# On the integration by parts infinitely many times

• mmzaj
In summary: 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 f^{(n+1)} (x) = 0 for some n < \infty if the function is analytic everywhere and the integral
mmzaj
greetings . it's known that if $g(x), f(x)$ are two functions ,and $f(x)$ is sufficiently differentiable , then by repeated integration by parts one gets :

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

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

$$\lim_{n \to \infty }\int\left[ \underbrace{\int...\int}g(x)dx^{n+1}\right ] f^{n+1}(x)dx=0$$
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 $g(x)$ - besides cauchy formula for repeated integration

Last edited:
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 :

$$\pm a \int f(x)g(x)dx$$

where$a$ 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 .

mmzaj said:
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 :

$$\pm a \int f(x)g(x)dx$$

where$a$ 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 .

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.

chiro said:
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.

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 $f^{(n+1)} (x) = 0$ for some $n < \infty$

## 1. What is the concept of integration by parts infinitely many times?

Integration by parts infinitely many times is a mathematical technique that involves repeatedly applying the integration by parts formula to an integral. This is often done in cases where the original integral cannot be solved using traditional methods.

## 2. How does integration by parts infinitely many times work?

The integration by parts formula expresses the integral of a product of two functions as the difference between the product of the first function and the integral of the product of the derivative of the first function and the second function. By repeatedly applying this formula, we can simplify and solve complicated integrals.

## 3. When should integration by parts infinitely many times be used?

Integration by parts infinitely many times should be used when traditional integration methods are not applicable or when the integral is too complex to be solved by hand. It is also useful in cases where the integral involves a product of functions with no obvious substitutions.

## 4. What are the limitations of integration by parts infinitely many times?

One limitation of integration by parts infinitely many times is that it can only be applied to integrals that can be written in the form of a product of functions. It also requires a good understanding of the integration by parts formula and its application. Additionally, in some cases, it may not yield a closed form solution.

## 5. Are there any alternatives to integration by parts infinitely many times?

Yes, there are alternative methods for solving complicated integrals, such as using substitution, partial fractions, or numerical methods. However, in some cases, integration by parts infinitely many times may be the most efficient and effective method for finding a solution.

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