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i found different results when integrating a special function (see below), depending where i place my assumption (x > 0, x0 > 0). The problem is that the two solutions are not compatible.

In fact, if I perform

FullSimplify[Integrate[(Exp[-((x + x0)/t)] Sqrt[x0/x] BesselI[1, (2 Sqrt[x x0])/t])/t, {t, 0, \[Infinity]}, Assumptions -> {x > 0, x0 > 0}]]

it gives as result

(x + x0 - Abs[x - x0])/(2 x)

While if I evaluate

FullSimplify[Integrate[(Exp[-((x + x0)/t)] Sqrt[x0/x] BesselI[1, (2 Sqrt[x x0])/t])/t, {t, 0, \[Infinity]}],Assumptions -> {x > 0, x0 > 0}]

then the result is

(x + x0)/(2 x)

Do you have any idea, why? Thanks

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    $\begingroup$ Can't reproduce it in version 12.0 under Windows 10 32-bit: I obtain $ \frac{x+\text{x0}}{2 x}$ in both the cases. $\endgroup$ – user64494 Jun 12 at 14:39
  • $\begingroup$ Can't reproduce it in version 12.0 under Windows 8.1 64-bit: I obtain $\frac{x+\text{x0}}{2 x}$ in both the cases. $\endgroup$ – Mariusz Iwaniuk Jun 12 at 16:23
  • $\begingroup$ In version 12.0.0 on MacOSX I confirm the OP's results. The first result seems wrong. $\endgroup$ – Roman Jun 12 at 18:49
  • $\begingroup$ I'm also using MacOSX (v10.14.5) and version 12.0.0. I tried Integrate with the Assumption and without and the results are different. Without the Assumption (putting it outside Integrate), I get ConditionalExpression[1/2 Sqrt[1/(x x0)] Sqrt[x0/x] (x + x0), Re[x + x0] >= 0], whereas with the Assumption inside Integrate, I get (x + x0 - Abs[x - x0])/(2 x). Does the condition Re[x + x0] >= 0 give a clue about why they are different? $\endgroup$ – Mark R Jun 14 at 1:35

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