1. ## Reduction formula problem

Hi,

Here is the problem. The integral $I_n$ is defined by

$I_n = \int_0^{\pi}(\pi / 2 - x)\sin(nx + x/2)\mbox{cosec}(x/2) \, \mbox{d}x$

where $n$ is a positive integer. Evaluate $I_n - I_{n-1}$, and hence evaluate $I_n$ leaving your answer in the form of a sum.

I was hoping someone could tell me if I was on the right track or not. Firstly, is this an improper integral? If it is, how would I start the problem.

If it's not, I did the following

$I_n - I_{n-1} = \int^{\pi}_0 (\pi/2 - x)\mbox{cosec}(x/2)(\sin(nx + x/2) - \sin(nx - x/2)) \,\mbox{d}x$

$= \int^{\pi}_0 (\pi/2 - x)\mbox{cosec}(x/2)(2\cos(nx)\sin(x/2)) \, \mbox{d}x$

I fiddled about with this for a while and managed to get something but of course it's meaningless if the first part is wrong.

Any help would be much appreciated.

Thanks

Stonehambey

2. Originally Posted by Stonehambey
Hi,

Here is the problem. The integral $I_n$ is defined by

$I_n = \int_0^{\pi}(\pi / 2 - x)\sin(nx + x/2)\mbox{cosec}(x/2) \, \mbox{d}x$

where $n$ is a positive integer. Evaluate $I_n - I_{n-1}$, and hence evaluate $I_n$ leaving your answer in the form of a sum.

I was hoping someone could tell me if I was on the right track or not. Firstly, is this an improper integral? If it is, how would I start the problem.

If it's not, I did the following

$I_n - I_{n-1} = \int^{\pi}_0 (\pi/2 - x)\mbox{cosec}(x/2)(\sin(nx + x/2) - \sin(nx - x/2)) \,\mbox{d}x$

$= \int^{\pi}_0 (\pi/2 - x)\mbox{cosec}(x/2)(2\cos(nx)\sin(x/2)) \, \mbox{d}x$

I fiddled about with this for a while and managed to get something but of course it's meaningless if the first part is wrong.

Any help would be much appreciated.

Thanks

Stonehambey

Yes, the integral is improper since $\csc x$ isn't defined at zero, but what you did is fine, imho:

$\displaystyle{I_n-I_{n-1}=\int\limits^\pi_0 \left(\frac{\pi}{2} - x\right)\csc\frac{x}{2}\cdot 2\cos nx\cdot\sin\frac{x}{2}\,dx=2\int\limits^\pi_0\left (\frac{\pi}{2}-x\right)\cos nx\,dx=}$

$\displaystyle{=\frac{\pi}{n}\sin nx|\limits^\pi_0-2\int\limits^\pi_0 x\cos nx\,dx=\frac{-2}{n^2}\cos nx|\limits^\pi_0=\frac{2}{n^2}(1-(-1)^n)=\left\{\begin{array}{ll}0&,\,\mbox{if n is even}\\{}\\\frac{4}{n^2}&,\,\mbox{if n is odd\end{array}\right.}$

...or I'm wrong. Check carefully the above.

Tonio

Originally Posted by tonio

$\displaystyle{=\frac{\pi}{n}\sin nx|\limits^\pi_0-2\int\limits^\pi_0 x\cos nx\,dx=\frac{-2}{n^2}\cos nx|\limits^\pi_0=\frac{2}{n^2}(1-(-1)^n)=\left\{\begin{array}{ll}0&,\,\mbox{if n is even}\\{}\\\frac{4}{n^2}&,\,\mbox{if n is odd\end{array}\right.}$

...or I'm wrong. Check carefully the above.

Tonio
This was exactly the result I got before I posted the question, which is somewhat comforting. So the two questions I have remaining would be

1) How does the fact that it's an improper integral change the result? (I did all the above working before I noticed this, d'oh!)

2) What is the question hinting at when it asks $I_n$ to be written in the form of a sum?

Stonehambey

4. Originally Posted by Stonehambey

This was exactly the result I got before I posted the question, which is somewhat comforting. So the two questions I have remaining would be

1) How does the fact that it's an improper integral change the result? (I did all the above working before I noticed this, d'oh!)

I supose you must check the trigonometric equality inside the integral sign, as this makes the problematic factor

$\csc(x/2)$ dissapear. Nevertheless I think it was correctly done and thus you changed an improper integral

into a proper one.

2) What is the question hinting at when it asks $I_n$ to be written in the form of a sum?

This one baffles me. This problem ressembles some calculations related to Fourier series of some function, so I

guess it could be possible that what you just evaluated are the coefficients of the cosine part of the Fourier series and

thus you must use this. I can't say for sure.

Tonio

Stonehambey
.

5. The question was lifted from the 1998 STEP II paper (Cambridge entrance exams) and since Fourier series are not on the A-level syllabus I don't think you need to use them in your answer, even if the problem is related to them (This is not uncommon).

Thanks for the help

Stonehambey

6. Originally Posted by Stonehambey
The question was lifted from the 1998 STEP II paper (Cambridge entrance exams) and since Fourier series are not on the A-level syllabus I don't think you need to use them in your answer, even if the problem is related to them (This is not uncommon).

Thanks for the help

Stonehambey

Oh, I think I got it! I just forgot that what I (we) calculated is $I_n-I_{n-1}$ , and not $I_n$ itself, so we got a recursive formula:

$\displaystyle{I_n-I_{n-1}=\left\{\begin{array}{ll}0&,\,\mbox{if n is even}\\{}\\\frac{4}{n^2}&,\,\mbox{if n is odd\end{array}\right.}$ , so we can try:

$I_1=I_1$

$I_2-I_1=0$

$\displaystyle{I_3-I_2=\frac{4}{9}}$

.........................................

$\displaystyle{I_n-I_{n-1}=\frac{4}{n^2}}$ . (** Odd n. For even n take the preceeding value of n)

And now add both columns above:

$\displaystyle{I_n=I_1+4\sum\limits^n_{k=2}\frac{1} {(2n-1)^2}}$ . (Read ** above)

Tonio