Odd function integrals

  • #1
Hello I'd first like to state I know how to solve and I know the answer to this integral however when I first looked at the integral my initial thought was that it was equal to zero. I'd like to explain why I thought it was equal to 0 and hopefully someone can tell me where I went wrong.

I used two different theorems to come to my conclusion:

1. The definite integral with limits [-a,a] of a function f(x) will always be equal to zero if the function is odd.

[itex]\int[/itex][itex]^{a}_{-a}[/itex]f(x)dx = 0

2. Suppose a function F(x) is T-periodic function then for any real number a we have:

[itex]\int[/itex][itex]^{T}_{0}[/itex]f(x)dx = [itex]\int[/itex][itex]^{T+a}_{a}[/itex]f(x)dx


I have a problem where I need to find the Fourier Series of the function f(x) = 2x2-1
on the interval 0[itex]\leq[/itex]x[itex]\prec[/itex]2L and it tells me that the function has a period of 2L

now when I use the equation to find the bn fourier coefficient I get:

bn=[itex]\frac{1}{L}[/itex][itex]\int[/itex][itex]^{2L}_{0}[/itex](2x2-1)sin([itex]\frac{xnπ}{L}[/itex])

When I multiply sine by the 2x2-1 you can easily see that I get two integrals of odd functions sin(xnπ/L)(2x2) and sin(xnπ/L)

If we plug in -x to both functions we see that we get the negative of the function therefore we know both are odd functions. Now because the problem tells us that the function is of period 2L and we're integrating from 2L I apply my second theorem and subtract L from the limits which now gives an integral that satisfy the first theorem and by which I then conclude that both integrals should be equal to 0. But after calculating the integral I see that it doesn't and I'm not sure where I'm making a false assumption.
 

Answers and Replies

  • #2
52
2
Well, for an odd function, f(-x) = -f(x). So, if f and g are odd functions, and h(x) = f(x)g(x), is h an odd function?
 
  • #3
That's not my question, I know that sine(xnpi/L)*((2x^2)-1) is an odd function. I'm wondering why the integral of this function from 0 to 2L is not equal to 0 based on the 2 theorems I've posted.
 
  • #4
jbunniii
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That's not my question, I know that sine(xnpi/L)*((2x^2)-1) is an odd function. I'm wondering why the integral of this function from 0 to 2L is not equal to 0 based on the 2 theorems I've posted.
Your function ##f(x) = \sin(xn\pi/L)(2x^2 - 1)## is an odd function, meaning that it satisfies ##f(-x) = -f(x)##. We also say that ##f## is antisymmetric around ##x=0##. It follows that ##\int_{-a}^{a} f(x) = 0##, where the integration is taken over any interval ##[-a,a]## which has midpoint ##0##.

However, you are integrating over the interval ##[0,2L]##. Note that ##f## is NOT periodic, because of the ##2x^2 - 1## factor. So the integral over this interval is not necessarily ##0##. This interval has midpoint ##L##, and it is easy to see that your function is NOT antisymmetric around ##x=L##. To see this, put ##g(x) = f(x-L)##, and check whether ##g(-x) = -g(x)##.
 

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