- #1
Mbert
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Homework Statement
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]
Homework Equations
see above
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:
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:
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?