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intlib::byparts -- performs integration by parts

Introduction

intlib::byparts(integral, du) performs on integral the integration by parts, where du is the part to be integrated.

Call(s)

intlib::byparts(integral, du)

Parameters

integral - integral: an expression of type "int" of the form int(du*v, x)
du - the part to be integrated: an arithmetical expression

Returns

an arithmetical expression containing the type "int" or the unevaluated function call.

Related Functions

subs, intlib::changevar

Details

Example 1

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.

Example 2

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

Example 3

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))

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