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

Question

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 $\Gamma [(\frac {m+1}{n})-p+1]$

@icurays1 i check several times it seems to me $\pi$ is missed. — Neo, Jan 30 '13 at 15:37
@L. F. my work is same but just there is a $\pi$ that i mention. — Neo, Jan 30 '13 at 15:43
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 — icurays1, Jan 30 '13 at 15:56
Alternately, if @neo is willing to share their work, we could check their work and then someone can edit the wiki. — icurays1, Jan 30 '13 at 16:03
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. — Willie Wong, Jan 30 '13 at 16:10
@nbubis Most of the time Jesus Wolfram Alpha says "calculation too complicated...give me your money and I will do it!" ;-) — Matemáticos Chibchas, Jan 30 '13 at 16:25

Nate Eldredge · Accepted Answer · 2013-02-07T04:36:11.260

8

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.

edited Feb 07 '13 at 04:36

answered Jan 30 '13 at 19:08

Nate Eldredge

97,710

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

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

1 Answers1