroup$
    @nbubis Most of the time Jesus Wolfram Alpha says "calculation too complicated...give me your money and I will do it!" ;-)
    $endgroup$
    – Matemáticos Chibchas
    Jan 30 '13 at 16:25
















$begingroup$
@icurays1 i check several times it seems to me $pi$ is missed.
$endgroup$
– Neo
Jan 30 '13 at 15:37




$begingroup$
@icurays1 i check several times it seems to me $pi$ is missed.
$endgroup$
– Neo
Jan 30 '13 at 15:37




4




4




$begingroup$
What would Jesus Wolfram Alpha say?
$endgroup$
– nbubis
Jan 30 '13 at 15:39




$begingroup$
What would Jesus Wolfram Alpha say?
$endgroup$
– nbubis
Jan 30 '13 at 15:39




1




1




$begingroup$
I've deleted my previous comment - @neo you might be correct, I've found this which agrees with you. Someone should verify with a published table though, for instance Gradshteyn-Ryzhik
$endgroup$
– icurays1
Jan 30 '13 at 15:56






$begingroup$
I've deleted my previous comment - @neo you might be correct, I've found this which agrees with you. Someone should verify with a published table though, for instance Gradshteyn-Ryzhik
$endgroup$
– icurays1
Jan 30 '13 at 15:56






1




1




$begingroup$
Well, something is wrong on that Wikipedia page. The second expression is a special case of the one you listed above, but the right hand sides do not agree.
$endgroup$
– Willie Wong
Jan 30 '13 at 16:10




$begingroup$
Well, something is wrong on that Wikipedia page. The second expression is a special case of the one you listed above, but the right hand sides do not agree.
$endgroup$
– Willie Wong
Jan 30 '13 at 16:10




3




3




$begingroup$
@nbubis Most of the time Jesus Wolfram Alpha says "calculation too complicated...give me your money and I will do it!" ;-)
$endgroup$
– Matemáticos Chibchas
Jan 30 '13 at 16:25




$begingroup$
@nbubis Most of the time Jesus Wolfram Alpha says "calculation too complicated...give me your money and I will do it!" ;-)
$endgroup$
– Matemáticos Chibchas
Jan 30 '13 at 16:25










1 Answer
1






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oldest

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8












$begingroup$

It is definitely wrong. Take $a=1, m=0, n=2, r=1$. Then it reads
$$int_0^infty frac{1}{x^2+1},dx = frac{pi}{2 sin(1/2)}$$
which is false. The correct value is $pi/2$, which agrees with your proposed $sinleft[frac{(m+1)pi}{n}right]$.



Taking the sine of a rational number is definitely a red flag.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    ok so wikipedia equation must be modified
    $endgroup$
    – Neo
    Jan 30 '13 at 21:11











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

It is definitely wrong. Take $a=1, m=0, n=2, r=1$. Then it reads
$$int_0^infty frac{1}{x^2+1},dx = frac{pi}{2 sin(1/2)}$$
which is false. The correct value is $pi/2$, which agrees with your proposed $sinleft[frac{(m+1)pi}{n}right]$.



Taking the sine of a rational number is definitely a red flag.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    ok so wikipedia equation must be modified
    $endgroup$
    – Neo
    Jan 30 '13 at 21:11
















8












$begingroup$

It is definitely wrong. Take $a=1, m=0, n=2, r=1$. Then it reads
$$int_0^infty frac{1}{x^2+1},dx = frac{pi}{2 sin(1/2)}$$
which is false. The correct value is $pi/2$, which agrees with your proposed $sinleft[frac{(m+1)pi}{n}right]$.



Taking the sine of a rational number is definitely a red flag.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    ok so wikipedia equation must be modified
    $endgroup$
    – Neo
    Jan 30 '13 at 21:11














8












8








8





$begingroup$

It is definitely wrong. Take $a=1, m=0, n=2, r=1$. Then it reads
$$int_0^infty frac{1}{x^2+1},dx = frac{pi}{2 sin(1/2)}$$
which is false. The correct value is $pi/2$, which agrees with your proposed $sinleft[frac{(m+1)pi}{n}right]$.



Taking the sine of a rational number is definitely a red flag.






share|cite|improve this answer











$endgroup$



It is definitely wrong. Take $a=1, m=0, n=2, r=1$. Then it reads
$$int_0^infty frac{1}{x^2+1},dx = frac{pi}{2 sin(1/2)}$$
which is false. The correct value is $pi/2$, which agrees with your proposed $sinleft[frac{(m+1)pi}{n}right]$.



Taking the sine of a rational number is definitely a red flag.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Feb 7 '13 at 4:36

























answered Jan 30 '13 at 19:08









Nate EldredgeNate Eldredge

63.1k682171




63.1k682171












  • $begingroup$
    ok so wikipedia equation must be modified
    $endgroup$
    – Neo
    Jan 30 '13 at 21:11


















  • $begingroup$
    ok so wikipedia equation must be modified
    $endgroup$
    – Neo
    Jan 30 '13 at 21:11
















$begingroup$
ok so wikipedia equation must be modified
$endgroup$
– Neo
Jan 30 '13 at 21:11




$begingroup$
ok so wikipedia equation must be modified
$endgroup$
– Neo
Jan 30 '13 at 21:11


















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