Finding the Limit of a Definite Integral in an Integral Problem

[Saitama]

Homework Statement

Let $\displaystyle f(r)=\int_0^{\pi/2} x^r\sin x \,\, dx$. Now match the following List-I with List-II.

$$
\begin{array} {|c| c | l c|}
\hline
& \text{List-I} & & \text{List-II} & & \\
\hline
\text{(P)} & \lim_{r\rightarrow \infty} r\left(\frac{2}{\pi}\right)^{r+1}f(r) & 1\cdot & 0 & & \\
\\
\\
\\
\text{(Q)} & \lim_{r\rightarrow \infty} \frac{f(r)}{f(r+1)} & 2\cdot & 1 & &\\
\\
\\
\\
\text{(R)} & \lim_{r\rightarrow \infty} \left(\frac{f(r)}{r\int_0^{\pi/2} x^r\cos x \,\, dx}\right) & 3\cdot & \frac{2}{\pi} & & \\
\\
\\
\\
\text{(S)} & \lim_{r\rightarrow \infty} \int_0^1 x^r\sin x \,\, dx & 4\cdot & \frac{\pi}{2} & &\\
\hline
\end{array}
$$

Homework Equations

The Attempt at a Solution

I haven't been able to make any useful attempt on this problem. I tried integrating by parts and reached the following:

$$f(r)=r\left(\frac{\pi}{2}\right)^{r-1}-r(r-1)f(r-2)$$
I am not sure if above is of any help.

Any help is appreciated. Thanks!

 

[jedishrfu]

In your f(r) function integral aren't you missing argument for the sin function?

 

[Saitama]

In your f(r) function integral aren't you missing argument for the sin function?

Woops, edited! Thanks. :)

 

[jedishrfu]

It seems that doing an integration may be too time consuming.

Have you looked at the questions and the potential answers.

For example, the last one where r goes to infinity with x ranging of 0 to 1, how would that curve look if you drew it for r=1 the r=2...?

then which answer would most likely be its answer?

 

[Saitama]

For example, the last one where r goes to infinity with x ranging of 0 to 1, how would that curve look if you drew it for r=1 the r=2...?

Can we please look for a proper method?

I usually sketch the curve when everything else fails. I doubt that there is a need to sketch the curve in this case.

 

[jedishrfu]

Can we please look for a proper method?

I usually sketch the curve when everything else fails. I doubt that there is a need to sketch the curve in this case.

Okay, this looks like a speed exam problem where you may need to improvise. Sometimes they will give you a really tough integral to put you off track wasting time trying to evaluate it. I agree that if you had the integral evaluated the answers would probably pop right out.

Okay so when I integrate by parts I define u=x^r and v'=sinx so that v=-cosx

and get integral f(r) = [-x^r*cos(x) + x^(r+1)*cos(x)*1/(r+1) ] 0 to pi/2

maybe by leaving it in that form the limits will be easier to evaluate.

 

[Saitama]

Okay, this looks like a speed exam problem where you may need to improvise. Sometimes they will give you a really tough integral to put you off track wasting time trying to evaluate it. I agree that if you had the integral evaluated the answers would probably pop right out.

Okay so when I integrate by parts I define u=x^r and v'=sinx so that v=-cosx

and get integral f(r) = [-x^r*cos(x) + x^(r+1)*cos(x)*1/(r+1) ] 0 to pi/2

maybe by leaving it in that form the limits will be easier to evaluate.

Yes, its an exam problem.

Before I proceed with your approach, I was wondering if you could comment on my thoughts about this problem.

In the first post, I found f(r) to be:
$$f(r)=r\left(\frac{\pi}{2}\right)^{r-1}-r(r-1)f(r-2)$$
I am thinking that if $r\rightarrow \infty$, I can take $f(r) \approx f(r-2)$, would this be valid?

 

[jedishrfu]

It doesn't seem right.

Wouldn't the pi/2 term be bigger than f(r-2) term?

You have an r factor that can be removed to make your relation more like:

f(r) ~ r*f(r-2)

How did you get your relation of f(r) = f(r-2) ?

We need a mathematician here...

 

[Saitama]

It doesn't seem right.

Wouldn't the pi/2 term be bigger than f(r-2) term?

You have an r factor that can be removed to make your relation more like:

f(r) ~ r*f(r-2)

How did you get your relation of f(r) = f(r-2) ?

We need a mathematician here...

My logic was that a function should have almost the same values near infinity. We have $f(\infty)$ and $f(\infty-2)$, both should have almost the same value as $r\rightarrow \infty$.

