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Please help with integration

  1. May 14, 2013 #1
    I am having some sort of blindness today!!! I just don't see how I can go from the first step to the second:

    [tex]2\int_0^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1-u^2} du = \int_{-1}^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1+u} du[/tex]
    Where "a" is just a constant.

    As you can see, the numerator does not change. so it really boils down to
    [tex]2\int_0^1\frac{1}{1-u^2} du = \int_{-1}^1\frac{1}{1+u} du[/tex]

    I tried using partial fraction where:
    [tex]\frac 1 { (1+u)(1-u)}=\frac A {(1+u)}+\frac B {(1-u)}[/tex]
    where A=B=##\frac 1 2##.
    [tex]\Rightarrow\; 2\int_0^1\frac{1}{1-u^2} du=\int_0^1\frac{1}{1+u} du+\int_0^1\frac{1}{1-u} du[/tex]

    I just don't see it. Please help.
     
    Last edited: May 14, 2013
  2. jcsd
  3. May 14, 2013 #2

    SammyS

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    Doing the pertial fraction decomposition may help -- probably does help, but you can't just say it boils down to [itex]\ \displaystyle 2\int_0^1\frac{1}{1-u^2} du = \int_{-1}^1\frac{1}{1+u} du\ .[/itex]

    One other thing also changes, the limits of integration.

    Also notice that the numerator is a perfect square.

    There is also a sum to product identity for cos(A) - cos(B) .

    http://en.wikipedia.org/wiki/Trigonometric_identities#Product-to-sum_and_sum-to-product_identities
     
  4. May 14, 2013 #3
    I get to this directly from what is given from the solution manual:

    [tex]2\int_0^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1-u^2} du = \int_{-1}^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1+u} du[/tex]
    Where "a" is just a constant.

    [tex]\Rightarrow\;2\int_0^1\frac{1}{1-u^2}[\cos^2 (au) +\cos^2a-2\cos (au) \cos a] du = \int_{-1}^1\frac{1}{1+u} [\cos^2 (au) +\cos^2a-2\cos (au) \cos a]du[/tex]

    As you can see, the numerator does not change. I just remove the whole numerator and boils down to
    [tex]2\int_0^1\frac{1}{1-u^2} du = \int_{-1}^1\frac{1}{1+u} du[/tex]

    I really did not change anything. I just don't get this and I show my work in the first post where I got stuck. I don't even see why I need to worry about sine and cosine terms in the numerator as it is not even affected. I only show the whole equation just to show where it comes from.
     
    Last edited: May 14, 2013
  5. May 14, 2013 #4

    Dick

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    You are barking up the wrong tree here. Saying that the integral of 1/g(x) is equal to the integral of 1/h(x) means that the integral of f(x)/g(x) equals the integral f(x)/h(x) is completely wrong. Not that I know how to solve your problem, but the integral of 1/(1+u) from -1 to 1 doesn't even exist. And the other one diverges as well. You can't 'cancel' the numerator.
     
    Last edited: May 14, 2013
  6. May 14, 2013 #5

    SammyS

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    Hint: What is ##\displaystyle \ \left( \cos(au)-\cos(a) \right)^2 \ ## when expanded ?

    After suitably changing the integrand and splitting the integral into the sum of two integrals, it looks like you will need to use a change of variable (substitution) with one of the integrals.

    I haven't done the problem either, but it looks to me that the integral does converge. (As u → 1 , we have the indeterminate form, 0/0 .)
     
    Last edited: May 15, 2013
  7. May 15, 2013 #6

    Dick

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    I don't think that helps. The original numerator is symmetric in u. I would try splitting 1/(1+u) into symmetric and antisymmetric parts. I think that will.
     
    Last edited: May 15, 2013
  8. May 15, 2013 #7
    Actually that's where it came from......##(\cos au -\cos a)^2##.

    I am still looking at this, I understand what I did wrong. But I still have no idea how to solve this.
     
  9. May 15, 2013 #8
    I see what you mean. But now I am even more lost. This is a scan of the solution manual. I use a=kl/2.
     

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    Last edited: May 15, 2013
  10. May 15, 2013 #9

    SammyS

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    Use the difference to product identity for ##\ \cos(\theta) - \cos(\varphi) \ .##

    ##\displaystyle
    \cos \theta - \cos \varphi = -2\sin\left( {\theta + \varphi \over 2}\right) \sin\left({\theta - \varphi \over 2}\right)
    ##

    In this case, ##\displaystyle \ {\theta + \varphi \over 2}\ ## and ##\displaystyle \ {\theta - \varphi \over 2}\ ## work out very nicely .

    Added in Edit:
    The only reason to change to the product of sines is to show that the limit of the integrand exists as u → ±1 .
     
    Last edited: May 15, 2013
  11. May 15, 2013 #10

    SammyS

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    Using the partial fraction decomposition you found for 1/(1-u2) , split the integral into the sum of two integrals, one with a denominator
    of 1 + u, the other with 1 - u.

    Each, of course, has a numerator of (cos(au)-cos(a))2 .

    Then, for the integral with a denominator of 1 - u, do a substitution such as x = -u .
     
  12. May 15, 2013 #11
    Thanks

    Let me work on this and come back later.
     
  13. May 15, 2013 #12
    Let me try this and see whether this is right

    [tex]\frac 2 { (1+u)(1-u)}=\frac A {(1+u)}+\frac B {(1-u)}=\frac 1 {(1+u)}+\frac 1 {(1-u)}[/tex]
    where A=B=##\frac 1 2##.

