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Can integration by parts (and other int related Q's)

  1. Nov 30, 2007 #1
    Be used to prove that a constant coefficiant can be taken out the integral? such as 2x just being 2 multiplied by the integral of x?

    [tex] \int{2x} = 2\int{x}[/tex]

    Also, what is the integral of 0? Is it 0 or "c"? I'm mainly asking because I recently got dropped with a fourier series homework and its difficult knowing when I can take coefficiants out of the integral and when It's wise to keep them inside, such as when you have Cos (x) c to take c out or keep it in.

    Thanks for the help
    -Abe
     
  2. jcsd
  3. Nov 30, 2007 #2
    You cant use integration by parts to prove this, as you don't have 2 functions of x. you have one function of x with two terms. Basically in an integration problem you want to take any constants outside the integral (constant is any function that is not to be evaluated (y etc in this case). In the general case i find it easier to leave numbers in as normally cancellation occurs, as it does in the above case, and is easier to spot in this situation.
    But as long as you follow the rules you are taught, its all preference from then on.

    EK
     
  4. Nov 30, 2007 #3

    HallsofIvy

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    First, if dF/dx= f(x), then [itex]\int f(x)dx= F(x)+ C[/itex] where C can be any constant. In particular, if f(x)= 0, then [itex]\int f(x)dx= C[/itex] because the derivative of any constant is 0. The derivative of cF(x) is c dF/dx. That's why [itex]\int cf(x)dx= c \int f(x)dx[/itex]. You certainly don't need integration by parts to prove that, just the definition of "integral".

    Quite frankly, if you don't already know this (a fundamental of calculus) well, you are going to have serious problems with Fourier series.
     
  5. Nov 30, 2007 #4
    I can do most integration A ok, I was asking for proofs unlike most math users I work with who will blindly go through the processes like beer drinking phrat boy calculators, and it's not hindering my fourier series at all.

    Just being able to explain the rules to the people who may want to learn it from me will come in handy rather than what i'm told by other students and lecturers alike "oh it just does, dont be silly and ask those questions".

    Thanks to erwin, and you ivy, but being frank with somone asking a more conceptual question than a technique one is a bit silly when it comes to the largely underexplained fourier series (at my level anyway).
     
  6. Nov 30, 2007 #5

    arildno

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    Okay, let's use integration by parts, then: :smile:
    [tex]\int{kx}dx, v'=k, u=x[/tex]
    [tex]\int{kx}dx=kx^{2}-\int{kx*1}dx\to{2}\int{kx}dx=kx^{2}\to\int{kx}dx=k*\frac{x^{2}}{2}=k*\int{x}dx}[/tex]
    Plus a C of course..

    Are you happy now?
     
  7. Dec 1, 2007 #6
    Yeah sort of, the fourier series i have in front of me just has little conceptual problems, its:

    [tex] f(x+8) = f(x)[/tex]

    Values:
    [tex]2, \ 0 \leq{x}<6[/tex]
    [tex]0, \ 6 \leq{x}<8[/tex]

    So the integral for [tex] a_{n}[/tex] is:

    [tex] a_{n} = \frac{1}{4}\int _{0}^{6} Cos(\frac{\pi nt}4) 2\ dt + \int _{6}^{8} 0\ dt [/tex]

    Evaluated:

    [tex]a_{n} = \frac{1}{4}\left[\frac{8}{\pi n}Sin(\frac{\pi nt}{4})\right]_{0}^{6}+\left[ c \right]_{6}^{8} [/tex]

    [tex]a_{n}=\frac{2}{\pi n} \left[ Sin(\frac{\pi nt}{4}) \right]_{0}^{6} + 2c [/tex]

    And so on to get a final number from there

    Is this going along the right lines? or shall I ignore the C in the fourier series?
     
    Last edited: Dec 1, 2007
  8. Dec 1, 2007 #7

    arildno

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    Your function's values do not match up with the periodicity requirement.

    Are you sure the period is 8, rather than 88??
     
  9. Dec 1, 2007 #8
    Changed it due to latex typo, its 8..
     
  10. Dec 1, 2007 #9

    arildno

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    Eeh, how do you get 2c???
     
  11. Dec 1, 2007 #10
    8c -6c, although the C's can just be merged into 1 C i guess since its an arbitary coefficiant...
     
  12. Dec 1, 2007 #11

    arildno

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    Completely wrong!
    Remember that [tex]f(x)\mid_{a}^{b}\equiv{f(b)}-f(a)[/tex]
    Hence, in the particular case f(x)=c, where c is some constant, you have:
    [tex]f(x)\mid_{a}^{b}={f(b)}-f(a)=c-c=0[/tex]
     
  13. Dec 1, 2007 #12
    Why are you math mentors so bitter these days ^^.

    But I didnt quite look at it like that ^^, could you have a look at the rest of the integral...I have a fourier homework in front of me so its nice to know if I'm doing the main bits wrong rather then being barked at for pressing 8 too many times or slight substitution problems.

    As for the other fourier series stuff, is it ok to move the integral boundaries in order to make it the integral more "Even" either side, such as:

    [tex] \int_{0}^{T} [/tex] goes to [tex] \int_{-\frac{1}{2}T}^{\frac{1}{2}T} [/tex]
     
  14. Dec 1, 2007 #13
    Am I wrong or can you compute the value of 'c' in each case by substituting the boundaries?
     
  15. Dec 1, 2007 #14

    arildno

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    Okay, the rest seems fine, however:
    [tex]\sin(\frac{\pi{nt}}{4}\mid_{0}^{6}=\sin(\frac{3\pi}{2}{n})[/tex]
    Therefore,
    [tex]a_{2m}=0[/tex]
    For the odd integers, we have that:
    [tex]\sin(\frac{3\pi}{2}(4m+1))=-1,\sin(\frac{3\pi}{2}(4m+3))=1, m=0,1...[/tex]
     
  16. Dec 1, 2007 #15
    Ah thanks alot arildno :D, so what your saying is that even n's will result in zeroes and odd ones will result in -1, 1, -1, 1 etc etc making the sum in the series cancel out?
     
  17. Dec 2, 2007 #16

    arildno

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    Not at all!

    Remember that for the even numbers, the an's will be the product of 0 and some other factors yielding 0, but for the odd integers, we have:
    [tex]a_{4m+1}=-\frac{2}{\pi(4m+1)},a_{4m+3}=\frac{2}{\pi(4m+3)}, m=0,1...[/tex]
     
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