# Homework Help: Definite integral problem

1. Nov 20, 2013

### Saitama

1. The problem statement, all variables and given/known data
Let $\displaystyle f(r)=\int_0^{\pi/2} x^r\sin x \,\, dx$. Now match the following List-I with List-II.

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

2. Relevant equations

3. 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!

Last edited: Nov 20, 2013
2. Nov 20, 2013

### Staff: Mentor

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

3. Nov 20, 2013

### Saitama

Woops, edited! Thanks. :)

4. Nov 20, 2013

### Staff: Mentor

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...?

5. Nov 20, 2013

### Saitama

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.

6. Nov 20, 2013

### Staff: Mentor

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.

Last edited: Nov 20, 2013
7. Nov 20, 2013

### Saitama

Yes, its an exam 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?

8. Nov 20, 2013

### Staff: Mentor

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...

9. Nov 21, 2013

### Saitama

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 $r\rightarrow \infty$.

10. Nov 21, 2013

### Staff: Mentor

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.

11. Nov 21, 2013

### Staff: Mentor

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

12. Nov 21, 2013

### vela

Staff Emeritus
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.

13. Nov 21, 2013

### vela

Staff Emeritus
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.

14. Nov 21, 2013

### 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 $s = \frac 2 \pi$

Last edited: Nov 22, 2013
15. Nov 22, 2013

### Saitama

Oh yes, that was easy. Thanks! :)
What is c? :uhh:

How?

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!

16. Nov 22, 2013

### pasmith

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.

17. Nov 22, 2013

### vela

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

18. Nov 22, 2013

### Saitama

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?

19. Nov 22, 2013

### haruspex

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.
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?

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.

20. Nov 22, 2013

### Ray Vickson

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.

21. Nov 22, 2013

### vela

Staff Emeritus
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.

22. Nov 23, 2013

### haruspex

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 rCk instead of r!/(r-k)!.
So $f_r = s^{-r}\Sigma_{k=0}^{2k+1 ≤ r}(-1)^ks^{2k+1}\frac{r!}{(r-2k-1)!} + Re((-1)^{\frac r2})r!$

23. Nov 23, 2013

### 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.

24. Nov 23, 2013

### Saitama

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! :)

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. :(

What is "asymptotic behaviour"?

Hi jackmell! :)

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

25. Nov 23, 2013

### vela

Staff Emeritus
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.