Comparing fractions with definite integrals

[Saitama]

Hello! I found the following problem on AOPS:

Which is larger,

$$\Large \frac{\int_{0}^{\frac{\pi}{2}}x^{2014}\sin^{2014}x\ dx}{\int_{0}^{\frac{\pi}{2}}x^{2013}\sin^{2013}x\ dx}\ \text{or}\ \frac{\int_{0}^{\frac{\pi}{2}}x^{2011}\sin^{2011}x\ dx}{\int_{0}^{\frac{\pi}{2}}x^{2012}\sin^{2012}x\ dx}\ ?$$

I don't really have much idea about the problem. I thought of defining:
$$I_n=\int_0^{\pi/2} x^n\sin^n x\,dx$$
From integration by parts, I got:
$$I_n=\left(\sin^n x \frac{x^{n+1}}{n+1}\right|_0^{\pi/2}-\int_0^{\pi/2} n\sin^{n-1} x \cos x \frac{x^{n+1}}{n+1}\,dx=-\frac{n}{n+1}\int_0^{\pi/2} x^{n+1}\sin^{n+1}x \frac{\cos x}{\sin^2 x}\,dx$$
but this doesn't help in obtaining a relation between $I_n$ and $I_{n-1}$. I am stuck here.

Any help is appreciated. Thanks!

 

[alyafey22]

I don't think that solving the integral will be helpful but I guess you should use approximations. Use the following two inequalities which hold correct on $[0,\pi /2]$

$$\sin(x)\leq x $$

$$\sin(x)\geq \frac{2}{\pi } x$$

 

[Saitama]

I don't think that solving the integral will be helpful but I guess you should use approximations. Use the following two inequalities which hold correct on $[0,\pi /2]$

$$\sin(x)\leq x $$

$$\sin(x)\geq \frac{2}{\pi } x$$

Using the inequalities:
$$\frac{2}{\pi}\int_0^{\pi/2} x^{2n}\,dx\leq I_n \leq \int_0^{\pi/2}x^{2n}\,dx$$
$$\Rightarrow \frac{1}{2n+1}\left(\frac{\pi}{2}\right)^{2n} \leq I_n \leq \frac{1}{2n+1}\left(\frac{\pi}{2}\right)^{2n+1}\,\,\,(*)$$
From above, we can write the bounds for $I_{n+1}$ i.e
$$\frac{1}{2n+3}\left(\frac{\pi}{2}\right)^{2n+2} \leq I_{n+1} \leq \frac{1}{2n+3}\left(\frac{\pi}{2}\right)^{2n+3}$$
$$\Rightarrow (2n+3)\left(\frac{2}{\pi}\right)^{2n+3}\leq \frac{1}{I_{n+1}} \leq (2n+3)\left(\frac{2}{\pi}\right)^{2n+2}\,\,\,\,(**)$$
Using (*) and (**),
$$\frac{2n+3}{2n+1}\left(\frac{2}{\pi}\right)^3 \leq \frac{I_n}{I_{n+1}} \leq \frac{2n+3}{2n+1}\left(\frac{2}{\pi}\right)$$
For $n=2011$,
$$\frac{4025}{4023}\left(\frac{2}{\pi}\right)^3 \leq \frac{I_{2011}}{I_{2012}}\leq \frac{4025}{4023} \left(\frac{2}{\pi}\right) \Rightarrow 0.2581 \leq \frac{I_{2011}}{I_{2012}} \leq 0.6369$$
For $n=2013$
$$\frac{4027}{4029}\left(\frac{\pi}{2}\right) \leq \frac{I_{2014}}{I_{2013}}\leq \frac{4027}{4029} \left(\frac{\pi}{2}\right)^3 \Rightarrow 1.57 \leq \frac{I_{2014}}{I_{2014}}\leq 3.87$$

This shows that the first ratio is greater. Thanks a lot ZaidAlyafey! :)

But I had to use a calculator to find those values, is their a way to do it without using a calculator?

 

[alyafey22]

You are using that

$$\sin^n(x)\geq \frac{2}{\pi } x^n$$

But I don't think that this inequality holds because

$$\left| \frac{2}{\pi } x^n \right| $$

is not bounded by 1.

 

[Saitama]

You are using that

$$\sin^n(x)\geq \frac{2}{\pi } x^n$$

But I don't think that this inequality holds because

$$\left| \frac{2}{\pi } x^n \right| $$

is not bounded by 1.

Do I have to use $\sin^n x \geq \left(\frac{2}{\pi}x\right)^n$? If so, the bounds would be very dirty. (Sweating)

 

[alyafey22]

Do I have to use $\sin^n x \geq \left(\frac{2}{\pi}x\right)^n$? If so, the bounds would be very dirty. (Sweating)

I don't think it is the best approach. The problem might be easier than I thought. Compare $I_n$ and $I_{n+1}$ and use that $x\sin x$ is increasing and positive on the interval.

 

[alyafey22]

Note that

Spoiler

$$\frac{I_n}{I_{n+1}}\leq 1 \leq \frac{I_{n+1}}{I_{n}} $$

 

[Saitama]

Note that

Spoiler

$$\frac{I_n}{I_{n+1}}\leq 1 \leq \frac{I_{n+1}}{I_{n}} $$

Sorry for the late reply.

I just noticed that what I wrote in the first post is incorrect. It should be

$$(n+1)I_n=\left(\frac{\pi}{2}\right)^{n+1}-n\int_0^{\pi/2} x^{n+1}\sin^{n+1}x \frac{\cos x}{\sin^2 x}\,dx$$
But I still do not see how you got those bounds.