# Why does Expectation function compute, while explicit integration does not?

The code:

Expectation[1 + 1/y, {y \[Distributed] ExponentialDistribution[1]}]


outputs

1 - EulerGamma


While,

Integrate[(1 + 1/y) Exp[-y], {y, 0, \[Infinity]}]


does not compute. Presumably Expectation is using something that I am not giving to Integrate. Something like domain specifications, like in this example. Maybe whether or not it includes y=0? Any suggestions for how to make this consistent? I presume Expectation is correct, but I was very surprised to see a result since Integrate[1/y Exp[-y], {y, 0, \[Infinity]}] of course does not converge for intervals starting at 0, so why would adding 1 help?!

• Integrate[(1 + 1/y) (E^(-y)), {y, 0, \[Infinity]}, GenerateConditions -> False] – ciao Jun 16 '16 at 21:54
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• Using NIntegrate, it is fairly clear that the integral does not converge. I guess that Expectation is using some formula that is outside its domain of validity. – mikado Jun 16 '16 at 22:34
• @mikado - I'd expect and expectation of infinity... and in fact Mean@TransformedDistribution[ 1 + 1/y, {y \[Distributed] ExponentialDistribution[1]}] gives just that. – ciao Jun 16 '16 at 23:20
• Thanks for the responses. I still don't know what the are conditions whose removal gives the 1 - EulerGamma answer, but common sense concensus is that it must be wrong. – puelmato Jun 17 '16 at 7:24