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The primitive function to e−
x2 .
Robert Maˇr´ık
February 27, 2006
The problem to find
Z
e
−x2 dx seems to be easy, but, surprisingly, the converse is true.
c
Robert Maˇr´ık, 2006 ×
Function y = e
−x2 :
The primitive function to e
−x2
exists, but it is not an elementary function.
We find at least Taylor expansion to this primitive function about x = 0.
Since all derivatives of y(x) = e
x are ex, we have
y(
n
)(0) = ex
x
=0
= e0 = 1,
and the n-degree Taylor polynomial is
Tn
(x) = e0 +
n
X
i
=1
y(
n
)(0)
n
!
x
n = 1 +
n
X
i
=1
1
n
!
x
n .
c
Robert Maˇr´ık, 2006 ×
e
x = 1 +
n
X
i
=1
1
n
!
x
n
(Taylor expansion)
e
−x2 = 1 +
n
X
i
=1
(−1)
n 1
n
!
x2
n
(replace x by x2)
Z
e
−x2 dx = x +
n
X
i
=1
(−1)
n
1
n
!(2n + 1)
x2
n
+1
(integrate)
It can be shown that the infinite series
F
(x) = x +
∞
X
i
=1
(−1)
n
1
n
!(2n + 1)
x
2n+1
is for every x ∈ R a well-defined differentiable function and F
′ (x) = e−
x2 . Hence F (x) is an
antiderivative to e
−x2 . This antiderivative cannot be written in closed finite form using basic
elemetary function. The function
2
√
π
Z
e
−x2 dx =
2
√
π
"
x
+
∞
X
i
=1
(−1)
n
1
n
!(2n + 1)
x2
n
+1
#
is one of the most famous nonelementary functions – the error function.
c
Robert Maˇr´ık, 2006 ×
e
x = 1 +
n
X
i
=1
1
n
!
x
n
(Taylor expansion)
e
−x2 = 1 +
n
X
i
=1
(−1)
n 1
n
!
x2
n
(replace x by x2)
Z
e
−x2 dx = x +
n
X
i
=1
(−1)
n
1
n
!(2n + 1)
x2
n
+1
(integrate)
It can be shown that the infinite series
F
(x) = x +
∞
X
i
=1
(−1)
n
1
n
!(2n + 1)
x
2n+1
is for every x ∈ R a well-defined differentiable function and F
′ (x) = e−
x2 . Hence F (x) is an
antiderivative to e
−x2 . This antiderivative cannot be written in closed finite form using basic
elemetary function. The function
2
√
π
Z
e
−x2 dx =
2