 

[jedishrfu]

Maybe you can send a PM to Mark44 or HallsofIvy who more versed in mathematics to look at your thread.

If there is a method to solving this kind of problem they would know it.

I am more of an amateur trying to relearn stuff I learned decades ago.

 

[Mark44]

Can we please look for a proper method?

I don't see anything wrong with sketching a curve.

I usually sketch the curve when everything else fails.

Sketching a graph is usually the first thing I do, not something that I try as a last resort. Getting a feel for the geometry of a problem brings in a whole different area of your brain than when you limit yourself to the area that works in symbols. In effect, you are tying one hand behind your back.

I doubt that there is a need to sketch the curve in this case.

I've just started looking at this problem, so I don't have much insight into it. However, the "Q" item in the first column is looking at the ratio of f(r) to f(r+1). I think you can rule out 0 as a potential answer.

As far as the integral itself, integration by parts is definitely something that I would try - maybe twice.

 

[vela]

In the first post, I found f(r) to be:
$$f(r)=r\left(\frac{\pi}{2}\right)^{r-1}-r(r-1)f(r-2)$$
I am thinking that if $r\rightarrow \infty$, I can take $f(r) \approx f(r-2)$, would this be valid?

This would probably be valid if f(r) converged to a finite value, but it doesn't appear to.

I'd go with the integral in S being equal to 0. Just sketch $x^r$ for increasing powers of $r$ on the interval [0,1] to see this.

 

[vela]

Since you know the limit in P is finite, you can deduce how f(r) behaves for large r. From that, you can get the likely answer for Q is #3.

 

[haruspex]

S is easy. Just use 0 <= sin(x) <= sin(1) < 1 and 0 <= x <= 1.
By elimination, that tells you P converges to a nonzero value. As vela suggests, it's then easy to find Q - just let c > 0 be the answer to P. That just leaves two choices for P and R. maybe it's easy to see which is larger?

Fwiw, I think I solved the recurrence relation for f. Something like $\frac 1{2i}s^{-r}((1+is)^r - (1-is)^{r})$, where $s = \frac 2 \pi$

 

[Saitama]

S is easy. Just use 0 <= sin(x) <= sin(1) < 1 and 0 <= x <= 1.

Oh yes, that was easy. Thanks! :)

As vela suggests, it's then easy to find Q - just let c > 0 be the answer to P.

What is c?

That just leaves two choices for P and R. maybe it's easy to see which is larger?

How?

Fwiw, I think I solved the recurrence relation for f. Something like $\frac 1{2i}s^{-r}((1+is)^r - (1-is)^{r})$, where $s = \frac 2 \pi$

Can you please post the steps to solve the recurrence relation? It's not going to help me at the moment but I will someday return to this thread when I am done with recurrence relations. Thanks!

 

[pasmith]

S is easy. Just use 0 <= sin(x) <= sin(1) < 1 and 0 <= x <= 1.
By elimination, that tells you P converges to a nonzero value. As vela suggests, it's then easy to find Q - just let c > 0 be the answer to P. That just leaves two choices for P and R. maybe it's easy to see which is larger?

Fwiw, I think I solved the recurrence relation for f. Something like $\frac 1{2i}s^{-r}((1+is)^r - (1-is)^{r})$, where $s = \frac 2 \pi$

You need $f(0) = \int_0^{\pi/2} \sin x\,dx = \cos 0 - \cos \frac\pi 2 = 1$. Your formula gives $f(0) = 0$.

I agree that $f$ is the imaginary part of
$$I(r) = \int_0^{\pi/2} x^r e^{ix}\,dx = \left( \frac \pi 2 \right)^r + irI(r-1)$$
subject to $I(0) = 1 + i$ but I don't think you've solved the recurrence correctly.

 

[vela]

Is there perhaps a typo in one of the limits? I find both Q and R converge to the same value.

 

[Saitama]

Is there perhaps a typo in one of the limits? I find both Q and R converge to the same value.

No there is no typo but according to the answer key, both Q and R share the same answer.

Any hints on how you determine that?

 

[haruspex]

What is c?

Any of answers 2, 3, 4 - it doesn't matter. Since S=0, P cannot be 0. So we know P converges to some constant c, 0 < c < ∞, giving us an expression for the asymptotic behaviour of f. Substitute that in Q.

How?

That was just a possibility, but I see another way.
Go back and look at your integration by parts. You did two steps of that to get your recurrence relation, right? The first step gave you something very close to the denominator in R. Can you see whether that makes R more or less than 1?

Can you please post the steps to solve the recurrence relation? It's not going to help me at the moment but I will someday return to this thread when I am done with recurrence relations. Thanks!

