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Double integral of a Log (natural)

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    I wish to find the following integral over the rectangle [-a,a] in u and [-b,b] in v using Mathematica. The constants a and b are positive (and non-zero). The variables x and y are in Reals.

    [tex]A(x,y)=\int_{-a}^a{\int_{-b}^b{\log{\left[(u-x)^2+(v-y)^2\right]}{\rm d}v}{\rm d}u}[/tex]

    2. Relevant equations

    see above

    3. The attempt at a solution

    I find using Mathematica a different integral depending if I integrate on u or v first. I use the following code:

    To integrate on v first:
    Code (Text):
    Integrate[Log[(u - x)^2 + (v - y)^2], v]
    (% /. v -> b) - (% /. v -> -b)
    Integrate[%, u]
    (% /. u -> a) - (% /. u -> -a)
    To integrate on u first:
    Code (Text):
    Integrate[Log[(u - x)^2 + (v - y)^2], u]
    (% /. u -> a) - (% /. u -> -a)
    Integrate[%, v]
    (% /. v -> b) - (% /. v -> -b)
    The difference between these two (definite) double integrals is:
    [tex](x-y) (x+y) \left(\text{ArcTan}\left[\frac{a-x}{b-y}\right]+\text{ArcTan}\left[\frac{a+x}{b-y}\right]+\text{ArcTan}\left[\frac{b-y}{a-x}\right]+\text{ArcTan}\left[\frac{b-y}{a+x}\right]+\text{ArcTan}\left[\frac{a-x}{b+y}\right]+\text{ArcTan}\left[\frac{a+x}{b+y}\right]+\text{ArcTan}\left[\frac{b+y}{a-x}\right]+\text{ArcTan}\left[\frac{b+y}{a+x}\right]\right)[/tex]

    Could someone please tell me what I am doing wrong?
     
  2. jcsd
  3. Oct 25, 2011 #2
    You know, there is a relation
    [tex] \arctan(\frac{1}{x}) = \frac{\pi}{2} sgn(x) - \arctan(x) [/tex]
    By looking at your fractions, you will see that there is always one that is reciprocal to another. So I'm guessing that under certain conditions, you will get 0.

    I'm saying under certain conditions, because I suspect the integral to diverge for -a<y<a and -b<x<b and then you can't expect to get 0 anymore, but it's just a guessing, I could be wrong, you should apply the relation and see if and when you get 0.
     
  4. Oct 25, 2011 #3
    Thank you.

    Wouldn't it diverge for y=+ or - b and x=+ or - a? Why would it diverge for values in between? I think this should be defined up to the boundary and the term sgn(x) depends on which side of the boundary we are approaching the limit.
     
  5. Oct 25, 2011 #4
    Well, my idea was just that, since
    [tex] \log 0 = -\infty [/tex]
    the integral might diverge in case the argument of the logarithm becomes zero. And as long as [itex]-a \leq y \leq a[/itex] and [itex]-b \leq x \leq b[/itex], you have this zero in your logarithm. Anyhow, I'm not saying that it is definately divergent. If you want to be sure, you should do the integration manually. Cheers
     
  6. Oct 26, 2011 #5
    It seems to give 2*pi inside (−b<y<b and −a<x<a) and Indeterminate on the boundary. Outside the boundary, it gives 0.
     
  7. Oct 26, 2011 #6
    Does that mean that if I am interested in the domain (−b<y<b and −a<x<a), I can just add the difference between the 2 definite double integrals (which is 2*pi*(x−y)*(x+y))? But then, if I integrate the other double integral first, the difference between them is 2*pi*(y−x)*(y+x)...

    How can I make the definite double integral give the same result for (−b<y<b and −a<x<a) whether I integrate u or v first?
     
  8. Oct 26, 2011 #7
    Well, it certainly seems like there is something fishy if x,y are inside the boundary.
    Can you check if you get a finite result for x,y inside the boundary?
     
  9. Oct 26, 2011 #8
    The result of both integrals inside the boundary is finite. The result inside the boundary of the difference between them is 2*pi*(x−y)*(x+y) or 2*pi*(y−x)*(y+x), depending on which is integrated first.
     
  10. Oct 26, 2011 #9
    Can you post what you got for one of the integrals?
     
  11. Oct 26, 2011 #10
    [tex]\begin{align*}A(x,y,a,b)=-12 a b &+(a-x) (b-y) \text{Log}\left[(a-x)^2+(b-y)^2\right]\\ &+(a+x) (b-y) \text{Log}\left[(a+x)^2+(b-y)^2\right]\\ &+(a-x) (b+y) \text{Log}\left[(a-x)^2+(b+y)^2\right]\\ &+(a+x) (b+y) \text{Log}\left[(a+x)^2+(b+y)^2\right]\\ &+(a-x)^2 \left(-\text{ArcTan}\left[\frac{a-x}{b-y}\right]+\text{ArcTan}\left[\frac{-a+x}{b+y}\right]\right)\\ &+(a+x)^2 \left(-\text{ArcTan}\left[\frac{a+x}{b-y}\right]-\text{ArcTan}\left[\frac{a+x}{b+y}\right]\right)\\ &-(b-y)^2 \left(\text{ArcTan}\left[\frac{b-y}{a-x}\right]+\text{ArcTan}\left[\frac{b-y}{a+x}\right]\right)\\ &-(b+y)^2 \left(\text{ArcTan}\left[\frac{b+y}{a-x}\right]+\text{ArcTan}\left[\frac{b+y}{a+x}\right]\right)\\ &+x^2\left(-\text{ArcTan}\left[\frac{a-x}{b-y}\right]-\text{ArcTan}\left[\frac{a+x}{b-y}\right]-\text{ArcTan}\left[\frac{b-y}{a-x}\right]-\text{ArcTan}\left[\frac{b-y}{a+x}\right]+\text{ArcTan}\left[\frac{-a+x}{b+y}\right]-\text{ArcTan}\left[\frac{a+x}{b+y}\right]-\text{ArcTan}\left[\frac{b+y}{a-x}\right]-\text{ArcTan}\left[\frac{b+y}{a+x}\right]\right)\end{align*}[/tex]
     
  12. Oct 26, 2011 #11
    If I integrate with respect to the other variable first, the last term is multiplied by y^2 instead of x^2.
     
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