Definite and indefinite integrals
This functionality does not run in MATLAB.
int(f
,x
) int(f
,x = a .. b
,options
)
int(f, x)
computes the indefinite integral
.
int(f, x = a..b)
computes the definite integral
.
int(f, x)
determines a function F such
that
.
The function F(x) is
called the antiderivative of f(x).
Results returned by int
do not include integration
constants.
For indefinite integrals, int
implicitly
assumes that the integration variable x
is real.
For definite integrals, int
restricts the integration
variable x
to the specified integration interval [a,
b]
of the type Type::Interval
.
If one or both integration bounds a
and b
are
not numeric, int
assumes that a <=
b
unless you explicitly specify otherwise. By default, int
does
not issue warnings about these assumptions. To display the warnings
about using implicit assumptions, set the value of intlib::printWarnings
to TRUE
.
In general, the result of int
is not required
to be valid for all complex values of x
. For example,
the identity
is
only valid for real values of x.
Therefore,
is
also valid only for real values of x.
You can specify your own assumptions of the integration variable.
If these assumptions do not conflict with the default assumption that
the variable is real or with the integration interval, then MuPAD^{®} uses
your assumptions. Otherwise, the int
function issues
a warning and changes the assumptions. Use intlib::printWarnings
to enable or disable
the warnings.
If you compute an indefinite integral and specify properties of the integration variable that describe a subset of the real numbers, MuPAD assumes that the variable is real. Otherwise, the system uses temporary assumption that the integration variable is complex. This assumption holds only during this particular integration.
If you compute a definite integral and specify properties that
conflict with the integration interval, int
uses
the integration interval.
int
can return results with discontinuities
even if the integrand is continuous.
Integration techniques, such as table lookup or Risch integration for an indefinite integral, can add new discontinuities during the integration process. These new discontinuities appear because the antiderivatives can require the introduction of complex logarithms. Complex logarithms have a jump discontinuity when the argument crosses the negative real axis, and the integration algorithms sometimes cannot find a representation where these jumps cancel.
If you compute a definite integral by first computing an indefinite integral and then substituting the integration boundaries into the result, remember that indefinite integration can produce discontinuities. If it does, you must investigate the discontinuities in the integration interval.
If MuPAD cannot find a closedform solution for the integral and cannot prove that such form does not exist, it returns an unresolved integral. In this case, you can approximate the integral numerically or try computing a series expansion of the integral. See Example 2 and Example 3.
You can approximate a definite integral numerically using numeric::int
or float
. Numeric approximation
of a definite integral only works when the float
function can convert the boundaries a
and b
of
the integration interval to floatingpoint
numbers. See Example 2.
int
might not find a closed form of a definite
integral because of singularities of the integrand in the interval
of integration. If the integral does not exist in a strict mathematical
sense, int
returns the value undefined
. In this case, try using assumptions.
Alternatively, use the PrincipalValue
option to
compute a weaker form of a definite integral called the Cauchy principal
value. This form of an integral can exist even though the standard
integral value is undefined. See Example 6.
In general, the derivative of the result coincides with f on a dense subset of the real numbers (or, if you use assumptions on the integration variable, the subset of real numbers specified by these assumptions).
It is not always possible to decide algorithmically if and f are equivalent. The reason is the socalled zero equivalence problem, which in general is undecidable.
int
is sensitive to properties of
identifiers set by assume
.
See Example 6.
Compute the indefinite integrals and :
int(1/x/ln(x), x)
int(1/(x^2  8), x)
Compute the definite integral over the interval [e, e^{2}]:
int(1/x/ln(x), x = exp(1)..exp(2))
When computing definite integrals, you can use infinities as the boundaries of the integration interval:
int(exp(x^2), x = 0..infinity)
You can compute multiple integrals. For example, compute the following definite multiple integral:
int(int(int(1, z = 0..c*(1  x/a  y/b)), y = 0..b*(1  x/a)), x = 0..a)
Use int
to compute this definite integral.
Since int
cannot find a closed form of this integral,
it returns an unresolved integral:
S := int(sin(cos(x)), x = 0..1)
Use the float
function
to approximate the integral numerically:
float(S)
Alternatively, use the numeric::int
function,
which is faster because it does not involve any symbolic preprocessing:
numeric::int(sin(cos(x)), x = 0..1)
Use int
to compute this indefinite integral.
Since int
cannot find a closed form of this integral,
it returns an unresolved integral:
int((x^2 + 1)/sqrt(sqrt(x + 1) + 1), x)
Use the series
function
to compute a series expansion of the integral:
series(%, x = 0)
Alternatively, compute a series expansion of the integrand, and then integrate the result. This approach is faster because it does not try to integrate the original expression. It integrates an approximation (the series expansion) of the original expression:
int(series((x^2 + 1)/sqrt(sqrt(x + 1) + 1), x = 0), x)
The IgnoreAnalyticConstraints
option applies
a set of purely algebraic simplifications including the equality of
sum of logarithms and a logarithm of a product. Using this option,
you get a simpler result, but one that might be incorrect for some
of the values of the variables:
int(ln(x) + ln(y)  ln(x*y), x, IgnoreAnalyticConstraints)
Without using this option, you get the following result, which is valid for all values of the parameters:
int(ln(x) + ln(y)  ln(x*y), x)
The results obtained with IgnoreAnalyticConstraints
might
be not generally valid:
f := int(ln(x) + ln(y)  ln(x*y), x): g := int(ln(x) + ln(y)  ln(x*y), x, IgnoreAnalyticConstraints): simplify([f, g]) assuming x = 1 and y = 1
By default, int
returns this integral as
a piecewise object where every branch corresponds to a particular
value (or a range of values) of the symbolic parameter t
:
int(x^t, x)
To ignore special cases of parameter values, use IgnoreSpecialCases
:
int(x^t, x, IgnoreSpecialCases)
Compute this definite integral, where the integrand has a pole in the interior of the interval of integration. Mathematically, this integral is not defined:
int(1/(x  1), x = 0..2)
However, the Cauchy principal value of the integral exists.
Use the PrincipalValue
option to compute the Cauchy
principal value of the integral:
hold(int)(1/(x  1), x = 0..2, PrincipalValue) = int( 1/(x  1), x = 0..2, PrincipalValue)
For integrands with parameters, int
might
be unable to decide if the integrand has poles in the interval of
integration. In this case, int
returns a piecewisedefined
function or an unresolved integral:
int(1/(x  a), x = 0..2)
int
does not call simplification functions
for its results. To simplify results returned by int
,
use eval
, simplify
, or Simplify
:
Simplify(eval(%))
The resulting piecewise expression has only one branch. If the
parameter a
does not satisfy this condition, the
integral is undefined
.

