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Integration in terms of exponential integrals and
incomplete gamma functions
Waldemar Hebisch
Wrocªaw University
[email protected]
13 July 2016
Introduction
What is FriCAS?
FriCAS is an advanced computer algebra system
forked from Axiom in 2007
about 30% of mathematical code is new compared to Axiom
about 200000 lines of mathematical code
math code written in Spad (high level strongly typed language
very similar to FriCAS interactive language)
runtime system currently based on Lisp
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Some functionality added to FriCAS after fork:
several improvements to integrator
limits via Gruntz algorithm
knows about most classical special functions
guessing package
package for computations in quantum probability
noncommutative Groebner bases
asymptotically fast arbitrary precision computation of elliptic
functions and elliptic integrals
new user interfaces (Emacs mode and Texmacs interface)
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FriCAS inherited from Axiom good (probably the best at that time)
implementation of Risch algorithm (Bronstein, ...)
strong for elementary integration
when integral needed special functions used pattern matching
Part of motivation for current work came from Rubi testsuite: Rubi
showed that there is a lot of functions which are integrable in terms
of relatively few special functions.
Adapting Rubi looked dicult
Previous work on adding special functions to Risch integrator
was widely considered impractical
My conclusion:
Need new algorithmic approach. Ex post I claim that for considered
here class of functions extension to Risch algorithm is much more
powerful than pattern matching.
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Informally
exp(u)/u
Γ(a, u ) = u u
exp(−u)
for a = (m + k )/k and −k < m < 0.
Our extension is very natural:
log(x − x ) = x −1 x
exp(x − x )
Ei(x − x ) =
x −x
Exponential integral plays the same role for integrals containing
exponential function as logarithm plays for rational functions!
Incomplete Gamma functions is has related role: it allows
integrating functions of form x exp(x ).
Ei (u )0 = u 0
0
0
0
0
/
m k
0
0
0
0
0
k
+m−1
k
Simple examples
integrate(exp(x)/x^2, x)
−e x + x Ei (x )
x
Needs preprocessing (Hermite reduction) to see Ei term.
integrate((x-1)*exp(x)/x^2, x)
e
x
x
Similar looking, but elementary.
integrate(exp(x)/(x^2 - 2), x)
2 + x e − Ei √2 + x 2 √2 e
Single exponential and denominator irreducible over integers, but
two Ei terms.
√
Ei −
√
2
2
√
2
x*exp(x^2)/(x^4-5)
√
5 Ei
5
5
5
√
√
√ 2 √
−
+ x2 e 5 −
Ei
+ x2
20 e
Similar, but relation between complex factors of denominator and
Ei terms is more complicated.
√
5
integrate(((x+1)*exp(x))/log(x*exp(x)), x)
li (x e )
x
This shows that result may appear as logarithmic integral.
integrate(((x+1)*exp(x))/(x + log(x)), x)
li (x e )
x
This shows that system can perform necessary simplications.
integrate(exp(-x^2+2*x), x)
1
e erf (x − )
2
D(%, x)
e e−
x
erf
term with a shift.
2 +2
x
√
π
−1
integrate((2*x+4)*exp(-x^2-2*x-1)/(x+1), x)
√
erf (x + ) π + Ei −x 2 − x −
2 1
This shows that Ei and erf may be mixed together.
1
integrate((1 + exp(x))*exp(exp(x)^2), x)
√
π er (e x ) + Ei e x 2
2
Similar, but no denominator to signal presence of Ei term. Also, we
get er when appropriate.
integrate(x^3*exp(-x^3), x)
−Γ
1
3,
Gamma plus elementary part.
3
x 3 − x e−
9
x
3
integrate((x+1)*exp(-x^3-3*x^2-3*x), x)
e Γ 23 , x 3 + x 2 + x +
−
Gamma with a shift.
3
3
3
1
integrate(x^2*exp(-(x+1)^3), x)
Γ 23 , x 3 + x 2 + x + − Γ 13 , x 3 +
3 3 1
3 x + 3 x + 1 − e
2
3
Single term in the input, two Gamma terms plus elementary part in
the output.
integrate(sin(x+1)/x^2, x)
2
−x 3 −3
x
2 −3
2 sin (x + 1) − 2 x sin (1) Si (x ) + x cos (1) Ci (x ) + x cos (1) Ci (−x )
2x
Si and Ci terms with shift.
−
integrate(x*sin(x^4), x)
fresnelS
can generate Fresnel integrals.
2
x2
q
2
π
q 2
π
integrate(exp(x + sqrt(x+1)), x)
2e
5
4
e
√
x
+1+x
−
√
π er
2
√
x
+1+1
2
2e
we can handle some algebraics by cheating (change of variable).
