8
$\begingroup$

Sorry if this is a duplicate, I've searched how to do this to no avail.

What I'd like to do is a function that integrates by parts $n$ times, i.e

$$ \int u(x) v(x) dx = u \left(\textstyle{\int}v\right) - \displaystyle{\int} u' \left(\textstyle{\int}v\right)dx $$ where $\textstyle{\int}v$ is the primitive of $v$.

I've done a very rustic function that does this,

parts[u_,v_]:=(#1 Integrate[#2,x] - Integrate[D[#1,x] Integrate[#2,x],x]&[u,v];

which performs well but, as you all can see has -at least- the mayor limitation that u and v should be given as functions of x.

At least it works, for example

In[1]=  parts[Exp[-x],1/x^2]
Out[1]= -Exp[-x]/x - ExpIntegralEi[-x]

The thing is, I'd like to tell parts to operate $n$ times, for example

\begin{align} \mbox{parts}\big(u,v,2\big) &= u \textstyle{\int}v - \mbox{parts}\left(u',\textstyle{\int}v,1\right) \\ \\ \mbox{parts}\big(u(x),v(x),3\big) &= u(x)\textstyle{\int}v(x) - u'\left(\textstyle{\int}\textstyle{\int}v\right) + \mbox{parts}\big(u''(x),\textstyle{\iint}v,1\big) \end{align} and so on.

I hope my question is clear.

$\endgroup$
4
$\begingroup$

You can specify the cases for when $u$ and $v$ are free of variable.

ByParts[u_, v_, t_] := 
 With[{w = Integrate[v, t]}, u w - Integrate[D[u, t] w, t]]
ByParts[u_, v_, t_] := u Integrate[v, t] /; FreeQ[u, t]
ByParts[u_, v_, t_] := v Integrate[u, t] /; FreeQ[v, t]
$\endgroup$
3
$\begingroup$

I think I can answer my question.

Mathematically, it makes sense to tell Mma which variable is the one to integrate by parts, like LaplaceTransform or D. Taking this into account, I redefine parts like this

parts[u_,v_,{x_,n_}]:= Sum[(-1)^m D[u,{x,m}] Nest[Integrate[#,x]&,v,m+1],{m,0,n-1}] +
        (-1)^n Integrate[D[u,{x,n}] Nest[Integrate[#,x]&,v,n],x]

Again, I believe is rather amateurish, and I'd appreciate comments and other answers.

$\endgroup$
  • $\begingroup$ D[parts[u[x], v[x], {x, 3}], x] != u[x] v[x] $\endgroup$ – Hector Nov 15 '13 at 19:58
  • $\begingroup$ @Hector Right. I had an error on the last term. I hadn't multiplied it by $(-1)^n$. Besides that, I don't know why it behaves as you are pointing out. $\endgroup$ – Pragabhava Nov 15 '13 at 20:10
  • $\begingroup$ Great, you fixed it. $\endgroup$ – Hector Nov 15 '13 at 20:29
0
$\begingroup$

Here is my solution:

Clear[int, integrate];

integrate[integrate[expr_, var1__], var2__] := integrate[expr, var1, var2]

Format@integrate[expr_, var__] := Integrate[expr, var]

int[deri_ rest_, var_, varrest___, deri_ -> oldtarget_] /; deri =!= oldtarget := 
 With[{intderi = Integrate[deri, var] /. Integrate -> integrate, 
   target = oldtarget /. Integrate -> integrate}, 
  int[intderi rest, varrest, intderi -> target] - 
   int[D[rest, var] intderi, var, varrest, intderi -> target]]

int[expr_, deri_] := expr

Format@int[expr_, var__, deri_] := Integrate[expr, var]

The endpoint of integration is specified by the last argument of int. For example, if I want v[t] to be integrated to Integrate[v[t], t], I just need to write:

int[u[t] v[t], t, v[t] -> Integrate[v[t], t]]

Mathematica graphics

If I want to integrate by parts twice:

int[u[t] v[t], t, v[t] -> Integrate[v[t], t, t]]

Mathematica graphics

Three times:

int[u[t] v[t], t, v[t] -> Integrate[v[t], t, t, t]]

Mathematica graphics

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.