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1 27th April 10:13
john
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Posts: 1
Default Help with integral.


Hi,

In a problem I've been working on recently, I have the following integral

int_0^{a} int _0^{b} { frac{y}{ x^2 + y^2 - 2xyc } } dx dy

where 0<=c<1

There is a singularity as x,y approach 0. I need to compute this
integral fairly accurately and quickly. I've spent a lot of time trying
to isolate the singularity in a convenient form, but have so far been
unsuccessful. Does anyone have any ideas about how to go about
computing this integral. I haven't been able to find a closed form
expression, but there might be one. Thanks in advance.

John
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2 27th April 10:13
spellucci
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Posts: 1
Default Help with integral.


In article <3281hrF3ivmvhU1@individual.net>,
John <gh14tq5@yahoo.com> writes:

int _0^b { frac{y}{ x^2 + y^2 - 2xyc } } dx =
= 1/sqrt(1-c^2)*(arctan( (b-y*c)/(y*sqrt(1-c^2)) ) + arctan(c/sqrt(1-c^2) )
but integrating this one from 0 to a seems not to be possible in closed form.
maybe this helps
(I hope I got it right)

peter
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3 27th April 10:13
rusin
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Posts: 1
Default Help with integral.


Polar coordinates leads to a rational function of sines and cosines;
the half-angle trick makes it the integral of a rational function;
use partial fractions.

dave
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4 4th May 00:40
paul abbott
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Posts: 1
Default Help with integral.


Mathematica gives

( a ArcTan[(b - a c)/(a Sqrt[1 - c^2])] +
b c ArcTan[(a - b c)/(b Sqrt[1 - c^2])] +
(a + b c) ArcSin[c])/Sqrt[1 - c^2] +
b/2 Log[(a^2 - 2 b c a + b^2)/b^2]
For example, with a -> 1, b -> 2, and c -> 1/2, one obtains

4 Pi/(3 Sqrt[3]) - Log[4/3]

which to 20 decimals is 2.1307170...

Doing the integral numerically,
NIntegrate[y/(x^2 + y^2 - 2 c y x) /. c -> 1/2, {x, 0, 2}, {y, 0, 1}]

2.1307170...

Cheers,
Paul

--
Paul Abbott Phone: +61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
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Crawley WA 6009 mailtoaul@physics.uwa.edu.au
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