Integration in Finite Terms

Integration in ﬁnite terms
Formalizing the question
Applications
Sources
Technicalities

2
Why can’t we do e −x dx?
Non impeditus ab ulla scientia

K. P. Hart

Faculty EEMCS
TU Delft

Delft, 10 November, 2006: 16:00–17:00

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K. P. Hart Why can’t we do e dx?

Applications
Sources
Technicalities

Outline

1 Integration in finite terms
2 Formalizing the question
Differential fields
Elementary extensions
The abstract formulation
3 Applications
Liouville’s criterion
2
e −z dz at last
Further examples
4 Sources
5 Technicalities

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Applications
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Technicalities
2
What does ‘do e −x dx’ mean?

To ‘do’ an (indeﬁnite) integral f (x) dx, means to ﬁnd a formula,
F (x), however nasty, such that F = f .

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2

What is a formula?

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2

What is a formula?
Can we formalize that?

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2

What is a formula?
2
How do we then prove that e −x dx cannot be done?

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What is a formula?

We recognise a formula when we see one.

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What is a formula?

2
E.g., Maple’s answer to e −x dx does not count, because

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What is a formula?

2

1√
π erf(x)
2

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What is a formula?

2

1√
π erf(x)
2
2
is simply an abbreviation for ‘a primitive function of e −x ’

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Technicalities

What is a formula?

2

1√
π erf(x)
2
2
is simply an abbreviation for ‘a primitive function of e −x ’
(see Maple’s help facility).

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What is a formula?

A formula is an expression built up from elementary functions
using only

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What is a formula?

using only
addition, multiplication, . . .

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Technicalities

What is a formula?

using only
other algebra: roots ’n such

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What is a formula?

using only
composition of functions

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Technicalities

What is a formula?

using only
composition of functions
Elementary functions: e x , sin x, x, log x, . . .

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Yes.

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Yes.
Start with C(z) the ﬁeld of (complex) rational functions and
add, one at a time,

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Yes.
add, one at a time,
algebraic elements

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Yes.
add, one at a time,
algebraic elements
logarithms

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Technicalities


Yes.
add, one at a time,
algebraic elements
logarithms
exponentials

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2

We do not look at all functions that we get in this way and check
2
that their derivatives are not e −x .

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2

2
We do establish an algebraic condition for a function to have a
primitive function that is expressible in terms of elementary
functions, as described above.

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Technicalities
2

2
We do establish an algebraic condition for a function to have a
primitive function that is expressible in terms of elementary
functions, as described above.
2
We then show that e −x does not satisfy this condition.

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Formalizing the question Diﬀerential ﬁelds
Applications Elementary extensions
Sources The abstract formulation
Technicalities

Outline

3 Applications
2
e −z dz at last
Further examples
4 Sources
5 Technicalities

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Technicalities

Definition

A differential field is a field F with a derivation, that is, a map
D : F → F that satisfies

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Technicalities

Deﬁnition

D(a + b) = D(a) + D(b)

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Technicalities

Deﬁnition

D(a + b) = D(a) + D(b)
D(ab) = D(a)b + aD(b)

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Technicalities

Main example(s)

The rational (meromorphic) functions on (some domain in) C,
with D(f ) = f (of course).

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Technicalities

Main example(s)

The rational (meromorphic) functions on (some domain in) C,
with D(f ) = f (of course).
We write a = D(a) in any diﬀerential ﬁeld.

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Technicalities

Easy properties

Exercises
(an ) = nan−1 a

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Technicalities

Easy properties

Exercises
(an ) = nan−1 a
(a/b) = (ab − a b)/b 2 (Hint: f = a/b solve (bf ) = a for f )

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Technicalities

Easy properties

Exercises
(an ) = nan−1 a
1 = 0 (Hint: 1 = (12 ) )

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Technicalities

Easy properties

Exercises
(an ) = nan−1 a
1 = 0 (Hint: 1 = (12 ) )
The ‘constants’, i.e., the c ∈ F with c = 0 form a subﬁeld

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Technicalities

Exponentials and logarithms

a is an exponential of b if a = b a

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Technicalities


b is a logarithm of a if b = a /a

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Technicalities


so: a is an exponential of b iﬀ b is a logarithm of a.

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Technicalities


‘logarithmic derivative’:

(am b n ) a b
m bn
=m +n
a a b

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Technicalities


‘logarithmic derivative’:

(am b n ) a b
m bn
=m +n
a a b
Much of Calculus is actually Algebra . . .

