Does the limit of this ratio of integrals exist?

Question

Does this limit exist?

$\lim \limits_{n\to \infty} \dfrac{\int^{\pi/2} \limits_{0}\cos ^{n+1}(x)\,\,dx}{\int^{\pi/2} \limits_{0}\cos ^{n}(x)\,\,dx}$

Answer 1

Here is a more general point of view ...

Consider a continuous and positive function $f:[a,b]\to\mathbb{R}$, and for all $n\in\mathbb{N}$ :

$$u_n=\int_a^bf(t)^n\,dt$$

It is well known that, if $M$ denotes the maximum of $f$ on $[a,b]$ :

$$\lim_{n\to\infty}(u_n)^{1/n}=M$$

It is maybe less known that :

$$\lim_{n\to\infty}\frac{u_{n+1}}{u_n}=M$$

Indeed, we observe via Cauchy-Schwarz inequality that the sequence $\displaystyle{\left(\frac{u_{n+1}}{u_n}\right)}$ is increasing :

$$\left(\int_a^bf(t)^n\,dt\right)^2=\left(\int_a^bf(t)^{(n-1)/2}f(t)^{(n+1)/2}\,dt\right)^2\le\left(\int_a^bf(t)^{n-1}\,dt\right)\left(\int_a^bf(t)^{n+1}\,dt\right)$$

Now, we conclude using a partial converse of Cesaro's lemma (see below).

Back to the OP :

$$\lim_{n\to\infty}\frac{\int_0^{\pi/2}\cos^{n+1}(t)\,dt}{\int_0^{\pi/2}\cos^n(t)\,dt}=\sup\left\{\cos(t);\,0\le t\le \frac\pi2\right\}=1$$

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Partial converse of Cesaro's lemma

If $(u_n)$ is a monotonic sequence of real numbers such that $\displaystyle{\frac 1n\sum_{k=0}^{n-1}u_k}=L\in\mathbb{R}\cup\{-\infty,+\infty\}$

then we also have $\displaystyle{\lim_{n\to\infty}u_n=L}$

Answer 2

One can write:

$$\int^{\pi/2}_{0}\cos ^{n}(\theta)d \theta=B(\frac12,\frac{n+1}{2}).$$

using Beta integral defined by

$$B(x,y)=\int_0^{\pi/2}(\sin \theta)^{2x-1} (\cos \theta)^{2y-1}d\theta$$

Thus, using formula $B(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$:

$$\tag{1}R_n:=\dfrac{\int^{\pi/2} \limits_{0}\cos ^{n}(x)\,\,dx}{\int^{\pi/2} \limits_{0}\cos ^{n-1}(x)\,\,dx}=\dfrac{B(\frac12,\frac{n+1}{2})}{B(\frac12,\frac{n}{2})}=\dfrac{\Gamma(\frac{n+1}{2})^2}{\Gamma(\frac{n}{2})\Gamma(\frac{n+2}{2})}=\dfrac{\Gamma(m+\dfrac12)^2}{\Gamma(m)\Gamma(m+1)}$$

with $m:=\frac{n}{2}$.

Using Gautschi's inequality (How do you prove Gautschi's inequality for the gamma function?) (as advised by Jack d'Aurizio) with $s=\dfrac12$, we have:

$$0<\sqrt{m}<\dfrac{\Gamma(m+1)}{\Gamma(m+\dfrac12)}<\sqrt{m+1}$$

Inverting and squaring these inequalities:

$$\tag{2}\dfrac{1}{m+1}<\dfrac{\Gamma(m+\dfrac12)^2}{\Gamma(m+1)\Gamma(m+1)}<\dfrac{1}{m}$$

Taking into account functional relationship: $\Gamma(m+1)=m\Gamma(m)$, (2) yields:

$$\tag{3}\dfrac{m}{m+1}<\dfrac{\Gamma(m+\dfrac12)^2}{\Gamma(m)\Gamma(m+1)}<1$$

proving that, when $m \to \infty$, $R_n \to 1.$

Remark: numerical tests show that the first inequality in (3) is not tight.

Answer 3

By convergence in distribution, if $f(x)$ is a continuous and bounded function on $(0,1)$, $$ \lim_{n\to +\infty}\frac{\int_{0}^{\pi/2} f(x)\cos(x)^n\,dx}{\int_{0}^{\pi/2} \cos(x)^n\,dx} = \lim_{x\to 0^+} f(x).$$ It follows that your limit is simply $\color{red}{1}$.

An alternative is to compute such integrals through Euler's Beta function, then exploit Gautschi's inequality and squeezing.

Answer 4

Ok here's my method:

Let $I_{n} = \int_{0}^{\frac{\pi}{2}} cos^{n}x dx$

Let $u = cos^{n-1}x \Rightarrow \frac{du}{dx} = -(n-1)cos^{n-2}xsinx$

Then $v = \int cosxdx = sinx$

We have $I_{n} = uv - \int v\frac{du}{dx}dx$

$\Rightarrow I_{n} = \left [ sinxcos^{n-1}x \right ]_{0}^{\pi/2} + (n-1)\int_{0}^{\pi/2} cos^{n-2}xsin^{2}xdx$

$I_{n} = (n-1)(I_{n-2} - I_{n})$ , $n\geq 2$

$I_{n} = \frac{n-1}{n} I_{n-2}$

$I_{0} = \frac{\pi}{2}$

$I_{1} = 1$

$I_{2} = \frac{2-1}{2}*\frac{\pi}{2}$

$I_{3} = \frac{3-1}{3} * 1$

$I_{4} = \frac{4-1}{4} * \frac{2-1}{2} *\frac{\pi}{2}$

$I_{5} = \frac{5-1}{5}*\frac{3-1}{3}$

Now you should see a pattern emerging, and then by using the formula for product of the first $n$ even numbers and the formula for the product of first $n$ odd numbers we get that for $n \geq 1$:

$I_{2n} = \frac{(2^{n})!}{(2^{n}n!)^{2}}*\frac{\pi}{2} $

$I_{2n+1} = \frac{2(n+1)(2^{n}n!)^{2}}{2^{n+1}!}$

Now I'm not sure how to compute the limit.