## Version 8.0 integrates but Version 9.0.1 doesn't

4

Bug introduced in 9.0 and fixed in 10.1

I am trying to run the following integral in version 9.0, but it fails:

Integrate[
2 I (-1 +
x) ((-8 + 12 x - 6 mπ^2 x - 2 mπ^2 x^2 -
4 x^3)/((1 - x)^2 +
mπ^2 x) + ((4 - 2 x + 4 mπ^2 x - 8 x^2 +
6 mπ^2 x^2 + 6 x^3) Log[
1/((1 - x)^2 + mπ^2 x)])/((1 - x)^2 + mπ^2 x)), {x, 0,
1}, Assumptions -> mπ > 0 && mπ < .1]


Version 9.0.1:

But in version 8.0, it works:

Is this a bug, or is the integral really non-convergent? What's a possible workaround, since I don't have easy v8 access?

@DanielLichtblau any news on this? – None – 2016-06-23T07:22:18.957

@Loius I get (2*I)*(1 - 3*mπ^2 + 3*mπ^2*(-2*Log[mπ] + mπ*(Sqrt[4 - mπ^2]*(ArcTan[mπ/Sqrt[4 - mπ^2]] - ArcTan[(-2 + mπ^2)/(mπ*Sqrt[4 - mπ^2])]) + mπ*Log[mπ]))) in my current build. – Daniel Lichtblau – 2016-06-23T15:49:12.470

10.1 gives a result about 6 lines long with a bunch of Log terms, that numerically seems to be equivalent to Dan's expression. – george2079 – 2016-11-16T20:06:51.257

BTW shouldn't there a space between m and pi, or is this intentional? – Sjoerd C. de Vries – 2013-03-08T22:59:23.510

$m\pi$ represents a single symbol. – QuantumDot – 2013-03-08T23:00:56.550

I can reproduce this, but in future, please do not use the bugs tag until a few users have confirmed the behaviour. – rm -rf – 2013-03-08T23:03:32.693

Thanks for reproducing; I have removed the bugs tag. – QuantumDot – 2013-03-08T23:04:09.590

Got the same as you both in 8 and 9, and the integral doesn't look non-convergent. – Sjoerd C. de Vries – 2013-03-08T23:13:51.703

1Thanks god 2πR can't be used as a symbol name – Dr. belisarius – 2013-03-08T23:21:14.380

Notice that the indefinite integration int = Integrate[..., {x,0,1}] works and then doing (int /. x -> 1) - (int /. x -> 0) will reproduce the result from version 8. – b.gates.you.know.what – 2013-03-09T07:58:31.987

Appears to be working again in the development Mathematica kernel. I filed a bug report so it should become a test in case. Which might be useful if it manages to rebreak itself. – Daniel Lichtblau – 2013-03-09T21:08:32.533