Lower bound of integral and approximate identity Problem: Let $\phi_n(t)= \frac{(1-t^2)^n}{c_n} \chi_{[-1,1]}(t)$  and $c_n= \int_{-1}^1 (1-t^2)^n \,\mathrm dt$. Show that $c_n \geq \frac{2}{n+1}$ and that $\{\phi_n\}_{n=0}^{\infty} $ is an approximate identity on $\mathbb{R}$ when $n \rightarrow \infty$.
For the first part of problem (evaluating integral), I tried to use binomial expansion. Is that correct approach? I get $c_n = 2((-1)^n + n (-1)^{n-1} +...+ n(-1))$ and I am not sure if this is correct. I would be thankful for any help.
 A: Partial answer: for $t\in(-1,1)$ we have that $t>t^2$ and therefore $(1-t^2)^n>(1-t)^n$.  We thus have that $$c_n\equiv\int_{-1}^1(1-t^2)^ndt>\int_{-1}^1(1-t)^ndt = \frac{2^{n+1}}{n+1}>\frac{2}{n+1},\quad n\ge0.$$
A: *

*The idea of WAH is good but the computation is false. If $t<0$, then surely $t > t^2$ is false ... it should be $t^2 < |t|$. Then one gets
$$
c_n > \int_{-1}^1 (1-|t|)^n\,\mathrm d t = 2 \int_0^1 (1-t)^n\,\mathrm d t = 2 \int_0^1 t^n\,\mathrm d t = \frac{2}{n+1}.
$$


*You can compute the integral using the Binomial formula, but indeed you will get an alternating sum. By symmetry
$$
c_n = 2 \int_0^1 (1-t^2)^n\,\mathrm d t = 2 \sum_{k=0}^n \binom{n}{k} (-1)^k \int_0^1 t^{2k}\,\mathrm d t = 2 \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{2k+1}
$$
but now the alternating sum is difficult to estimate below.

For more advanced readers: This integral can indeed be computed in terms of the Gamma function, and more precisely the Beta function, as
$$
c_n = 2\int_0^1 (1-t^2)^n\,\mathrm d t = \int_0^1 s^{-1/2} (1-s)^n\,\mathrm d s = B(\tfrac{1}{2},n+1) = \frac{\sqrt{\pi}\,n!}{\Gamma(n+3/2)}.
$$
From this and the Stirling formula, it is immediate that $c_n \sim \sqrt{\pi/n}$ for large $n$. By the duplication formula for the Gamma function and the fact that $2^{2n} (n!)^2 = 2\cdot 2\cdot 4\cdot 4\cdot\dots (2n)\cdot(2n)$,
$$
c_n = \frac{2^{2n+1}\,(n!)^2}{(2n+1)!} = 2 \prod_{k=1}^{2n} \frac{k+\mathbf{1}_{k \text{ is odd}}}{k+1} = 2 \prod_{k=1}^{n} \frac{2k}{2k+1}
$$
Therefore
$$
\ln(2/c_n) = \sum_{k=1}^n \ln(1+\tfrac{1}{2k}) \leq \tfrac{1}{2}\sum_{k=1}^n \tfrac{1}{k}
$$
Comparing the sum with an integral gives
$$
\ln(2/c_n) \leq \tfrac{1}{2}\int_{1/2}^{n+1/2} \frac{\mathrm dt}{t} = \tfrac{1}{2} \ln(2n+1)
$$
leading to
$$
c_n \geq \frac{2}{\sqrt{2n+1}}
$$
which implies that $c_n \geq \frac{2}{n+1}$ since $\sqrt{2n+1} \leq \sqrt{n^2+2n+1} = n+1$.
