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ostensibly discontinuous antiderivative∗
pahio†
2013-03-22 2:11:30
The real function
x 7→
1
5 − 3 cos x
(1)
is continuous for any x (the denominator is always positive) and therefore it has
an antiderivative, defined for all x. Using the universal trigonometric substitution
1−t2
2dt
x
cos x :=
,
dx =
,
t = tan ,
2
2
1+t
1+t
2
we obtain
5(1+t2 ) − 3(1−t2 )
2(1+4t2 )
=
,
5 − 3 cos x =
1+t2
1+t2
whence
Z
Z
dt
1
1
x
dx
=
=
arctan
2t
+
C
=
arctan
2
tan
+ C.
5 − 3 cos x
1+4t2
2
2
2
This result is not defined in the odd multiples of π, and it seems that the function
(1) does not have a continuous antiderivative.
However, one can check that the function
x 7→
x 1
sin x
+ arctan
+C
4 2
3 − cos x
(2)
is everywhere continuous and has as its derivative the function (1); one has
sin x 1
1
π
3 − cos x 5 3−1 = 2 < 2 .
References
[1] Ernst Lindelöf: Johdatus korkeampaan analyysiin. Fourth edition.
Werner Söderström Osakeyhtiö, Porvoo ja Helsinki (1956).
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