In Sage 8.6 (Mac OS):
sage: integral(ln(1 + sqrt(x))/sqrt(x), x, 1, 4)
4*log(3) + 2*log(4/3) - 2
sage: n(_)
2.96981329957600
while
sage: numerical_integral(ln(1 + sqrt(x))/sqrt(x), 1, 4)
(1.8190850097688764, 2.0195900615958844e-14).
By hand, the correct result is 6*ln(3) - 4*ln(2) - 2.
Thanks for reporting, this is indeed a bug.
Sage's default integration engine is Maxima and gets this wrong, maybe because we set up maxima to work in the complex domain.
sage: integral(ln(1 + sqrt(x))/sqrt(x), x, 1, 4)
4*log(3) + 2*log(4/3) - 2
sage: integral(ln(1 + sqrt(x))/sqrt(x), x, 1, 4, algorithm='maxima')
4*log(3) + 2*log(4/3) - 2
You can specify which engine you want Sage to call to compute the integral, and the other engines get this right:
sage: integral(ln(1 + sqrt(x))/sqrt(x), x, 1, 4, algorithm='sympy')
6*log(3) - 4*log(2) - 2
sage: integral(ln(1 + sqrt(x))/sqrt(x), x, 1, 4, algorithm='giac')
6*log(3) - 4*log(2) - 2
sage: integral(ln(1 + sqrt(x))/sqrt(x), x, 1, 4, algorithm='fricas')
6*log(3) - 4*log(2) - 2
(the last one depends on installing FriCAS, which can be done by running
sage -i fricas
in a terminal).
[ Not really an answer, but a complement to Serge's answer, in order too pinpoint the answer ]
A quick check:
f(x)=log(sqrt(x)+1)/sqrt(x)
def tst1(alg="maxima"):
ad=f(x).integrate(x, algorithm=alg)
check_ad=bool(ad.diff(x)==f(x))
ub=ad.limit(x=4)
lb=ad.limit(x=1)
di=(ub-lb).simplify()
did=f(x).integrate(x,1,4, algorithm=alg)
check_di=bool(di==did)
ndi=di.n()
# return [ad, check_ad, lb, ub, di, did, check_di, ndi]
return [alg, check_ad, di, did, check_di]
T=table(rows=[tst1(alg=u) for u in ["maxima", "sympy", "giac", "fricas"]],
header_row=["Algorithm", " AD checks", "Def. int. through AD",
"Def. int. directly", "Def. int. checks"])
shows that the antiderivative (aka primitive, AD) is correct (i.e. derivates back to f(x)) in all cases (and that its limits are correctly evaluated at each bound), but that the direct computation of the definite integral is wrong in Maxima :
Algorithm AD checks Def. int. through AD Def. int. directly Def. int. checks
+-----------+------------+-------------------------+---------------------------+------------------+
maxima True 6*log(3) - 4*log(2) - 2 4*log(3) + 2*log(4/3) - 2 False
sympy True 6*log(3) - 4*log(2) - 2 6*log(3) - 4*log(2) - 2 True
giac True 6*log(3) - 4*log(2) - 2 6*log(3) - 4*log(2) - 2 True
fricas True 6*log(3) - 4*log(2) - 2 6*log(3) - 4*log(2) - 2 True
This problem has been encountered more than once, is known to upstream, but does not seem to have been formally reported per se.
Thanks for your answers. But now, I just wonder if I shall always use sympy algorithm when computing integrals !
Asked: 2019-03-11 05:37:06 +0100
Seen: 457 times
Last updated: Mar 17 '19
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Does 6n(3) - 4ln(2) - 2 mean 6ln(3) - 4ln(2) - 2? If so, it will be same with 1.8190850097688764.