- #1

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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?