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:

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

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

  • $\begingroup$ @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

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 $\sin\left[\frac{(m+1)\pi}{n}\right]$.

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

| cite | improve this answer | |
  • $\begingroup$ ok so wikipedia equation must be modified $\endgroup$ – Neo Jan 30 '13 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.