It can be solved by using First mean value theorem for definite integrals, but I have no idea about the exact answer, only have a answer range. There is a similar question: $\lim _{n\to \infty }\left(\int _0^1x^n\sqrt{1+x^2}\right)$ By using First mean value theorem for definite integrals, it turns into: $\lim _{n\to \infty }\sqrt{1+ξ^2}\int _0^1\:x^ndx$ Because $\int _0^1\:x^ndx=\frac{1}{n+1}$→0, $1<\sqrt{1+ξ^2}<\sqrt{2}$ so the original formula=0.
Asked
Active
Viewed 40 times
0
-
1First, change $x \mapsto 1 - x$ and then Laplace’s Method. – Felix Marin May 24 '24 at 14:16
-
This is a special case of https://math.stackexchange.com/q/1563666/42969. See also https://math.stackexchange.com/q/1303672/42969 and the linked questions. – Martin R May 24 '24 at 14:23