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Technicalities

Definition

A simple elementary extension of a differential field F is a field
extension F (t) where t is
algebraic over F ,

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Technicalities

Deﬁnition

algebraic over F ,
an exponential of some b ∈ F , or

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Technicalities

Deﬁnition

algebraic over F ,
a logarithm of some a ∈ F

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Technicalities

Deﬁnition

algebraic over F ,
a logarithm of some a ∈ F
G is an elementary extension of F is G = F (t1 , t2 , . . . , tN ), where
each time Fi (ti+1 ) is a simple elementary extension of
Fi = F (t1 , . . . , ti ).

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Technicalities

Elementary integrals

We say that a ∈ F has an elementary integral if there is an
elementary extension G of F with an element t such that t = a.

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Technicalities

Elementary integrals

We say that a ∈ F has an elementary integral if there is an
elementary extension G of F with an element t such that t = a.
The Question: characterize (of give necessary conditions for) this.

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Technicalities

A characterization

Theorem (Rosenlicht)
Let F be a differential field of characteristic zero and a ∈ F . If a
has an elementary integral in some extension with the same
field of constants then there are v ∈ F , constants c1 , . . . , cn ∈ F
and elements u1 , . . . un ∈ F such that

u1 u
a = v + c1 + · · · + cn n .
u1 un

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Technicalities

A characterization

Theorem (Rosenlicht)
Let F be a differential field of characteristic zero and a ∈ F . If a
has an elementary integral in some extension with the same
field of constants then there are v ∈ F , constants c1 , . . . , cn ∈ F
and elements u1 , . . . un ∈ F such that

u1 u
a = v + c1 + · · · + cn n .
u1 un

The converse is also true: a = (v + c1 log u1 + · · · + cn log un ) .

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Technicalities

Comment on the constants

1
Consider 1+x 2
∈ R(x)

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Technicalities


1
Consider 1+x 2
∈ R(x)
an elementary integral is

1 x −i
ln ,
2i x +i

using a larger ﬁeld of constants: C

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Technicalities


1
Consider 1+x 2
∈ R(x)
an elementary integral is

1 x −i
ln ,
2i x +i

using a larger ﬁeld of constants: C
there are no v , ui and ci in R(x) as in Rosenlicht’s theorem.

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Formalizing the question Liouville’s criterion
R −z 2
Applications e dz at last
Sources Further examples
Technicalities

Outline

3 Applications
2
e −z dz at last
Further examples
4 Sources
5 Technicalities

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R −z 2
Technicalities

When can we do f (x)e g (x) , dx?

Let f and g be rational functions over C, with f nonzero and g
non-constant.

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Technicalities


non-constant.
fe g belongs to the ﬁeld F = C (z, t), where t = e g (and t = gt).

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Technicalities


non-constant.
F is a transcendental extension of C(z).

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R −z 2
Technicalities


non-constant.
F is a transcendental extension of C(z).
If fe g has an elementary integral then in F we must have

u1 u
ft = v + c1 + · · · + cn n
u1 un
with c1 , . . . , cn ∈ C and v , u1 , . . . , un ∈ C(z, t).

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R −z 2
Technicalities

The criterion

Using algebraic considerations one can then get the following
criterion.
Theorem (Liouville)
The function fe g has an elementary integral iﬀ there is a rational
function q ∈ C(z) such that

f = q + qg

the integral then is qe g (of course).

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R −z 2
Technicalities
2
e −z dz

In this case f (z) = 1 and g (z) = −z 2 .

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R −z 2
Technicalities
2
e −z dz

Is there a q such that 1 = q (z) − 2zq(z)?

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R −z 2
Technicalities
2
e −z dz

Is there a q such that 1 = q (z) − 2zq(z)?
Assume q has a pole β and look at principal part of Laurent series
m
αi
(z − β)i
i=1

Its contribution to the right-hand side should be zero.

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R −z 2
Technicalities
2
e −z dz

We get, at the pole β:
m
iαi 2zαi
0= − i+1
−
(z − β) (z − β)i
i=1

Successively: α1 = 0, . . . , αm = 0.

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R −z 2
Technicalities
2
e −z dz

We get, at the pole β:
m
iαi 2zαi
0= − i+1
−
(z − β) (z − β)i
i=1

Successively: α1 = 0, . . . , αm = 0.
So, q is a polynomial, but 1 = q (z) − 2zq(z) will give a mismatch
of degrees.

