Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does anyone know how to approximate this infinite integral $$ H(\beta, i, l) = \iint_{1 \leq x \leq y < \infty} (1-x^{-\beta})^{i-1} (x^{-\beta} - y^{-\beta})^{2(l-i)} x^{-\beta} y^{-\beta i} dxdy, $$ where $l, i$ are both positive integers with $ i<l$ and $\beta > 2$?

I tried to reduce this integral to Gamma functions but failed. I also tried Mathematica but still got nothing. So I guess I should alternatively investigate it by figuring out a reasonable approximation. Even just knowing the leading orders of $\beta, i, l$ in the function $H$ would be helpful.

share|improve this question

1 Answer 1

up vote 0 down vote accepted

I think you can calculate as following, it can be reduced to Gamma functions and can represented into elementary expression if you want.

$$\begin{align*} H(\beta,i,l) &=\iint_{1\leq x\leq y<\infty}(1-x^{-\beta})^{i-1}(x^{-\beta}-y^{-\beta})^{2(l-i)}x^{-\beta} y^{-\beta i}dxdy\\ &=\int_{1}^{\infty}(1-x^{-\beta})^{i-1}x^{-\beta}dx \int_{x}^{\infty}(x^{-\beta}-y^{-\beta})^{2(l-i)}y^{-\beta i}dy\\ (\text{set}\,\,y=xt) &=\int_{1}^{\infty}(1-x^{-\beta})^{i-1}x^{-\beta}dx \int_{1}^{\infty}x^{1-\beta(2l-i)}(1-t^{-\beta})^{2(l-i)}t^{-\beta i}dt\\ &=\int_{1}^{\infty}(1-x^{-\beta})^{i-1}x^{1-\beta(2l-i+1)}dx \int_{1}^{\infty}(1-t^{-\beta})^{2(l-i)}t^{-\beta i}dt\\ (\text{set}\,\,x=u^{-\frac{1}{\beta}},t=v^{-\frac{1}{\beta}}) &=\frac{1}{\beta^{2}}\int_{0}^{1}(1-u)^{i-1}u^{2l-i-\frac{2}{\beta}}dx \int_{0}^{1}(1-v)^{2l-2i}v^{i-1-\frac{1}{\beta}}dt\\ &=\frac{1}{\beta^{2}}B(i,2l-i+1-\frac{2}{\beta})B(2l-2i+1,i-\frac{1}{\beta})\\ &=\frac{1}{\beta^{2}} \frac{\Gamma(i)\Gamma(2l-i+1-\frac{2}{\beta})}{\Gamma(2l+1-\frac{2}{\beta})}\frac{\Gamma(2l-2i+1)\Gamma(i-\frac{1}{\beta})}{\Gamma(2l-i+1-\frac{1}{\beta})}\\ &=\frac{1}{\beta^{2}}(i-1)!(2l-2i)!\\ &\cdot\frac{1}{(2l-\frac{2}{\beta})(2l-\frac{2}{\beta}-1)\cdots(2l-\frac{2}{\beta}-(i-1))}\\ &\cdot\frac{1}{((2l-2i)+(i-\frac{1}{\beta}))((2l-2i-1)+(i-\frac{1}{\beta}))\cdots(i-\frac{1}{\beta})} \end{align*}$$

share|improve this answer
    
Great, many thanks! –  RichardKwo Oct 9 '12 at 1:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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