Upon evaluating the integral
Integrate[Exp[-x^(2 n)] x^(2 k), {x, -\[Infinity], \[Infinity]},
Assumptions -> {n > 0, k > 0, (n | k) \[Element] Integers}]
I get the result ($Version
="11.1.0 for Mac OS X x86 (64-bit) (March 16, 2017)"
)
ConditionalExpression[
(((-1)^(2 n))^(-((1 + 2 k)/(2 n))) ((-1)^(2 k) + ((-1)^(2 n))^((1 + 2 k)/(2 n)))
Gamma[(1 + 2 k)/(2 n)])/(2 n),
C[1] \[Element] Integers &&
n < 1/4 + C[1] && C[1] >= 1 && C[1] < 1/4 + n]
My question is: what is C[1]
doing there? It's a definite integral, so there should be no constant of integration. Moreover, it doesn't show up in the "value" part of the ConditionalExpression
. Putting the inequalities with n
together just gives n < 1/2 + n
, which is always true, so I think the only thing that it enforces is that (saturating the bound on C[1]
), 1 ≤ 1/4 + n
so that n
must be at least 1 (if it's going to be an integer). But I already told Mma that it's a positive integer, so in the end those conditionals should all be satisfied. So: what's the point of C[1]
in this result?
As an added (but not super-important) follow-up, if I tell Mma that n
and k
are integers, why doesn't it simplify (-1)^(2 k)
or the other squares of -1?
Update as of Version 13.0
As of $Version == "13.0.0 for Mac OS X x86 (64-bit) (December 3, 2021)"
the result I get is
(((-1)^(2 n))^(-((1+2 k)/(2 n))) ((-1)^(2 k)+((-1)^(2 n))^((1+2 k)/(2 n))) Gamma[(1+2 k)/(2 n)])/(2 n)
which, at least, has eliminated the unneeded mystery constant. Of course, it's still not leveraging all of the assumptions, since
FullSimplify[%,
Assumptions -> {n > 0, k > 0, (n | k) \[Element] Integers}]
yields Gamma[(1+2 k)/(2 n)]/n
.
So, why doesn't Integrate
actually leverage the assumptions?
C[1]
is not an integration constant, it is part of the conditional onn
$\endgroup$n
that's redundant with what I already specified? $\endgroup$n
must be +/- 1/4 of an integer >=1 , ie true for all integers > 0. FWIW v10.1 just gives the (unsimplified) expression, not conditional. $\endgroup$