OK, but not right now. I did say 'something like'; I knew I had not started it off correctly at f(0), as pasmith notes.

 

[Ray Vickson]

S is easy. Just use 0 <= sin(x) <= sin(1) < 1 and 0 <= x <= 1.
By elimination, that tells you P converges to a nonzero value. As vela suggests, it's then easy to find Q - just let c > 0 be the answer to P. That just leaves two choices for P and R. maybe it's easy to see which is larger?

Fwiw, I think I solved the recurrence relation for f. Something like $\frac 1{2i}s^{-r}((1+is)^r - (1-is)^{r})$, where $s = \frac 2 \pi$

I don't think it is this simple. For integer r the recursions for even r and for odd r separate. For even r, the recursion for $g(m) = f(2m)$ is
$$g(m) = \frac{2}{c} p^m - 2m (2m-1) g(m-1), \; g(0) = 1,$$
where $c = \pi/2$ and $p = c^2 = \pi^2/4$. Maple gets the solution
$$g(m) = \frac{(-1)^m 4^m}{\sqrt{\pi}} m! (m \, - \, 1/2)! \left( 1 - \frac{\sqrt{\pi}}{2c} \sum_{j=0}^{m-1} (-1)^j \frac{(j+1)p^{j+1}}{4^j (j+1)! (j \,+\,1/2)!} \right)$$
where $u! \equiv \Gamma(u+1)$ for any $u$.

We can get something similar for odd r.

 

[vela]

No there is no typo but according to the answer key, both Q and R share the same answer.

Any hints on how you determine that?

Oh, okay. For Q, I think the simplest way to see it is to make the limits look like the one in P. Start with
$$\lim_{r \to \infty} \frac{f(r)}{f(r+1)} = \lim_{r \to \infty} \frac{r\left(\frac{2}{\pi}\right)^{r+2}f(r)}{r\left(\frac{2}{\pi}\right)^{r+2}f(r+1)}$$ and go from there.

For R, use integration by parts to write the denominator in terms of f(r+1). Then it looks kinda like Q.

 

[haruspex]

I agree that $f$ is the imaginary part of
$$I(r) = \int_0^{\pi/2} x^r e^{ix}\,dx = \left( \frac \pi 2 \right)^r + irI(r-1)$$
subject to $I(0) = 1 + i$ but I don't think you've solved the recurrence correctly.

You're right, I blundered.
The recurrence relation leads to $I_r = s^{-r}\Sigma_{k=0}^r(is)^k\frac{r!}{(r-k)!} + i^{r+1}r!$, but I erroneously wrote down ^rC_k instead of r!/(r-k)!.
So $f_r = s^{-r}\Sigma_{k=0}^{2k+1 ≤ r}(-1)^ks^{2k+1}\frac{r!}{(r-2k-1)!} + Re((-1)^{\frac r2})r!$

 

[jackmell]

Guys, isn't it obvious the last one is zero? I'm almost sure it is without doing any work on it. If indeed so, we could at least get rid of that answer.

 

[Saitama]

Oh, okay. For Q, I think the simplest way to see it is to make the limits look like the one in P. Start with
$$\lim_{r \to \infty} \frac{f(r)}{f(r+1)} = \lim_{r \to \infty} \frac{r\left(\frac{2}{\pi}\right)^{r+2}f(r)}{r\left(\frac{2}{\pi}\right)^{r+2}f(r+1)}$$ and go from there.

Since I don't know the answer to P, I assume it is c.

As per your suggestion, I write Q as
$$\frac{f(r)}{f(r+1)}\cdot \frac{r}{r} \cdot \frac{r+1}{r+1} \cdot \frac{(2/\pi)^{r+1}}{(2/\pi)^{r+1}} \cdot \frac{(2/\pi)^{r+2}}{(2/\pi)^{r+2}}$$
Taking the limit, I get $(2/\pi)$ as the answer.

Thanks a lot vela! :)

For R, use integration by parts to write the denominator in terms of f(r+1). Then it looks kinda like Q.

I have
$$f(r+1)=(r+1)\left(\frac{\pi}{2}\right)^r-r(r+1)f(r-1)$$

I used integration by parts on the denominator as you say,
$$\int_0^{\pi/2} x^r\cos x \,dx=\left(\frac{\pi}{2}\right)^r-r(r-1)\int_0^{\pi/2}x^{r-2}\cos x\, dx$$
but I don't see how this is similar to Q. :(

Any of answers 2, 3, 4 - it doesn't matter. Since S=0, P cannot be 0. So we know P converges to some constant c, 0 < c < ∞, giving us an expression for the asymptotic behaviour of f. Substitute that in Q.