    [tex]\Rightarrow\;2\int_0^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1-u^2} du = \int_{0}^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1+u} du\;+\; \int_{0}^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1-u} du [/tex]



    [tex]\hbox {Let}\;v=-u\Rightarrow\; du=-dv[/tex]
    [tex]\int_{0}^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1-u} du\;=\;-\int_{0}^{-1}\frac{[\cos^2 (-av) +\cos^2a-2\cos (-av) \cos a]}{1+v} dv\;=\;\int_{-1}^0\frac{[\cos^2 (av) +\cos^2a-2\cos (av) \cos a]}{1+v} dv[/tex]


    [tex]\hbox {Let}\;u=v\Rightarrow\; du=dv[/tex]
    [tex]\int_{-1}^0\frac{[\cos^2 (av) +\cos^2a-2\cos (av) \cos a]}{1+v} dv\;=\;\int_{-1}^0\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1+u} du[/tex]

    Therefore
    [tex]2\int_0^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1-u^2} du\;=\;\int_{0}^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1+u} du\;+\;\int_{-1}^0\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1+u} du[/tex]

    [tex]\Rightarrow\;2\int_0^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1-u^2} du\;=\;\int_{-1}^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1+u} du[/tex]

    How does this look?
     
    Last edited: May 15, 2013
  14. May 15, 2013 #13

    SteamKing

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    In the first line, if A = B = 1/2, shouldn't the fractions be (1/2) / (1 + u) and (1/2) / (1 - u)?
     
  15. May 15, 2013 #14

    SammyS

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    He is decomposing ##\displaystyle \ \frac{2}{1-u^2}\ ## not ##\displaystyle \ \frac{1}{1-u^2}\ .##
     
  16. May 15, 2013 #15

    SammyS

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    It looks good to me.
     
  17. May 15, 2013 #16

    SteamKing

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    I agree that 1/(1+u) + 1/(1-u) = 2 / (1-u^2), but his partial fraction decomp says that his unknowns A = B = 1/2, which is incorrect according to his initial assumption. It is a minor point, but one should endeavor to be correct in all details.
     
  18. May 15, 2013 #17
    Thanks for the help from everybody.
     
  19. May 15, 2013 #18
    It was late, I was sloppy, it should be ##2\frac 1 {1-u^2}## and A=B=##\frac 1 2##.
     
  20. May 15, 2013 #19
    This is a new question all together, but it is on the same page of the derivation. I cannot follow the derivation of the notes. Here is my derivation starting from (1) or the notes.

    [tex]I_{int}=\frac 1 2 \int_{-1}^1\frac{[1+\cos(klu)+1+\cos(kl)]}{(1+u)}du\;-\;\int_{-1}^1\frac{\cos\left (\frac{kl}{2}(1+u)\right)+\cos\left (\frac{kl}{2}(1-u)\right) }{(1+u)}du[/tex]
    [tex]\hbox {Let }\;(1+u)kl=v\;\Rightarrow\;v=(kl+klu),\;u=(\frac {v}{kl}-1),\;(1+u)=\frac {v}{kl}, \;\hbox { and } \;du=\frac {dv}{kl}[/tex]

    [tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 + \cos (v-kl)+\cos(kl)]}{v} dv-\int_0^{2kl}\frac{\cos\left(\frac v 2 \right)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
    [tex]\hbox {For the second term, let }\; w=\frac v 2 \;\hbox { and then Let } v=w \hbox { later}[/tex]
    [tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 + \cos (v-kl)+\cos(kl)]}{v} dv-\int_0^{kl}\frac{\cos(v)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
    [tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 +\cos(kl)]}{v} dv+ \frac 1 2 \int_0^{2kl} \frac { \cos (v-kl)}{v} dv -\int_0^{kl}\frac{\cos(v)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]


    Now compare the line (2) in the scanned notes:
    [tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 + \cos (klv)+\cos(kl)]}{v} dv-\int_0^{kl}\frac{\cos(v)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]
    [tex]\Rightarrow\; I_{int}=\frac 1 2 \int_0^{2kl} \frac {[2 +\cos(kl)]}{v} dv+ \frac 1 2 \int_0^{2kl} \frac { \cos (klv)}{v} dv -\int_0^{kl}\frac{\cos(v)}{v}dv-\int_{-1}^1 \frac{\cos\left[\frac{kl}{2}(1-u)\right]}{1+u}du[/tex]

    If you compare the equation I derive with (2) of the notes, the only terms that is different is:
    [tex]\frac 1 2 \int_0^{2kl} \frac { \cos (v-kl)}{v} dv\;\hbox { from my equation to }\;\frac 1 2 \int_0^{2kl} \frac { \cos (klv)}{v} dv\;\hbox { from the notes.}[/tex]

    I just don't see how I can get the answer from the notes. As you see, I separate the longer integration into two parts. I am pretty sure I did this right this time.....hopefully. Please help.

    I attach the whole page of the notes as I might have more question coming and I might have typos. This is a difficult derivation that involve sine and cosine integral and I am rusty in math!!! I am not math major, I do this as part of the derivation of radiating power of antenna. I am not like you guys that are in "mid season condition"!!!:rofl:. Everything is rusty for me and need some kick start!!! What I posted is from equation (1) to (2) in the notes.
     

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    Last edited: May 15, 2013
  21. May 15, 2013 #20

    SammyS

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    It's generally the case that a new question should be posted in a new thread. This is no exception.

    If this question were very closely related to the one that you started this thread with, it might be justified to tack this on to the the thread, but that is not the case here.
     
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