# Vanishing Integral

1. Dec 28, 2007

### ehrenfest

[SOLVED] Vanishing Integral

1. The problem statement, all variables and given/known data
http://mikef.org/files/phys_4242_hw5.pdf [Broken]
When does the integeral at the bottom of page 2 vanish? He says it vanishes when n'_y and n'_y have oppositive parity, but I think it always vanishes because n'_y - n'_y and n'_y + n'_y always have the same parity, so the integral over

cos(pi (n'_y - n'_y) y) and cos(pi (n'_y + n'_y) y)

should be the same.

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 3, 2017
2. Dec 29, 2007

### HallsofIvy

Staff Emeritus
The anti-derivative of cos is sin, which is 0 at any multiple of $\pi$. At the upper limit, the b in the denominator is cancelled by the upper limit b, leaving an integeer multiple of $\pi$. Of course, at the lower limit, 0, sin(0)= 0.

3. Dec 29, 2007

### ehrenfest

Notice there is a y/2 in front. Could you explain what your answer means for the result of the integral? It IS always 0, correct? So, the last sentence is wrong?

4. Dec 29, 2007

### HallsofIvy

Staff Emeritus
You are right, I missed the y/2 completely! Okay, do an "integration by parts". Let u= y and dv= (cos(py)- cos(q))dy. Then du= dy, v= -(sin(py)/p+ sin(qy)/q).

The integeral is "$uv\left|_0^b- \int_0^b vdu[/tex]" . As before The first term will be 0 at both y= 0 and y= b. Now the problem is just $$\int_0^v sin(py)/p+ sin(qy)/q)dy= -cos(py)/p^2- cos(qy)/q^2$$ evaluated between 0 and b. Now, if [itex]n_y$ and $n_y'$ differ by an even number[/b] (i.e. are either both even or both odd) those reduce to cosine at 0 and an even multiple of $\pi$ and so the difference is 0. That is the reason for the condition "unless $n_y' = n_y$+ an odd positive integer".

Last edited: Dec 29, 2007
5. Dec 29, 2007

### ehrenfest

I think it should be

$$\int_0^v sin(py)/p- sin(qy)/q)dy= -cos(py)/p^2+ cos(qy)/q^2$$

but now I see why it is 0 unless n_y and n'_y have opposite parity. Thanks.