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.

I would like to get an asymptotic limit at the following integral: for $p\ge 2, n \in N$, $t \ge 0$ $$ \int_{0}^{\frac 12 \sqrt{(n+1)!}}\left(1-\frac{t^2}{2^2(n+1)!}\right)^p \mathrm{d} t $$ I think substitution $t=\frac 12 \sqrt{(n+1)!}y$ should work. But after the substittution, I don't know what to do.

Thank you for your help.

share|improve this question
    
$n$ is an integer? –  J. M. May 3 '12 at 16:30
    
yes, $n \in N$. –  Alex K. May 3 '12 at 16:35
add comment

1 Answer

up vote 5 down vote accepted

Performing the substitution $ t = \frac{y}{2} \sqrt{(n+1)!}$, the integral becomes: $$ \frac{\sqrt{(n+1)!}}{2} \int_0^1 \left( 1 - \frac{y^2}{16} \right)^p \mathrm{d} y = \frac{\sqrt{(n+1)!}}{2} \sum_{k=0}^\infty \frac{1}{2k+1} \binom{p}{k} \frac{(-1)^k}{16^k} $$ So as $n$ grows, so does the magnitude of the integral.

share|improve this answer
add comment

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.