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Please help with integration again

  1. May 15, 2013 #1
    This is a derivation of radiating power of dipole antenna, 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.
     
  2. jcsd
  3. May 16, 2013 #2

    haruspex

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    Your version looks right to me, the scanned notes wrong.
     
  4. May 16, 2013 #3
    Thanks, I spent a few hours on this!!!!
     
  5. May 16, 2013 #4
    it looks right to me too.

    I believe the notes would not contradict your answer unless the term that was used in the notes in the first line was not cos(klu). Check what the theory says for that

    Thanks :)
     
  6. May 16, 2013 #5
    It is klu because it is from ##\cos^2(\frac {KL}{2}u)=\frac 1 2[1+ \cos(klu)]## of the line above.

    Thanks
     
  7. May 16, 2013 #6
    I have a related question. As you can see in the equation, there is a term
    [tex] \frac 1 2 \int_0^{2kl} \frac {[2 +\cos(kl)]}{v} dv[/tex]
    Where ##2+\cos(kl)## are constant with respect to ##v##. This means
    [tex] \frac {[2 +\cos(kl)]} 2 \int_0^{2kl} \frac {1}{v} dv=\frac {[2 +\cos(kl)]} 2 ln(v)|_0^{2kl}[/tex]

    As you all know, ##ln(0)## is undefined. This means there is no solution for this equation!!! Please advice.
     
  8. May 16, 2013 #7

    haruspex

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    None of the four integrals converge, but if you look at the region near the singularity in each case you find that the four leading terms asymptotically cancel. This means that the problem is a result of breaking the integral up in this way. (The first line of your post, with just two integrals, has the same feature.)
     
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