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Intergrating Functions

  1. Jul 9, 2012 #1
    Hi, I'm trying to understand how you integrate fractions, I have a couple of examples which I was given, unfortunately the lecturer skipped the actual method and just gave us the answers.
    By any chance could someone go through the method to solve these? (I think you use U substitution?)
    Thanks in advance

    (a) integral ((1-z)/(z2 +1))dz

    (b) integral ((1)/(z+z2)) dz

    (c) integral ((z)/(1-3z2))dz

    I tried taking taking the fractions up and solving by parts but that just seems to complicate things further
     
  2. jcsd
  3. Jul 9, 2012 #2

    Curious3141

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    You're supposed to show more work, but I can give you some hints to get you started.

    (a) Split up the numerator. The first fraction [itex]\frac{1}{z^2 + 1}[/itex] can be immediately integrated if you recognise it as a special form. Alternatively, substitute [itex]z = \tan \theta[/itex]. The second fraction [itex]-\frac{z}{z^2 + 1}[/itex] can be integrated by substituting [itex]y = z^2 + 1[/itex].

    (b) Factorise the denominator. Do a partial fraction decomposition.

    (c) Substitute [itex]y = 1 - 3z^2[/itex].
     
  4. Jul 9, 2012 #3

    SammyS

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    Hello MickGrif. Welcome to PF !

    As Curious3141 points out, the rules of this Forum require you to show your work before we can help you.
     
  5. Jul 9, 2012 #4
    Is z real or complex? Which class are you getting these integrals?

    z usually means α+jω (where jj=-1) in Complex Analysis. The integrals you listed would be typical elementary examples of application of Cauchy's residue theorem meant to show how to recognize the order of poles.

    On the other hand, if z is real (weird convention to me), you have been given good hints.



     
  6. Jul 9, 2012 #5

    Ray Vickson

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    If z2 is supposed to be z2, you should either write it as z^2 or use the "X2" button on the palette at the top of this panel; that will give you z2.

    RGV
     
  7. Jul 10, 2012 #6
    Can I first thank you for the responses.
    In regards showing the work done before this I have posted pics, the reason that I didnt do this in the first place was because the problem I was working on does not really tie into integrating up to this point. (As you will see below)

    The course is on ODEs (Ordinary Differential Equations) these problems in particular relating to 1st order homogenous ODEs

    I'm sorry it was meant to come out as z^2, I could have swore I clicked the button but must have double clicked or something, my bad.


    Work done up to this point; (alot of my own notes are in the margins so please excuse any thing that seems like obvious material :|)

    (a) P1040407.jpg

    (b) P1040408.jpg
    P1040409.jpg


    (c) P1040411.jpg
     
    Last edited: Jul 10, 2012
  8. Jul 10, 2012 #7

    SammyS

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    For (a), split into two integrals.
    [itex]\displaystyle \int\frac{1}{z^2+1}\,dz[/itex] is a standard integral.

    For [itex]\displaystyle \int\frac{z}{z^2+1}\,dz\,,[/itex] use a substitution.​

    For (b), use partial fraction decomposition.
    [itex]\displaystyle \frac{1}{z+z^2}=\frac{A}{1+z}+\frac{B}{z}[/itex]​

    (c) can be done with a substitution.
     
  9. Jul 10, 2012 #8
    For (a) could you use partial fractions as well? Or does the constant have to be positive, for example you can do ∫[itex]\frac{x+1}{(x+3)(x+2}[/itex] with partial fractions, can you also do ∫[itex]\frac{x-1}{(x+3)(x+2)}[/itex] using this method and thus use the same method for (a)?
     
  10. Jul 10, 2012 #9

    HallsofIvy

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    If you are doing these as part of a Differential Equations class, you should be able to do these integrals easily. The questions you are asking are basic integral Calculus. For one thing, do you know what the derivative of arctan(x) is?
     
  11. Jul 10, 2012 #10
    That is extremely condescending. Clearly I can't do these easily, hence why I came looking for help. I didnt choose to do this module, it was not an option in my degree. All I want to do is to do ok in it.
    Also I believe the derivative of arctan(x) is[itex]\frac{1}{1+x^{2}}[/itex]
     
  12. Jul 10, 2012 #11

    SammyS

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    Yes, that's the derivative of the arctangent function .
     
  13. Jul 10, 2012 #12

    SammyS

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    The problem with using partial fractions for (a), is that z2+1 is only factorable over the complex numbers:
    [itex]z^2+1=(z-i)(z+i)\,.[/itex]​
    That would leave you with the logarithm of complex quantities.
     
  14. Jul 11, 2012 #13
    Take the principal branch of the logarithm and the definition of the complex logarithm yields the integral correctly (if it is definite.) Partial fractions decomposition is still applicable. The problem with it is that you would have to do some working with Euler's identity and the complex logarithm to get the expression in terms of real functions. With definite integrals, since you get a scalar, there is no such problem.
     
  15. Jul 11, 2012 #14

    HallsofIvy

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    Okay, what does that tell you about the integral of [itex]1/(1+ x^2)[/itex]?
     
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