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On the integration by parts infinitely many times

  1. Mar 14, 2012 #1
    greetings . it's known that if [itex] g(x), f(x)[/itex] are two functions ,and [itex]f(x)[/itex] is sufficiently differentiable , then by repeated integration by parts one gets :

    [tex] \int f(x)g(x)dx=f(x)\int g(x)dx -f^{'}(x)\int\int g(x)dx^{2}+f^{''}(x)\int \int \int g(x)dx^{3} - .... +(-1)^{n+1}f^{(n)}(x)\underbrace{\int.....\int}g(x)dx^{n+1}+(-1)^{n}\int\left[ \underbrace{\int.....\int}g(x)dx^{n+1}\right ]f^{n+1}(x)dx[/tex]

    now, if [itex]f(x) [/itex] is a smooth function , one would expect the formula above to be repeatable infinitely man times . therefore :

    [tex] \lim_{n \to \infty }\int\left[ \underbrace{\int.....\int}g(x)dx^{n+1}\right ] f^{n+1}(x)dx=0[/tex]
    is a necessary but not sufficient condition for the summation to converge . my question is , what are the conditions needed to extend the scope of the formula !?!?
    also, are there any theorems on the multiple integrals - the ones containing [itex]g(x)[/itex] - besides cauchy formula for repeated integration
     
    Last edited: Mar 14, 2012
  2. jcsd
  3. Mar 15, 2012 #2
    i should have mentioned that another necessary condition for the formula above to be carried out infinitely many times, is that none of the terms in the expansion/summation should be equal to :

    [tex] \pm a \int f(x)g(x)dx [/tex]

    where[itex] a [/itex] is a constant .

    a precise formulation of the question would be : what are the conditions for the formula above to be repeatable infinitely many times, and for the summation to converge .

    i also should mention that this formula could come in handy when doing some nonelementary integral .
     
  4. Mar 15, 2012 #3

    chiro

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    Hey mmzaj.

    The key things to look out for is that a) the integral converges and b) that derivative information makes sense under the type of integration that you are using.

    If your functions are analytic continuous (derivatives are also continuous) in the region of integration and the integral itself is finite, then any valid transformation of the integral preserving these properties will also have these properties.
     
  5. Mar 15, 2012 #4
    thanks for the prompt reply... i am trying to use this formula to do some mellin-type integrals appearing in analytic number theory , where little is known about singularities- hence residues - of a function. the integration is finite , and the integrand is well defined , and analytic everywhere. however , residue calculus is rendered useless when trying to get an exact-explicit formula . hence the proposition of using the infinite integral by parts .

    looking at the expansion as a conventional series; convergence implies the limit, unless repeated IBP is stopped when [itex] f^{(n+1)} (x) = 0 [/itex] for some [itex] n < \infty [/itex]
     
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