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Mathematica 10.1.0 returns a strange result for the following double integral

f = Integrate[ 1/(1 + x^2 y^2), {x, 0, ∞}, {y, 0, ∞}]

(* Out[1407]= \!\(
\*SubsuperscriptBox[\(∫\), \(0\), \(∞\)]\(
\*FractionBox[\(π\), \(2\ x\)] \[DifferentialD]x\)\) *)

in Latex:

$$\int_0^{\infty } \frac{\pi }{2 x} \, dx$$

There is no warning that the result might be infinite or not existent. Applying Simplify[] or FullSimplify[] to $f$ does not help. Numerically, however, the divergence is revealed

f // N

During evaluation of In[1418]:= NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. >>

and further error Messages which I have dropped here.

(* Out[1418]= 366.404 *)

I suggest to consider this behaviour a bug.

Taking finite integration regions reveals the logarithmic divergnce of the double integral:

fi = Integrate[ 1/(1 + x^2 y^2), {x, 0, t}, {y, 0, t}, Assumptions -> t > 0]

(* Out[1411]= 1/2 I (PolyLog[2, -I t^2] - PolyLog[2, I t^2]) *)

Series[fi, {t, ∞, 2}] // Normal

(* Out[1416]= 1/t^2 - π Log[1/t] *)

Limit[fi, t -> ∞]

(* Out[1400]= ∞ *)
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  • 1
    $\begingroup$ I do see an Integral::idiv message (in fact two of them). $\endgroup$ – Szabolcs Feb 20 '17 at 12:30
  • $\begingroup$ Mathematica 11.0.1 gives "Integral of π/(2 x) does not converge on {0,∞}." $\endgroup$ – user36273 Feb 20 '17 at 12:33
  • $\begingroup$ 10.1 gives error message too. Curious, whatever version does not give an error, what does it do if you directly evaluate Integral[1/x,{x,0,Infinity]? $\endgroup$ – george2079 Feb 20 '17 at 15:45
  • $\begingroup$ @george2079 1) Again: my version "10.1.0 for Microsoft Windows (64-bit) (March 24, 2015)" does NOT give errors message. If your 10.1 does, is must be a different version. 2) on your test case it says "Integral of 1/x does not converge on {0,[Infinity]}." 3) version 5.2 as well as 8 correctly give this error message (besides other messages) on the original double integral. Hence it seems to be just "bad luck" in version 10.1.0. $\endgroup$ – Dr. Wolfgang Hintze Feb 20 '17 at 23:02
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One may use the definition of the double improper integral (see Multivariable improper integrals in https://en.wikipedia.org/wiki/Improper_integral ). Let us calculate

r = Integrate[1/(1 + x^2*y^2), {x, 0, a}, {y, 0, b},Assumptions -> a > 0 && b > 0]

1/2 I (PolyLog[2, -I a b] - PolyLog[2, I a b]

If the improper integral under consideration exists, the one equals $$\lim\limits_{a\to \infty,\,b\to \infty} r $$ and this limit is finite and is equal to the iterated limit $$\lim_{b\to\infty}(\lim_{a\to\infty} r).$$ But

Limit[ r, a -> Infinity, Assumptions -> b > 0] 

Infinity

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In:

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

Out:

Infinity

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  • $\begingroup$ Or Integrate[1/(1 + x^2 y^2), {x, y} [Element] Disk[{0, 0}, Infinity, {0, Pi/2}]] $\endgroup$ – user64494 May 6 '17 at 16:55

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