With $\alpha>0$, showing that $\int_0^1 t^\alpha(1-t^2)^kdt$ goes to $0$ as $k\to+\infty$ My idea was to compute the derivative of the integrand, in order to find its maximum in $[0,1]$. It turns out it is attained at $t=\sqrt{\frac\alpha{\alpha+2k}}$. Then the claim follows from $$0<\int_0^1 t^\alpha(1-t^2)^kdt\le\left(\frac\alpha{\alpha+2k}\right)^{\alpha/2}\left(1-\frac\alpha{\alpha+2k}\right)^k\to0.$$
Is this correct? How else could one do it?
 A: Your method seems fine to me.
Another way is as follow :
$$\forall t \in  [0,1], 0 \le t^\alpha(1-t^2)^k \le 1$$
But $1 \in L^1([0,1])$. So by dominated convergence theorem :
$$\lim_{k \to \infty} \int_0^1 t^\alpha(1-t^2)^kdt=\int_0^1 \lim_{k \to \infty} t^\alpha(1-t^2)^kdt=\int_0^10~dt=0$$
A: I suppose your $\alpha $ is positive. Your proof is correct. If you know measure theory you can  get it immediately from DCT: the integrand tends to $0$ because $1-t^{2}<1$ and the integrand is dominated by $t^{\alpha}$ which is integrable. 
A: Alternative method.
Solution. $\blacktriangleleft$ Consider the upper limit. Compute as follows.
\begin{align*}
\varlimsup_k \left|\int_0^1 t^\alpha (1 - t^2)^k\mathrm d t\right| &\leqslant \varlimsup_k \left(\int_0^c + \int_c^1\right)t^\alpha (1 - t^2)^k \mathrm dt \quad [\text {the function} \geqslant 0, c\in (0,1)] \\
&\leqslant \varlimsup_k \int_0^c (1-t^2)^k\mathrm dt + \varlimsup_k \int_c^1 (1- t^2)^k \mathrm d t \quad [t^\alpha \leqslant1]\\
&\leqslant \int_0^c \mathrm dt + \varlimsup_k \int_c^1 (1-c^2)^k \mathrm dt \quad [1 \geqslant 1 -t^2 \searrow 0]\\
&=c + \varlimsup_k(1-c)(1 -c^2)^k\\
&= c + 0 \quad[1-c^2<1, (1-c^2)^k \to 0 \text{ as }k \to \infty]\\
&= c \to 0 \quad [c \to 0^+]. \blacktriangleright
\end{align*}
A: Another way via Beta and Gamma  functions $$\int_0^1 t^\alpha(1-t^2)^kdt= \int_0^1 t^{\alpha/2+1/2-1}(1-t)^{k+1-1}dt \\=B(\alpha/2+1/2,k+1)=\frac{\Gamma(\alpha/2+1/2)\Gamma(k+1)}{\Gamma(\alpha/2+k+3/2)}\to 0$$
