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This Integration does not return for hours, though I don't know why as the graph doesn't look all that complicated. Also NIntegrate returns the result instantly... but I need the exact value!

Integrate[1/((1 + x^2015) (1 + x^2)), {x, 0, Infinity}]

Is there any way to simplify it or tell Mathematica to use a more efficient memory map, etc.?

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    $\begingroup$ $\frac{\pi }{4}$ $\endgroup$
    – azerbajdzan
    Commented Apr 28 at 21:02
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    $\begingroup$ But if your original Integrate returned $\pi/4$, would you consider that a proof? $\endgroup$
    – JimB
    Commented Apr 28 at 21:22
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    $\begingroup$ Not complicated? There are 2,017 poles....1,009 real factors...It may not be complicated, but it's quite a large computation. For instance, this quits after 4 sec: MemoryConstrained[Integrate[1/((1 + x^2015) (1 + x^2)), {x, 0, 1, Infinity}], 10^7] $\endgroup$
    – Goofy
    Commented Apr 28 at 22:10
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    $\begingroup$ int = RootApproximant[NIntegrate[1/((1 + x^2015) (1 + x^2)), {x, 0, Infinity}] / Pi] * Pi evaluates to Pi/4 $\endgroup$
    – Bob Hanlon
    Commented Apr 29 at 1:54
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    $\begingroup$ I think you are right about the integral, because I can't reproduce my results from yesterday. I may have had the value a assigned without realizing it. $\endgroup$
    – Bill Watts
    Commented Apr 30 at 4:30

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