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While studying the $O(n)$-model I found myself in the need of integrating $$\frac{\int_0^{\pi}\text{d}\theta\sin(\theta)^{n-2}\cos(\theta)^{2r}}{\int_0^\pi\text{d}\theta\sin(\theta)^{n-2}},$$ where $n\in\mathbb{N}^+$ and $r\in\mathbb{N}$. I imagine however that the author (Brézin, Introduction to Statistical Field Theory) is only considering $n>1$. He claims that this can be done through Beta function integrals. Following his suggestion, I noticed that the integrands are even around $\pi/2$. Then the integral can be reduced to
$$\frac{\int_0^{\pi/2}\text{d}\theta\sin(\theta)^{n-2}\cos(\theta)^{2r}}{\int_0^{\pi/2}\text{d}\theta\sin(\theta)^{n-2}}.$$ We have the beta functions $$\frac{(x-1)!(y-1)!}{(x+y-1)!}=B(x,y)=2\int_0^{\pi/2}\text{d}\theta\sin(\theta)^{2x-1}\cos(\theta)^{2y-1},$$ for $x,y\in\mathbb{N}^+$. Then the integral I'm looking for is $$\frac{\left(\frac{n-3}{2}\right)!\left(\frac{2r-1}{2}\right)!\left(\frac{n-2}{2}\right)!}{\left(\frac{n-3}{2}\right)!\left(-\frac{1}{2}\right)!\left(\frac{2r+n}{2}\right)!}$$ This doesn't seem correct though. All I know is that $$\sum_{r=0}^\infty\frac{x^r}{r!}\frac{\int_0^{\pi}\text{d}\theta\sin(\theta)^{n-2}\cos(\theta)^{r}}{\int_0^\pi\text{d}\theta\sin(\theta)^{n-2}}=1+\sum_{r=1}^\infty\frac{n^r}{2^rr!n(n+2)\cdots(n+2r-2)}x^{2r}.$$ I suspect there is however an error in the book and the correct statement is $$\sum_{r=0}^\infty\frac{(\sqrt{n}x)^r}{r!}\frac{\int_0^{\pi}\text{d}\theta\sin(\theta)^{n-2}\cos(\theta)^{r}}{\int_0^\pi\text{d}\theta\sin(\theta)^{n-2}}=1+\sum_{r=1}^\infty\frac{n^r}{2^rr!n(n+2)\cdots(n+2r-2)}x^{2r}.$$ This means of course that, the integral I'm looking for equals $$\frac{\int_0^{\pi}\text{d}\theta\sin(\theta)^{n-2}\cos(\theta)^{2r}}{\int_0^\pi\text{d}\theta\sin(\theta)^{n-2}}=\frac{(2r)!}{2^rr!n(n+2)\cdots(n+2r-2)}$$ Can somebody help me understand this integral?

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$$\begin{align} \frac{\left(\frac{n-3}{2}\right)!\left(\frac{2r-1}{2}\right)!\left(\frac{n-2}{2}\right)!}{\left(\frac{n-3}{2}\right)!\left(-\frac{1}{2}\right)!\left(\frac{2r+n}{2}\right)!} &=\frac{\left(\frac{2r-1}{2}\right)!\left(\frac{n-2}{2}\right)!}{\left(-\frac{1}{2}\right)!\left(\frac{2r+n}{2}\right)!}\\ &=\frac{\left(r-\frac{1}{2}\right)!\left(\frac{n}{2}-1\right)!}{\left(-\frac{1}{2}\right)!\left(\frac{n}{2}+r\right)!}\\ &=\frac{\overbrace{(r-\frac12)(r-\frac32)\dots(\frac32)(\frac12)\left(-\frac{1}{2}\right)!}^{(r-\frac12)!}\left(\frac{n}{2}-1\right)!}{\left(-\frac{1}{2}\right)!\underbrace{(\frac{n}2+r)(\frac{n}2+r-1)\dots(\frac{n}2+1)(\frac{n}2)\left(\frac{n}{2}-1\right)!}}_{(\frac{n}2+r)!}\\ &=\frac{(r-\frac12)(r-\frac32)\dots(\frac32)(\frac12)}{(\frac{n}2+r)(\frac{n}2+r-1)\dots(\frac{n}2+1)(\frac{n}2)}\\ &=\frac{(2r-1)(2r-3)\dots(3)(1)}{(n+2r)(n+2r-2)\dots(n+2)(n)}\\ &=\frac{(2r)!}{(2r)(2r-2)\dots(4)(2)(n+2r)(n+2r-2)\dots(n+2)(n)}\\ &=\frac{(2r)!}{2^rr!(n+2r)(n+2r-2)\dots(n+2)(n)}\\ \end{align}$$ Which is the answer required, so you were not wrong.

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  • $\begingroup$ Thanks! I guess I just got scared by the non integer factorials $\endgroup$ – Iván Mauricio Burbano Apr 17 at 18:41

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