The integrand: an arithmetical expression representing a function
in 

The integration variable: an identifier 

The boundaries: arithmetical expressions 

When you use this option,
Using this option, you can get simpler solutions for some integrals for which the direct call of the integrator returns complicated results. With this option the integrator does not verify the correctness and completeness of the result. See Example 4. 

If integration requires case analysis, ignore cases that require one or more parameters to be elements of a comparatively small set, such as a fixed finite set or a set of integers. With this option, For example, if the property mechanism cannot prove that a denominator is equal to zero, MuPAD regards this denominator as nonzero. This option can significantly reduce the number of piecewise objects in the result. See Example 5. 

Compute the Cauchy principal value of the integral. If the interior of the integration interval contains poles of
the integrand or the boundaries are a =
 ∞ and b =
∞, then the definite integral might not
exist in a strict mathematical sense. However, if the integrand changes
sign at all poles in the integration interval, you can compute a weaker
form of a definite integral called the Cauchy principal
value. In this form, the socalled infinite parts of the
integral to the left and to the right of a pole cancel each other.
When you use the 
f
[1] Bronstein, M. "A Unification of Liouvillian Extension." AAECC Applicable Algebra in Engineering, Communication and Computing. 1: 5–24, 1990.
[2] Bronstein, M. "The Transcendental Risch Differential Equation." Journal of Symbolic Computation. 9: 49–60, 1990.
[3] Bronstein, M. "Symbolic Integration I: Transcendental Functions." Springer. 1997.
[4] Epstein, H. I. and B. F. Caviness. "A Structure Theorem for the Elementary Functions and its Application to the Identity Problem." International Journal of Computer and Information Science. 8: 9–37, 1979.
[5] Fakler, W. "Vereinfachen von komplexen Integralen reeller Funktionen." mathPAD 9 No. 1: 59, 1999.
[6] Geddes, K. O., S. R. Czapor and G. Labahn. "Algorithms for Computer Algebra." 1992.