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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.

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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.

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