intlib::byparts
-- performs
integration by partsintlib::byparts
(integral, du)
performs on
integral
the integration by parts, where du
is the part to be integrated.
intlib::byparts(integral, du)
integral |
- | integral: an expression of type "int" of
the form int(du*v, x) |
du |
- | the part to be integrated: an arithmetical expression |
an arithmetical expression containing the type "int"
or
the unevaluated function call.
int(u'(x)*v(x), x) = u(x)*v(x) - int(u(x)*v'(x), x)and for definite integrals as
int(u'(x)*v(x), x = a..b) = u(b)*v(b) - u(a)*v(a) - int(u(x)*v'(x), x = a..b)
intlib::byparts
(integral, du)
performs in
integral
the integration by parts where du
is
the part to be integrated and returns an expression containing the
unevaluated partial integral.intlib::byparts
works for indefinite as well as for
definite integrals.du
in
case of definite integration, the function call is returned
unevaluated."int"
. This can be obtained with hold
or freeze
(cf. example 1).du
should typically be a partial
expression of the integrand in integral
.As a first example we apply the rule of integration by
parts to the integral int(x*exp(x), x=a..b). By using the
hold
function we secure
that the first argument is of type "int"
:
>> intlib::byparts(hold(int)(x*exp(x), x = a..b), exp(x))
b exp(b) - a exp(a) - int(exp(x), x = a..b)
In this case the ansatz is choosen as u'(x)=exp(x) and thus v(x)=x.
In the following we give a more advanced example using
the method of integration by parts for solving the integral
int(exp(a*x)*sin(b*x), x). For this we have to prevent that
the integrator already evaluates the integrals. Thus we first
inactivate the requested integral with the function freeze
>> F := freeze(int)(exp(a*x)*sin(b*x), x)
int(sin(b x) exp(a x), x)
and apply afterwards partial integration with u'(x)=exp(ax):
>> F1 := intlib::byparts(F, exp(a*x))
sin(b x) exp(a x) / b cos(b x) exp(a x) \ ----------------- - int| -------------------, x | a \ a /
To this result again we can apply integration by parts.
But to avoid evaluating that integral we have to be very carefully. In
order to get it we must use the function level
:
>> F2 := -op(level(F1, 1), 2)
/ b cos(b x) exp(a x) \ int| -------------------, x | \ a /
With that we can now calculate the requested integral:
>> F3 := expand(simplify(op(F1, 1) - intlib::byparts(level(F2, 1), exp(a*x))))
sin(b x) exp(a x) b cos(b x) exp(a x) ----------------- - ------------------- - a 2 a 2 b int(sin(b x) exp(a x), x) ---------------------------- 2 a
As we can see the both integration by parts steps lead to same integral but with a different factor. Therefore we can solve it for the requested integral and we finally get:
>> F = normal(1/(1 + b^2/a^2)* _plus(op(level(F3, 1), [1..2])))
int(sin(b x) exp(a x), x) = a sin(b x) exp(a x) - b cos(b x) exp(a x) ----------------------------------------- 2 2 a + b
Here we demonstrate the difference between indefinite and definite integration by parts. If in the indefinite case the partial part cannot be solved, simply the unevaluated integral is plugged into the integration rule:
>> intlib::byparts(hold(int)(x*f(x), x),f(x))
x int(f(x), x) - int(int(f(x), x), x)
This is no longer true for the definite case:
>> intlib::byparts(hold(int)(x*f(x), x=a..b),f(x))
Warning: found no closed form for int(f(x), x) [intlib::bypart\ s] intlib::byparts(int(x f(x), x = a..b), f(x))