What is "asymptotic behaviour"?

Guys, isn't it obvious the last one is zero? I'm almost sure it is without doing any work on it. If indeed so, we could at least get rid of that answer.

Hi jackmell! :)

A few posts before, everyone already agreed that answer to S is zero. We are discussing about P, Q and R.

 

[vela]

I have
$$f(r+1)=(r+1)\left(\frac{\pi}{2}\right)^r-r(r+1)f(r-1)$$

I used integration by parts on the denominator as you say,
$$\int_0^{\pi/2} x^r\cos x \,dx=\left(\frac{\pi}{2}\right)^r-r(r-1)\int_0^{\pi/2}x^{r-2}\cos x\, dx$$
but I don't see how this is similar to Q. :(

You can't use the recurrence relation you derived for f(r) because the denominator isn't equal to f(r). It has a cosine in it, not a sine.

Use integration by parts to raise the power of x from r to r+1.

 

[haruspex]

I have
$$f(r+1)=(r+1)\left(\frac{\pi}{2}\right)^r-r(r+1)f(r-1)$$

I used integration by parts on the denominator as you say,
$$\int_0^{\pi/2} x^r\cos x \,dx=\left(\frac{\pi}{2}\right)^r-r(r-1)\int_0^{\pi/2}x^{r-2}\cos x\, dx$$
but I don't see how this is similar to Q. :(

I think Vela was making the same suggestion I posted previously, but wrote Q by mistake for R?

Go back and look at your integration by parts. You did two steps of that to get your recurrence relation, right? The first step gave you something very close to the denominator in R.

$f(r) = \int_0^{\pi/2} x^r\sin x \,dx=\left[-x^r\cos x\right]_0^{\pi/2}+r\int_0^{\pi/2}x^{r-1}\cos x\, dx = r\int_0^{\pi/2}x^{r-1}\cos x\, dx$
So R becomes
$\frac{\int_0^{\pi/2}x^{r-1}\cos x\, dx}{\int_0^{\pi/2}x^{r}\cos x\, dx}$
It's not hard to show that, for the purposes of the limit, you can change the ranges of integration to be from 1 instead of 0. From that you can see the limit cannot exceed 1.

What is "asymptotic behaviour"?

It's the behaviour as r → ∞. But you already used Vela's more informative version of this suggestion.

 

[Saitama]

You can't use the recurrence relation you derived for f(r) because the denominator isn't equal to f(r). It has a cosine in it, not a sine.

Use integration by parts to raise the power of x from r to r+1.

Very sorry, I don't know why I wrote f(r+1) there.

Increase the power? Usually, while integrating by parts, the trigonometric expression is integrated but I follow your advice.

$$\int_0^{\pi/2} x^r\cos x\, dx=\left(\cos x \cdot \frac{x^{r+1}}{r+1}\right|_0^{\pi/2}+\frac{1}{r+1}\int_0^{\pi/2}x^{r+1} \sin x\,dx $$
$$\Rightarrow \int_0^{\pi/2} x^r\cos x\, dx=\frac{1}{r+1}\int_0^{\pi/2}x^{r+1} \sin x\,dx $$

Wow, great vela, thanks a lot! :)

How did you think of integrating x^r instead of cos x which is usually done while integrating by parts? Some insight would be helpful. :)

The only question remaining is P, can I get some hints?

 

[Saitama]

It's not hard to show that, for the purposes of the limit, you can change the ranges of integration to be from 1 instead of 0. From that you can see the limit cannot exceed 1.

Nice explanation, this is a good way of finding the answers, thanks haruspex. :)

But is it possible to continue from your expression and reach the answer? vela's suggestion helped me to convert the limit in R similar to the one in Q and that immediately gave the answer. I am sorry if I am missing something obvious.

 

[vela]

How did you think of integrating x^r instead of cos x which is usually done while integrating by parts? Some insight would be helpful. :)

Trial and error. I originally tried the usual approach, but it didn't seem to help. Also, I could see if I kept the cosine around for the surface term, that term would disappear because of the limits of integration, so it seemed like trying it was worth a shot.

 

[Saitama]

Trial and error. I originally tried the usual approach, but it didn't seem to help. Also, I could see if I kept the cosine around for the surface term, that term would disappear because of the limits of integration, so it seemed like trying it was worth a shot.

Thanks!

Can I please have a few hints for P?

 

[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 $uv\bigl|_0^{\pi/2} = 0$.