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Formalizing the question R −z 2 criterion
Liouville’s
Technicalities

Outline

3 Applications
2
e −z dz at last
Further examples
4 Sources
5 Technicalities

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Liouville’s
Technicalities

ez
z dx

Here f (z) = 1/z and g (z) = z, so we need q(z) such that

1
= q (z) + q(z)
z
Again, via partial fractions: no such q exists.

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Liouville’s
Technicalities

ez
z dx

Here f (z) = 1/z and g (z) = z, so we need q(z) such that

1
= q (z) + q(z)
z
Again, via partial fractions: no such q exists.
z u
e e dz = eu du = ln v dv1

(substitutions: u = e z and u = ln v )

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Liouville’s
Technicalities

sin z
z dx

e z −e −z
In the complex case this is just z dz.

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Liouville’s
Technicalities

sin z
z dx

z −z
In the complex case this is just e −e dz.
z
Let t = e z and work in C(z, t); adapt the proof of the main
1
theorem to reduce this to z = q (z) + q(z) with q ∈ C(z), still
impossible.

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Technicalities

Light reading

These slides at: fa.its.tudelft.nl/~hart
J. Liouville.
M´moire sur les transcendents elliptiques consid´r´es comme
e ee
´
functions de leur amplitudes, Journal d’Ecole Royale
Polytechnique (1834)
M. Rosenlicht,
Integration in ﬁnite terms, American Mathematical Monthly,
79 (1972), 963–972.

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Applications
Sources
Technicalities

A useful lemma, I

Lemma
Let F be a differential field, F (t) a differential extension with the
same constants, with t transcendental over F and such that t ∈ F .

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A useful lemma, I

Lemma
Let f (t) ∈ F [t] be a polynomial of positive degree.

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A useful lemma, I

Lemma
Let f (t) ∈ F [t] be a polynomial of positive degree.
Then f (t) is a polynomial in F [t] of the same degree as f (t) or
one less, depending on whether the leading coeﬃcient of f (t) is
not, or is, a constant.

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A useful lemma, II

Lemma
same constants, with t transcendental over F and such that
t /t ∈ F .

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A useful lemma, II

Lemma
t /t ∈ F . Let f (t) ∈ F [t] be a polynomial of positive degree.

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A useful lemma, II

Lemma
for nonzero a ∈ F and nonzero n ∈ Z we have (at n ) = ht n
for some nonzero h ∈ F ;

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A useful lemma, II

Lemma
for nonzero a ∈ F and nonzero n ∈ Z we have (at n ) = ht n
for some nonzero h ∈ F ;
if f (t) ∈ F [t] then f (t) is of the same degree as f (t) and
f (t) is a multiple of f (t) iﬀ f (t) is a monomial (at n ).

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Applications
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Technicalities

sin x
x dz, I

Write F = C(z) and t = e z .

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Technicalities

sin x
x dz, I

If sin z dz were elementary then
z

t2 − 1 u u
= v + c1 1 + · · · + cn n
tz u1 un
with c1 , . . . , cn ∈ C and v , u1 , . . . , un ∈ F (t).

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Technicalities

sin x
x dz, I

If sin z dz were elementary then
z

t2 − 1 u u
= v + c1 1 + · · · + cn n
tz u1 un
By logarithmic diﬀerentiation: the ui ’s not in F are monic and
irreducible in F [t].

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Applications
Sources
Technicalities

sin x
x dz, II

sin z
If z dz were elementary then

t2 − 1 u u
= v + c1 1 + · · · + cn n
tz u1 un

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Applications
Sources
Technicalities

sin x
x dz, II

sin z

t2 − 1 u u
= v + c1 1 + · · · + cn n
tz u1 un
By the lemma just one ui is not in F and this ui is t.

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Applications
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Technicalities

sin x
x dz, II

sin z

t2 − 1 u u
= v + c1 1 + · · · + cn n
tz u1 un
By the lemma just one ui is not in F and this ui is t.
u1
So c1 u1 + · · · + cn un is in F .
un

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Applications
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Technicalities

sin x
x dz, III

Finally, in
t2 − 1 u u
= v + c1 1 + · · · + cn n
tz u1 un
1
we must have v = bj t j and from this: z = b1 + b1 .

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Integration in Finite Terms

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