Finding the Limit of a Definite Integral in an Integral Problem

[jedishrfu]

don't forget to use the thanks button to thank those who helped.

 

[vela]

I haven't figured out P yet, though I know the likely answer from plotting f(r). One thing you can do is use the fact that $\sin x \le 1$ to find an upper bound on f(r). By doing that, you can eliminate one of the answers. Based on the answer to Q, I think you can eliminate another possibility as well.

 

[haruspex]

Thanks!

Can I please have a few hints for P?

We know P by process of elimination, yes? Are you just looking for some independent way of finding the value?
Edit: hmmm... I was reading the question as implying a 1-1 matching between the integrals and the answers. maybe that was wrong.
$r \left(\frac{2}{\pi}\right)^{r+1}f(r) = r \left(\frac{2}{\pi}\right)^{r+1}\int_0^{\frac{\pi}{2}} x^r \sin(x).dx$
$= r \left(\frac{2}{\pi}\right)^{r+1}\int_0^{\frac{\pi}{2}} (\frac{\pi}{2}-x)^r \cos(x).dx = r \frac{2}{\pi}\int_0^{\frac{\pi}{2}} (1-x\frac{2}{\pi})^r \cos(x).dx$
$= r \int_0^1 (1-x)^r \cos(x\frac{\pi}{2}).dx$
I'm running out of time to write this all out, but i think if you then integrate by parts the 'wrong' way you'll get an r/(r+1) term plus an integral with (1-x)^(r+1) sin(αx). If you break the range of integral at some small c > 0, you can take an upper bound for 1-x in one range and for sine in the other range such that both integrals can be shown to tend to 0 as r→∞.
But I was wrong last time.

 

[Saitama]

We know P by process of elimination, yes? Are you just looking for some independent way of finding the value?
Edit: hmmm... I was reading the question as implying a 1-1 matching between the integrals and the answers. maybe that was wrong.
$r \left(\frac{2}{\pi}\right)^{r+1}f(r) = r \left(\frac{2}{\pi}\right)^{r+1}\int_0^{\frac{\pi}{2}} x^r \sin(x).dx$
$= r \left(\frac{2}{\pi}\right)^{r+1}\int_0^{\frac{\pi}{2}} (\frac{\pi}{2}-x)^r \cos(x).dx = r \frac{2}{\pi}\int_0^{\frac{\pi}{2}} (1-x\frac{2}{\pi})^r \cos(x).dx$
$= r \int_0^1 (1-x)^r \cos(x\frac{\pi}{2}).dx$
I'm running out of time to write this all out, but i think if you then integrate by parts the 'wrong' way you'll get an r/(r+1) term plus an integral with (1-x)^(r+1) sin(αx). If you break the range of integral at some small c > 0, you can take an upper bound for 1-x in one range and for sine in the other range such that both integrals can be shown to tend to 0 as r→∞.
But I was wrong last time.

Integrating by parts,
$$r\left(\frac{1}{r+1}-\frac{\pi}{2(r+1)}\int_0^1 \sin\left(\frac{\pi}{2}x\right)\cdot(1-x)^{r+1} \, \,dx \right)$$
It is similar to S now, right?

 

[D H]

$r \left(\frac{2}{\pi}\right)^{r+1}f(r) = r \left(\frac{2}{\pi}\right)^{r+1}\int_0^{\frac{\pi}{2}} x^r \sin(x).dx$
$= r \left(\frac{2}{\pi}\right)^{r+1}\int_0^{\frac{\pi}{2}} (\frac{\pi}{2}-x)^r \cos(x).dx = r \frac{2}{\pi}\int_0^{\frac{\pi}{2}} (1-x\frac{2}{\pi})^r \cos(x).dx$
$= r \int_0^1 (1-x)^r \cos(x\frac{\pi}{2}).dx$

Too complicated!

There's no need for the change of variables here.

All that is needed is a relationship between $f(r) \equiv \int_0^{\pi/2} x^r \sin x\,dx$ and $g(r) \equiv \int_0^{\pi/2} x^r \cos x\,dx$. Integration by parts will give that relationship. It's best to choose u and v such that $uv\bigl|_0^{\pi/2} = 0$.

 

[Saitama]

Too complicated!

There's no need for the change of variables here.

All that is needed is a relationship between $f(r) \equiv \int_0^{\pi/2} x^r \sin x\,dx$ and $g(r) \equiv \int_0^{\pi/2} x^r \cos x\,dx$. Integration by parts will give that relationship. It's best to choose u and v such that $uv\bigl|_0^{\pi/2} = 0$.

Hi D H! :)

I used integration by parts and got the following relations:

$$f(r)=rg(r-1)$$
$$g(r)=\frac{f(r+1)}{r+1}$$

How should I use the above?

 

[D H]

Hi D H! :)

I used integration by parts and got the following relations:

$$f(r)=rg(r-1)$$
$$g(r)=\frac{f(r+1)}{r+1}$$

Very good.

How should I use the above?

For one thing, you can use it to show that Q and R are the same question. If you can solve one you can solve the other.

 

[Saitama]

For one thing, you can use it to show that Q and R are the same question. If you can solve one you can solve the other.

D H, I already solved Q and R, I still haven't been able to find the correct approach for P. Can you please look at my post #34? I used integration by parts as haruspex suggested and now I think the limit goes to zero for the new integral (as $r \rightarrow \infty$ and $0 < 1-x < 1$). Is that correct?

 

[D H]

Suppose you ignore the sin(x) term in the integral: $\int_0^{\pi/2} x^r dr$ . Since 0 < sin(x) < 1 for all x in (0,pi/2), this gives an upper bound on f(r). Substituting that sin(x) term with sin(1) leads to another simple integral. Show that this is a lower bound on f(r).

Now use these bounds to show that $f(r)=c_r \int_0^{\pi/2} x^r dr$ where c_r is some value between sin(1) and 1. What is the behavior of c_r as r→∞?