$I_n = \int_{0}^{\frac{\pi}{2}}(\cos t)^n \ dt$ converges to 0?

Question

How one can prove that the sequence $\left ( I_n \right )$ defined as $$ I_n = \int_{0}^{\frac{\pi}{2}}(\cos t)^n \ dt, $$ $n \in \{ 0,1,2,...\}$ converges to $0$?

Is easy to show, by the way, that the sequence is decreasing because, for $t \in (0, \pi/2)$, $$(\cos t)^{n+1}<(\cos t)^n \Rightarrow I_{n+1} < I_{n}, \ \forall n $$

Answer 1

HINT: $\forall\varepsilon>0 \exists\delta>0: \cos x<1-\varepsilon$ on $(\delta,\pi/2]$ and $\delta$ decreases to $0$ when $\varepsilon$ tends to $0$.

Answer 2

Your observation in the comments also makes Dominated Convergence work.

It remains to observe that on $(0, \frac \pi 2)$, the sequence converges pointwise to $0$. So pointwise a.e. convergence + dominated convergence implies the result.

Answer 3

BTW the integrals can be done explicitly: $$ \int_0^{\pi/2} (\cos t)^n \; dt = \dfrac{\sqrt{\pi}\; \Gamma((n+1)/2)}{2 \;\Gamma(1+n/2)} \sim \sqrt{\frac{\pi}{2n}} \ \text{as}\ n \to \infty$$

Answer 4

With integation by parts one may show that $$I_{n}=\dfrac{n-1}{n}I_{n-2}$$ then from $I_0=\dfrac{\pi}{2}$ and $I_1=1$, both odd and even terms go $0$ as $n\to\infty$

Answer 5

Suppose $a\in \Big(0,\dfrac{\pi}{2}\Big]$, let $f_{n}(t)=\cos(t)^{n}$ for all $t\in \Big[a,\dfrac{\pi}{2}\Big]$. If $\{f_{n}\}_{n\in\mathbb{N}}$ converges uniformly to the zero function, we can assure that

$$\lim_{n\to\infty}\int_{a}^{\dfrac{\pi}{2}}f_{n}=\int_{a}^{\dfrac{\pi}{2}}\lim_{n\to\infty}f_{n}=\int_{a}^\dfrac{\pi}{2}0dt=0$$

And the function does converge uniformly to $f(t)=0$, because we have that $$ |f_{n}(t)|=|\cos(t)^{n}|\leq \cos(a)^{n} $$ for all $t$, since $\cos$ is a positive decreasing function in $\Big[a,\dfrac{\pi}{2}\Big]$. Therefore, due to the fact that $\cos(a)<1$, we have that $\lim \cos(a)^{n}=0$, which then implies uniform convergence to $f(t)=0$ in $\Big[a,\dfrac{\pi}{2}\Big]$, so the equality showed earlier is true for every choice of $a$.

This alone doesn't imply that the limit we're dealing with is $0$, but now it will be much easier to prove that it is.

Let $\varepsilon>0$, and WLOG I'll assume that $\varepsilon < \pi$. If we choose $a$ such that $0<a<\dfrac{\varepsilon}{2}$, and $N\in\mathbb{N}$ which verifies $$n\geq N\implies \Bigg|\int_{a}^{\dfrac{\pi}{2}}\cos(t)^{n}dt\Bigg|=\int_{a}^{\dfrac{\pi}{2}}\cos(t)^{n}dt<\dfrac{\varepsilon}{2}$$ we have that, given $n\in\mathbb{N}, n\geq N$,

$$\Bigg|\int_{0}^{\dfrac{\pi}{2}}\cos(t)^{n}dt\Bigg|=\int_{0}^{\dfrac{\pi}{2}}\cos(t)^{n}dt=\int_{0}^{a}\cos(t)^{n}dt+\int_{a}^{\dfrac{\pi}{2}}\cos(t)^{n}dt\leq \int_{0}^{a}1dt+\int_{a}^{\dfrac{\pi}{2}}\cos(t)^{n}dt=a+\int_{a}^{\dfrac{\pi}{2}}\cos(t)^{n}dt<\dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon$$

So we can conclude that $$\lim_{n\to\infty}\int_{0}^{\dfrac{\pi}{2}}\cos(t)^{n}dt=0$$

Answer 6

Let $I_n=\int_0^{\pi/2}(\cos x)^ndx.$

For $x\in [0,\pi/2]$ we have $0\leq (\cos x)^{n+1}\leq (\cos x)^n\leq 1.$ So $0\leq I_{n+1}\leq I_n\leq \pi/2.$.... So $$L=\lim_{n\to \infty}I_n$$ exists and is not negative.

For $k\geq 1$ we have $$I_n=\int_0^{\pi/2k}(\cos x)^ndx +\int_{\pi/2k}^{\pi/2}(\cos x)^ndx\leq$$ $$\leq \int_0^{\pi/2k}(1)^ndx+\int_{\pi/2k}^{\pi/2}(\cos \pi/2k)^ndx=$$ $$=\pi/2k+(\pi/2-\pi/2k)(\cos \pi/2k)^n.$$ In this last expression, keeping $k$ fixed and letting $n \to \infty,$ we have $(\cos \pi/2k)^n\to 0.$ Therefore $$L\leq \pi/2k +(\pi/2-\pi/2k)(0)=\pi /2k.$$ That is, $$\forall k\geq 1\;(0\leq L\leq \pi/2k).$$ So $L=0.$