I would like to move from Mathematica to Sympy. Unfortunately, I couldn't
calculate several integrals on Sympy, which Mathematica calculates easily. For
example, Mathematica analytically integrate the Green function (gf) over X
and Y
norm=sqrt(X**2+Y**2); gf=1/ norm
giving simple expression
expr=Y*log(X+ norm) + X*log(Y+ norm).
I was lucky enough to prove this result on Sympy, using differentiation and
simplifications:
import sympy
X, Y = sympy.symbols("X, Y", real=True)
from sympy import log, sqrt, diff
norm=sqrt(X**2+Y**2)
expr=Y*log(X+ norm) + X*log(Y+ norm)
sympy.radsimp(sympy.simplify(diff(expr, X, Y)))-1/norm =0
My naïve attempt to calculate the integral
sympy.integrate(1/norm, X, Y)
failed, since Sympy return the following expression
X*Integral(Abs(Y)*asinh(Abs(X)/Abs(Y))/Y, Y)/Abs(X)
Then I tried to simplify calculation of the integral, subdividing it to
two following integrals
Int1= sympy.integrate(1/norm, X)
Int2= sympy.integrate(Int1, Y).
Sympy calculated Int1 as
X*Abs(Y)*asinh(Abs(X)/Abs(Y))/(Y*Abs(X))
And Int2 returned unevaluated in the form
X*Integral(Abs(Y)*asinh(Abs(X)/Abs(Y))/Y, Y)/Abs(X)
I recalculate Int1 by Mathematica and have got results expressed in terms
of logarithmic function Int1=log(X+ norm)
(rather than asinh as Sympy does). And, finally, I have calculated
sympy.integrate(log(X+ norm), Y)
and have got results in the form
Y*log(X + sqrt(X**2 + Y**2)) + Y*Abs(X)*asinh(Abs(Y)/Abs(X))/Abs(Y) - Y
I feel that this result is equal to the Mathematica result Y*log(X+ norm) +
X*log(Y+ norm), but failed to prove it.
May be somebody knows, how to calculate integrals
sympy.integrate(1/norm, X), sympy.integrate(1/norm, X, Y)
in terms of logarithmic functions?
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