10
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Integrate[ ArcTan[x]/(1 + x) Log[(1 + x^2)/2], {x, -1, 1}]

I used Mathematica 9.0.1 on Windows7 32bit, Mathematica 9 cannot compute this, but Mathematica 8 gives Pi^3/96, is this a bug?

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  • $\begingroup$ I get the same result with Mathematica 9, but, admittedly, it took 3 times as long as with v8... $\endgroup$
    – sebhofer
    Jun 13, 2013 at 8:32
  • $\begingroup$ I have to correct myself, I now (after quitting the kernel) get Infinity (as pointed out by Artes)... No clue what went wrong before. $\endgroup$
    – sebhofer
    Jun 13, 2013 at 16:21
  • $\begingroup$ @explorer Couldn't Mathematica 9 really return any result or colud it return Infinity as I demonstrated in my answer? Have you tried to integrate TrigToExp@ArcTan[x] ? $\endgroup$
    – Artes
    Jun 16, 2013 at 17:19
  • $\begingroup$ Has this been fixed in Mathematica 10.0.2? Anyone? $\endgroup$
    – user58955
    Dec 29, 2014 at 16:43
  • $\begingroup$ Not fixed in 13. $\endgroup$ Dec 29, 2021 at 3:08

1 Answer 1

9
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This issue reminds many similar problems with Integrate. We have in Mathematica 8.0.4:

Integrate[ ArcTan[x]/(1 + x) Log[(1 + x^2)/2], {x, -1, 1}]
Pi^3/96

However in Mathematica 9.0.1 it takes quite a long time yielding a different result:

Integrate[ ArcTan[x]/(1 + x) Log[(1 + x^2)/2], {x, -1, 1}]
Infinity

This is a bug, we can compare it with the NIntegrate result which yields the number numerically the same as in ver. 8.

Of course the result should be the same if we substitute ArcTan by its equivalent:

TrigToExp[ ArcTan[x] ]
1/2 I Log[1 - I x] - 1/2 I Log[1 + I x]

but now the result is the same as in ver. 8:

Integrate[ TrigToExp[ ArcTan[x]]/(1 + x) Log[(1 + x^2)/2], {x, -1, 1}]
Pi^3/96
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