5
4
f := (2*x^4-x^3+3*x^2+2*x+2)*exp(x/(x^2+2))/(x^3+2*x)
integrate(f, x)
2
x
2
+
2
x + e
+ Ei
x2 +
2
Hard to separate elementary and Ei part.
x
x
2
Peek inside
)trace INTEF )math
integrate(sin(x^2), x)
1<enter ElementaryIntegration.lfintegrate,43 :
2 2
%i x
- %i (%e
) + %i
arg1= -------------------2
%i x
2%e
arg2= x
1>exit ElementaryIntegration.lfintegrate,43 :
+---+
| 2
fresnelS(x |--- )
\|%pi
Theoretical background
If K is a dierential eld, v , u ∈ K and v = u /u we say that u is
exponential element and that v is logarithmic element. The set of
exponential elements in K is an abelian group with respect to
multiplication. The set of logarithmic elements in K is an abelian
group with respect to addition.
Main examples:
explicit exponential is exponential element
argument of logarithm is exponential element
if a is an exponential element, then a is also an exponential
element
Example: If K = F (θ), θ is Lambert W over F , that is θ is
transcendental and there exist z ∈ F such that θ = , then
θ/z is an exponential element in K and −θ is corresponding
logarithmic element.
0
0
I
I
I
k
0
z
0
z θ
(θ+1)
Theorem
If K is nitely generated dierential eld, then group of logarithmic
elements modulo additive constants is a nite rank free abelian
group. The same for group of exponential elements modulo
multiplicative constants.
In our elds we can eectively test if element is an exponential
element and nd associated logarithmic element and vice versa.
Gamma extensions
We say that a dierential eld E is a gamma extension of F if there
exists θ , . . . , θ ∈ E such that E = F (θ , . . . , θ ) and for each i ,
1 ≤ i ≤ n, one of the following holds
1. θ is algebraic over F
2. = u for some u ∈ F
3. θ = for some u ∈ F
4. θ = for some u, v ∈ F such that u = v u
5. there are w , u, v ∈ F and integers k , m such that θ = wu ,
u = v u , w = v , −k < m < 0.
where F = F (θ , . . . , θ ).
1
1
n
i
n
i
θi0
θi
0
0
u
i
0
i
u
0
u
i
v
i
0
i
i
0
0
i
k
1
m
i
−i
0
0
0
i
0
Liouville principle
Let f ∈ K (θ) where θ is a exponential or a primitive. Under
appropriate technical assumptions, if f has integral in a Gamma
extension of K (θ), then there is an algebraic extension K̄ of K such
that
X
X v
u X
cv +
cvwu
(1)
f =v +
c +
v
v
where c ∈ K are constants, u = v u , v ∈ K (θ) for i ∈ I ∪ I ,
u ∈ K (θ) for i ∈ I , u , w ∈ K̄ (θ), w u ∈ K (θ) for i ∈ I ,
w =v .
0
0
0
i
i
2
ki
mi
∈I1
i
i
i
i
i
i
i
0
i
i
i
i
i
∈I2
0
0
i
i
0
i
i
i
i
i
i
i
i
i
i
∈I3
2
3
3
The main case is integration of functions of form f θ where θ is an
exponential. We get equation:
f θ = (g θ)0 +
X
v0
i
i
∈I2
uθ X 0
+
vwuθ
v
∈
i
i
i
i
i
I3
i
where v u θ = (u θ) .
We will denote logarithm of θ as v , that is θ = v θ. Then we can
write
X u
X
vwu
(2)
f =g +v g +
v +
v
We will call terms of the sum over I Ei terms and terms of the sum
over I gamma terms. Formula 2 generalises Risch dierential
equation (RDE).
0
i
i
i
0
0
0
0
0
0
i
i
i
i
i
∈I2
2
3
0
i
∈I3
i
i
Parametric RDE
Problem: given f , u nd set of g , c such that
i
i
f0 = g 0 + ug +
X
cf
i i
and c are constants.
We assume availability of solver for parametric RDE.
In generic case we need to nd nite number of candidates for
possible Ei and gamma terms and use solver for parametric RDE.
In few special cases we have innite number of candidate Ei or
gamma terms and determine which terms actually occur during
solving RDE.
i
The following cases may appear:
regular Ei terms, depend only on denominator of f
special Ei terms, depend only on structure of dierential eld
innite family of Ei terms in case of v linear in logarithm
regular gamma terms, depend only on structure of dierential
eld
innite families of gamma (erf ) terms
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General setup
Let ψ be top transcendental in K and M be a dierential subeld
such that K is algebraic over M (ψ). By assumption we can write
v = v + r ψ + a where a ∈ M and r ∈ Q 6= 0 only when ψ is a
logarithm.
