Re: Another strange bug in Mathematica 4.0's Integrate

*To*: mathgroup at smc.vnet.net*Subject*: [mg25438] Re: Another strange bug in Mathematica 4.0's Integrate*From*: Hendrik van Hees <h.vanhees at gsi.de>*Date*: Sun, 1 Oct 2000 02:44:22 -0400 (EDT)*References*: <8q21lp$hoh@smc.vnet.net> <ztgz5.89556$Zh6.152771@ralph.vnet.net> <8qlo62$q42@smc.vnet.net> <8r19m2$hv3@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Ok, then please tell me, _how_ Integrate is defined in Mathematica because this will help me a lot in more difficult problems. Let me again cite the really strange behaviour for this very simple Integral: Mathematica 4.0 for Students: Linux Version Copyright 1988-1999 Wolfram Research, Inc. -- Motif graphics initialized -- In[1]:= Integrate[Sqrt[(r-x)(r+x)],{x,-r,r},Assumptions->{r>0}] 2 Pi r Sqrt[r ] Out[1]= ------------- 4 Of course the result should be twice as big and this result you get indeed from a _mathematically identical expression for the integrand_: In[2]:= In[2]:= Integrate[Sqrt[r^2-x^2],{x,-r,r},Assumptions->{r>0}] 2 Pi r Out[2]= ----- 2 Ok. Let's look what happens with the indefinite integrals: x Sqrt[(r - x) (r + x)] Out[3]= ----------------------- - 2 2 Sqrt[r - x] x r Sqrt[(r - x) (r + x)] ArcTan[--------------------] (-r + x) Sqrt[r + x] > ----------------------------------------------------- 2 Sqrt[r - x] Sqrt[r + x] In[4]:= Integrate[Sqrt[r^2-x^2],x,Assumptions->{r>0}] 2 2 2 x Sqrt[r - x ] r ArcTan[---------------] 2 2 2 2 x Sqrt[r - x ] -r + x Out[4]= --------------- - -------------------------- 2 2 This explains a lot! As you see in the first expression Mathematica splits the square root! I guess for the definite integral the boundaries of the integral are naively put in and of course at the boundaries in the first expression you are running in trouble because you are evaluating the square root along the branch cut which runs along the negative real axis. If Mathematica would handle branch cuts and so on correctly it should at least print a warning that it is running into that problems. My only way out in really serious calculations is to take the indefinite integrals (checking by differentiation if they are correct) and then to take the correct expression by putting it on the right Riemann sheet by hand. -- Hendrik van Hees Phone: ++49 6159 71-2751 c/o GSI-Darmstadt SB3 3.183 Fax: ++49 6159 71-2990 Planckstr. 1 mailto:h.vanhees at gsi.de D-64291 Darmstadt http://theory.gsi.de/~vanhees/index.html