Two small problems compute OK, but not their sum.

*To*: mathgroup at smc.vnet.net*Subject*: [mg70901] Two small problems compute OK, but not their sum.*From*: aaronfude at gmail.com*Date*: Wed, 1 Nov 2006 03:55:18 -0500 (EST)

Hi, I needed to intergate a linear function times Log of something, and Mathematica would give me a too-compicated answer. So I broke it up into two smaller programs and it worked much better as you can see below. Although the answer is probably correct in both cases, I'm just trying to understand why the difference in appearance is quite so significant. Thanks! Aaron Fude In[31]:= Zero=A*Log[c^2+x^2]/2; One=(B-A)/a*x*Log[c^2+x^2]/2; Assuming[a>0&&c>0, Integrate[Zero, {x, 0, a}]] Assuming[a>0&&c>0, Integrate[One, {x, 0, a}]] Assuming[a>0&&c>0, Integrate[Zero+One, {x, 0, a}]] Out[33]= \!\(1\/2\ A\ \((2\ c\ ArcTan[a\/c] + a\ \((\(-2\) + Log[a\^2 + c\^2])\))\)\) Out[34]= \!\(\(-\(\(\((A - B)\)\ \((\(-a\^2\) - 2\ c\^2\ Log[ c] + \((a\^2 + c\^2)\)\ Log[a\^2 + c\^2])\)\)\/\(4\ a\)\)\)\) Out[35]= \!\(\(\(1\/\(4\ a\ c\)\)\((\(-\[ImaginaryI]\)\ a\ A\ \((a\^2 + c\^2)\)\ HypergeometricPFQ[{1\/2, 1, 1}, {2, 2}, 1 + a\^2\/c\^2] + c\ \((\(( a + \[ImaginaryI]\ c)\)\ \((a\ \((A + B)\) + \[ImaginaryI]\ \((A - \ B)\)\ c)\)\ Log[1 + a\^2\/c\^2] + a\ \((a\ \((A - B)\) - 2\ \[ImaginaryI]\ A\ c\ \((\(-2\) + Log[4])\) + 2\ a\ \((A + B)\)\ Log[c])\))\))\)\)\)