Help with integral question.

Kuma

1. The problem statement, all variables and given/known data

Here is the question given:

2. Relevant equations

3. The attempt at a solution

SO i used the product to sum angle formula:
sin u sin v = 1/2[cos(u-v) - cos (u + v)]

so I get
from ∏ to -∏

1/2 ∫cos(k-n)x dx - ∫cos(k+n)x dx

= 1/2 [1/k-n (sin (k-n)∏ - sin(k-n)-∏) -1/k+n(sin(k+n)∏ - sin(k+n)-∏)]

so when k isn't equal to ±n, i think the second term becomes 0, but the first term doesn't have to be 0 as well (for the first condition).

micromass

What is [itex]\sin(n\pi)[/itex] if [itex]n\in \mathbb{Z}[/itex]???

Also note that your method does not work if n=k. Indeed, you seem to have a term [itex]\frac{1}{n-k}[/itex] there. But if n=k, then you will divide by zero. So if n=k, then you'll need to do something different.

SammyS

Quote by Kuma

1. The problem statement, all variables and given/known data

Here is the question given:

b]3. The attempt at a solution[/b]

SO i used the product to sum angle formula:
sin u sin v = 1/2[cos(u-v) - cos (u + v)]

so I get
from ∏ to -∏

1/2 ∫cos(k-n)x dx - ∫cos(k+n)x dx

= 1/2 [1/(k-n) (sin (k-n)∏ + sin(k-n)∏) -1/(k+n)(sin(k+n)∏ + sin(k+n)∏)]

so when k isn't equal to ±n, i think the second term becomes 0, but the first term doesn't have to be 0 as well (for the first condition).

sin(mπ) = 0 for all integers, m. So when k≠n, all those terms are zero.

When k=n, the last term is zero. The first term must be evaluated as a limit n → k .

Kuma

Yes you are right about the second condition when k = n i will get 1/0, but how do I eliminate that term then

SammyS

Quote by Kuma

Yes you are right about the second condition when k = n i will get 1/0, but how do I eliminate that term then

Actually it's of the form 0/0 .

micromass

Quote by Kuma

Yes you are right about the second condition when k = n i will get 1/0, but how do I eliminate that term then

If k=n, then

[tex]\int_{-\pi}^\pi \cos(n-k)x dx=\int_{-\pi}^\pi \cos(0)dx=\int_{-\pi}^\pi dx[/tex]

SammyS

If n=k, the integrand becomes sin²(kx).

Kuma

Quote by SammyS

sin(mπ) = 0 for all integers, m. So when k≠n, all those terms are zero.

When k=n, the last term is zero. The first term must be evaluated as a limit n → k .

I see what you did but k+n doesn't necessarily have to be an integer does it? What if k+n= 0.5, it doesn't make it 0.

micromass

Quote by Kuma

I see what you did but k+n doesn't necessarily have to be an integer does it? What if k+n= 0.5, it doesn't make it 0.

I think the question implies that both n and k are integers. If n and k are not integers, then the result is not true.

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micromass Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus	What is [itex]\sin(n\pi)[/itex] if [itex]n\in \mathbb{Z}[/itex]??? Also note that your method does not work if n=k. Indeed, you seem to have a term [itex]\frac{1}{n-k}[/itex] there. But if n=k, then you will divide by zero. So if n=k, then you'll need to do something different.

Kuma	Help with integral question. Yes you are right about the second condition when k = n i will get 1/0, but how do I eliminate that term then