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Incorrect answers for improper integrals
Hi guys,thank you for using Hilbert.
You need to execute "postulate zfc_analysis" if you wanna do real analysis.
Enjoy! -> postulate zfc_analysis
success! :)
Enjoy! -> S(x dx)[-oo..oo]
0.0
Enjoy! ->
Since S(x dx)[-oo..oo]
is improper at both endpoints (and the integrand has no discontinuities), the correct way to evaluate this integral is to split up the interval into two pieces: S(x dx)[-oo..0] + S(x dx)[0..oo]
. Neither of these integrals converge, therefore S(x dx)[-oo..oo]
diverges as well.
In a similar vein, we have this:
Hi guys,thank you for using Hilbert.
You need to execute "postulate zfc_analysis" if you wanna do real analysis.
Enjoy! -> postulate zfc_analysis
success! :)
Enjoy! -> S(1/x dx)[-1..1]
oo
Enjoy! ->
This is also incorrect: the integral diverges.
Likewise:
Hi guys,thank you for using Hilbert.
You need to execute "postulate zfc_analysis" if you wanna do real analysis.
Enjoy! -> postulate zfc_analysis
success! :)
Enjoy! -> S(1/x dx)[1..oo]
6.90916718
Enjoy! ->
The most general antiderivative of 1/x
is ln(|x|) + C
which, of course, goes to infinity as x
goes to infinity.
Oh dear...
Hi guys,thank you for using Hilbert.
You need to execute "postulate zfc_analysis" if you wanna do real analysis.
Enjoy! -> postulate zfc_analysis
success! :)
Enjoy! -> S(sin(x) dx)[0..oo]
0.43777752000000003
Enjoy! ->
Hi guys,thank you for using Hilbert.
You need to execute "postulate zfc_analysis" if you wanna do real analysis.
Enjoy! -> postulate zfc_analysis
success! :)
Enjoy! -> S(x^(-0.5) dx)[0..1]
oo
Here's another wrong one:
Hi guys,thank you for using Hilbert.
You need to execute "postulate zfc_analysis" if you wanna do real analysis.
Enjoy! -> postulate zfc_analysis
success! :)
Enjoy! -> S(log(x)/x^(0.5) dx)[0..1]
-oo
Here's another:
Hi guys,thank you for using Hilbert.
You need to execute "postulate zfc_analysis" if you wanna do real analysis.
Enjoy! -> postulate zfc_analysis
success! :)
Enjoy! -> S(1/(1-x^2)^0.5 dx)[0..1]
oo
The correct answer is, of course, Math::PI / 2
.