Odd Function Integrals: Exploring Answers and Assumptions

In summary: Since ##g## is antisymmetric around ##x=0##, the answer is yes. This means that the integral over this interval is not necessarily zero.
  • #1
AerospaceEng
28
0
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)dxI 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.
 
Physics news on Phys.org
  • #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
AerospaceEng said:
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)##.
 

1. What is an odd function integral?

An odd function integral is the integral of an odd function, which is a function that satisfies the condition f(-x) = -f(x). This means that the graph of an odd function is symmetric about the origin. The integral of an odd function over a symmetric interval is always equal to zero, since the positive and negative areas cancel each other out.

2. How do you solve an odd function integral?

To solve an odd function integral, you can use the fundamental theorem of calculus, which states that the integral of a function is equal to the difference between the values of the antiderivative of that function at the upper and lower bounds of the integral. Since the integral of an odd function over a symmetric interval is zero, you can simply evaluate the antiderivative at the upper and lower bounds to find the solution.

3. What are some examples of odd functions?

Some examples of odd functions include sine, cosine, and tangent. Other examples include the reciprocal of an odd function, such as 1/x, and the product of an odd and an even function, such as x*sin(x).

4. Are there any special properties of odd function integrals?

Yes, odd function integrals have a few special properties. One is that the integral of an odd function over a symmetric interval is always equal to zero. Another is that the integral of the product of two odd functions is always an even function, meaning that it is symmetric about the y-axis. Additionally, if the limits of integration are both negative or both positive, the integral of an odd function is equal to twice the integral from 0 to the upper limit.

5. What assumptions are made when evaluating an odd function integral?

When evaluating an odd function integral, it is assumed that the function is continuous and has a well-defined antiderivative. It is also assumed that the interval of integration is symmetric about the origin, so that the integral is equal to zero. Additionally, the fundamental theorem of calculus is applied, which assumes that the function is integrable and that the antiderivative exists.

Similar threads

  • Calculus
Replies
6
Views
1K
Replies
8
Views
310
  • Calculus
Replies
25
Views
910
Replies
3
Views
1K
Replies
1
Views
821
Replies
1
Views
832
Replies
3
Views
1K
Replies
20
Views
2K
Replies
2
Views
831
Back
Top