Bug in integrate in in 13.1.0?

Question

Bug introduced in 13.1, and fixed in 13.2.

I'm trying the following locally (13.1.0 for Mac) and getting incorrect result

Quit[];
Integrate[1/(x + 1)^2 Exp[-t 1/(x + 1)^2], {x, 0, \[Infinity]}] /. 
 t -> 3.  (* 0.0575796 *)

This answer is wrong. I get the correct answer 0.504344 when running same expression on Wolfram Cloud (13.2.0).

Is this a bug in 13.1, corrupted local state, or something else?

Because of this, I can't rerun the notebook from this post anymore, any workarounds appreciated.

v11.3 and v13.2.1 get 0.504344, and v13.1 get 0.0575796. — cvgmt, Commented Mar 2, 2023 at 0:18
In 13.2.1 (Windows) I get 1/2 Sqrt[\[Pi]/3] Erf[Sqrt[3]] which evaluates to 0.504344. — ulvi, Commented Mar 2, 2023 at 0:54
The tag numerical-integration suggests you should be using NIntegrate instead of Integrate....Perhaps this should be tagged [tag[calculus-and-analysis]? And if so, perhaps the question should be about the general result, not the numerical value at the floating-point number 3.? — Michael E2, Commented Mar 2, 2023 at 1:51
Why 3. ??? That is float, and will switch to NIntegrate and not produce nice symbolic answer. @MichaelE2 NIntegrate outright gives a warning. 13.2.1 is all good. Fixed bug. — Валерий Заподовников, Commented Mar 2, 2023 at 2:04
The symbolic result of Integrate is incorrect, 3. is there to show how far off it is — Yaroslav Bulatov, Commented Mar 2, 2023 at 2:32

Bob Hanlon · Accepted Answer · 2023-03-02 03:20:55Z

6

$Version

(* "13.1.0 for Mac OS X x86 (64-bit) (June 16, 2022)" *)

Clear["Global`*"]

g[x_] = 1/(x + 1)^2;

The incorrect result is

Integrate[1/(x + 1)^2 Exp[-t 1/(x + 1)^2], {x, 0, ∞}]

(* E^(2 - t)/(-1 + E^2) *)

To get the correct result, do a change of variables

IntegrateChangeVariables[
  Inactive[Integrate][1/(x + 1)^2 Exp[-t 1/(x + 1)^2], {x, 0, ∞}], 
  u, u == x + 1]

% // Activate

(* (Sqrt[π] Erf[Sqrt[t]])/(2 Sqrt[t]) *)

Compare with,

WolframAlpha["Integrate[1/(x+1)^2 Exp[-t 1/(x+1)^2], {x, 0, ∞}]", \
{{"Input", 1}, "Content"}]

answered Mar 2, 2023 at 3:20

Bob Hanlon

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$\begingroup$ It is simpler to just update to 13.2.1. $\endgroup$
– Валерий Заподовников
Commented Mar 2, 2023 at 3:32
2

$\begingroup$ @ВалерийЗаподовников while this is one way to think about the situation, finding work-arounds is good practice. Also, Bob gave a nice and very simple answer, so this comment is largely immaterial $\endgroup$
– bmf
Commented Mar 2, 2023 at 4:29

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bmf · Accepted Answer · 2023-03-02 04:23:56Z

3

Another way of doing the integral.

g[x_] = 1/(x + 1)^2;

Integrate`InverseIntegrate[
  1/(x + 1)^2 Exp[-t 1/(x + 1)^2], {x, 0, ∞}] /. t -> 3.

0.504344

which means that the above gives the correct symbolic result

Integrate`InverseIntegrate[
 1/(x + 1)^2 Exp[-t 1/(x + 1)^2], {x, 0, ∞}]

answered Mar 2, 2023 at 4:23

bmf

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2 Answers 2