Partial integration with maxima



There is also ".../share/integration/bypart.mac" for indefinite
integration by parts. 

load("bypart.mac");
byparts(log(x),x,log(x),1);   i,e., byparts(integrand,variable,u,dv)

In other words, the user has to determine u and dv, no Maxima magic
there. The actual "byparts" function is simple, you may want to take a
look yourself.

Milan

* Michel Gosse <michel.gosse@free.fr> [2005-09-14 21:21:06 +0200]:

> What do you think of :
> 
>  (C1) intpart(u,v,a,b):=subst(x=b,integrate(u,x))*subst(x=b,v)-
> subst(x=a,integrate(u,x))*subst(x=a,v)-integrate(integrate(u,x)*diff(v,x),x,a,b);
> 
> (D1) intpart(u,v,a,b):=SUBSTITUTE(x=b,INTEGRATE(u,x)) SUBSTITUTE(x=b,v) - 
> SUBSTITUTE(x=a,INTEGRATE(u,x)) SUBSTITUTE(x=a,v) - INTEGRATE(INTEGRATE(u,x) 
> DIFF(v,x),x,a,b)
> 
> Examples :
> 
> (C2) 'integrate(x*log(x),x,1,%e)=intpart(x,log(x),1,%e);
> 
> (C4) 'integrate(x*exp(2*x),x,0,1)=intpart(x,exp(2*x),0,1);
> 
> (D5) 'integrate(x^n*sin(x),x,0,%pi/2)=intpart(x^n,sin(x),0,%pi/2);
> 
> Maxima will not do the job alone...
> 
> Le mercredi 14 Septembre 2005 13:32, Marc Hodapp a écrit :
> > Hi everybody,
> >
> > how can I perform an integration by parts like
> >
> >  b                               b
> >  /                      b        /
> >
> >  | u'*v dx = [u*v]      -  |  u*v' dx
> >
> > /                       a       /
> > a                              a
> >
> > with maxima?  u and v are functions of x by the way.
> >
> > Many thanks in advance.
> >
> > Marc Hodapp
> >
> > _______________________________________________
> > Maxima mailing list
> > Maxima@math.utexas.edu
> > http://www.math.utexas.edu/mailman/listinfo/maxima
> 
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