When ψ is not an exponential let S be integral closure of M [ψ] in
K . When ψ is an exponential let S be integral closure of
M [ψ, ψ ]. (In general we put inverses of all special elements in
S ). We can write elements of f as quotient with numerator in S
and denominator in M [ψ]. When ψ is an exponential we
additionally may assume that denominator is relatively prime to ψ.
i
i
−1
i
i
i
Notation:
v , = v + rψ + a
r a
is norm of numerator of v .
is norm of numerator of ∂ v .
is norm of numerator of ∂ v .
N,
L
P
,
r a
r
r a
ψ
2
ψ
,
r a
Ei terms
Theorem
If v 0 is an Ei term appearing in formula 2 and p is a prime divisor
such that ord (v ) > . Then ord (E ) > .
Res(N , , E ) =
ui
0
0
Corollary:
0.
Problem: If ψ is a logarithm we need more equations.
Solution: If denominator of v is divisible by two prime divisors we
combine constraints from both divisors.
In transcendental case we just can use Res(N , E ) = 0 and
∂ Res(N , E ) = 0.
i vi
p
i
p
r a
,
r a
ψ
,
r a
Ei cases
Algebraic constants
exp(x )/(x − 5) = e Ei(x − √5) − e Ei(x + √5)
In general, if p is irreducible polynomial with constant
coecients, c are roots of p, d are nonzero
constants and Ei(φ) has denominator
P
Q D , then
d Ei(φ + c ) has denominator (D + c ).
In our method we get polynomial equation q(a) = 0
for possible shifts. Each irreducible factor q of q
corresponds to orbit of single logarithmic element by
algebraic shifts. The second highest coecient of q
gives trace ã of a. Shifting arguments of q by −ã we
get equation for algebraic shifts.
√
2
i
5
0
√
− 5
0
i
0
i
i
0
i
i
i
i
Factors in higher powers
x exp(log(x ) )/(log(x ) + 1) = Ei((log(x ) + 1) ) /(2e )
extra equation comes from condition that v is
divisible by a square.
Overlapped factors
(x (2log(x ) + 3log(x )) + (2log(x ) + 5log(x ) + 2)) exp(log(x ) )
=
log(x )(log(x ) + 1)(log(x ) + 2)
(Ei((log(x ) + 1)(log(x ) + 2)) /e + Ei(log(x )(log(x ) + 1)) .
Actually, no problem for our method but may confuse systems using
partial fraction decomposition.
2
2 0
,
r a
2
2
2
0
2
2
0
Special terms
1
exp(exp(x )/(exp(x )+1)) = Ei e + 1
e +1
When ψ is an exponential than we check if v + a can
reduce to a monomial.
exp(x − log(x )) = Ei (x )
When ψ is a logarithm degree at innity can drop to
0. In transcendental case this can only happen if v is
a polynomial of degree 1 in ψ with integer leading
coecient. In such case we can reduce integral to
integral with v ∈ M . In algebraic case there are
nitely many values of r which may lead to zeros at
innity, we need to handle them separately.
e
x
0
x
0
−e Ei −
x
0
Innite family
exp((c + k ) log(x )) = x exp(c log(x ))
Ei((c + k ) log(x )) =
x log(x )
log(x )
so we get innite family of Ei terms with the same denominator.
Can happen only if v is linear in ψ. We can reduce the problem to
f ∈ M and then to equation
k
0
f˜ =
with constant c .
k
X
c z
k
k
−1
Completing powers
To nd Gamma terms we search for r , v such that v is a power
modulo multiplication by exponentials. Main idea is that
exponentials have no nontrivial divisors, so that if v is divisible by
a nontrivial divisor p and v is a k -th power, then k is divisible by
p . To test for divisibility by power of divisors we use GCD with
derivative. More precisely, if v is divisible by p than N and L
have common factor, that is Res(N , L ) = 0. This works if ψ is
not a logarithm. For k ≥ 3 we get second equation using second
derivative. For k = 2 (that is erf terms) we need to check for
presence of two zeros of multiplicity 2. In case of single zero of
multiplicity 2 there is possibility for innite family ("movable zero").
,
r a
,
r a
,
r a
k
2
,
r a
,
r a
r
,
r a
r
exp(−x
2
− erf (x )2 )
,
,
2,
4
,
0 i
, so
and the corresponding term is
.
ψ = erf (x ) va = −ψ 2 − x 2 + a ∂ψ va = − ψ
Res(va , ∂ψ va ) = (a − x 2 ) Res(va , ∂ψ va ) =
a = x2
2
va = −ψ
erf (erf (x ))0
It may happen that there are no nontrivial zeros:
f = Γ(1/3, exp(x )) = exp(− exp(x ) + x /3).
We have u = exp(− exp(x )), v = − exp(x ) and w = exp(x /3).
Variation: f = erf (exp(3x /2)) = exp(− exp(x ) + 3x /2).
The a term is uniquely determined by condition that v + a contains
only one term. Consequently the gamma term is uniquely
determined and can appear only if v is a monomial or a binomial.
√
√
√
π erf ( x ) = exp(−x )/ x = exp(−x − log(x )/2)
Non reduced case.
0
1
0
0
1
3
1
Movable zeros
There are four cases where we get innite families divisible by a
square of prime divisor:
v = bψ + c ψ + d is quadratic polynomial
p
v = bψ + c + dy where y = e ψ + f
p
v = bψ + c + dy where y = e ψ + f ψ + g
I
2
I
2
I
2
I
v=
2
bψ2 + c ψ + d
ψ+e
In integration context there are extra constraints and they eliminate
the second family. For the three other families we show via
examples that they may appear during integration.
Innite families
log(x ) + k ) = 2x exp(− log(x ) − 2k log(x ) − k )
= 2exp(−k )x
exp(− log(x ) ).
This one is due to Cherry and he also gave method to handle it.
√
π erf (
−1
0
2
2k −1
2
2
2
log(plog(x )) + log(x )/2 ) =
log(log(x ))
2(log(x ) + k ) log(logp(x )) − log(x ) ×
2x log(x ) log(log(x )) log(log(x ))
4
k log(log(x )) + 4k log(x ) log(log(x )) + log(x )
exp
=
4log(log(x ))
((log(x ) + k ) log(log(x )) − log(x )/2) log(x )
x ×
(x )
exp 4loglog(log
− 3log(log(log(x )))/2
(x ))
√
2
π er(
k
0
2
2
k
2
2 −1
k
−1
In general can happen when ψ is a logarithm and
v = (bψ + c ψ + d )/ψ − 3/2log(ψ). Then b is a constant. We
can reduce problem to
2
Aψ + B =
X
v 0w u .
i
i
i
I3
This is the most complicated case. We get a quadratic relation
between a and r . Depending on this relation we have further
subcases:
change of variables reduces problem to dierent case
a is uniquely determined by r . Can use method similar to
quadratic case
P
a is a constant. Need to solve C = d z where d are
unknown constants and z and C are known
can use onePcoordinate of a to parametrise solutions. Need to
solve C = d z z where d are unknown constants, z and
z are known and r is quadratic function of k .
I
I
I
r
r
r
I
k
2
r
k
1 2
r
1
The last subcase is most tricky to solve. We use a single pole of z
to get bound on k . More precisely, the problem is that there may
be two values of k , k such that z (k )z have pole of the same
maximal order and the constants are such that
1
1
r
2
1
ki
2
i
d 1 z1 (k1 )z2 1 + d 2 z1 (k2 )z2 2
r
k
k
r
k
k
is of lower order. However we can look at highest order term in
such combination dierence between order of linear combination
and order of terms is bounded by a constant. Since orders of terms
grow quadratically there is gap between orders of consecutive terms
growing linearly in |k |. So for |k | large enough the can be no
cancellation between highest order term of linear combination
above and lower order terms.
Examples were known, but there were no theory how to handle
them.
If y = plog(x ) − 1, φ = log(x ) + y , v = y , γ = √k − 1/k , then
√
k (φ + γ)
π er( √
) = (k γ(log(x ) − 1) − k (γ log(x ) − 1)y ) ×
φ
2
2
0
exp((2k − 1) log(x ) + 2k γ + y )/pφ =
(k γ(log(x ) − 1) − k (γ log(x ) − 1)y ) x
×
exp(2k γ) exp(y + log(φ)/2)
so we get innite family of erf terms. The example above contains
new constants γ and exp(2k γ). exp(2k γ) is a multiplicative
factor so we can simply omit it. γ plays nontrivial role but taking
trace with respect to γ we get example with rational coecients:
k (−1 + y )x
exp(y + log(φ)/2).
2
2
2k 2 −1
2
2
2k 2 −1
2
In general we have
log 2 with
,
, 1,
. Method of
solving is similar to previous examples. In particular, innite family
is possible only when b, d e and (f − 4eg )/e are constants and
we have linear relation between r and a .
v = bφ + c + dy + (φ)/
y 2 = e ψ 2 + f ψ + g φ = h(ψ + ty ) et 2 = h ∈ M
2
2
2
0
Thank you