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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,and none of the terms in the expansion/summation is equal to $\pm\int f(x)g(x)dx$ , one would expect the formula above to be repeatable infinitely many 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 - to perform the IBP infinitely many times - !?!? also, are there any theorems on the multiple integrals - the ones containing $g(x)$ - besides cauchy formula for repeated integration

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I don't see how the repeated IBP formula can imply the limit you mention is $0$, even when the integrand is smooth. For example, with $f(x) = e^x$ and $g(x) = \sin x$, then the limit does not tend (even pointwise) to $0$. Instead, there is a "repetition" in the formula that allows one to carry out the integration:

$$\int e^x \sin x \;dx = e^x(-\cos x) - e^x(-\sin x) + \int e^x(-\sin x) \;dx$$

$$ 2\int e^x \sin x \; dx = e^x(\sin x - \cos x) $$

$$ \int e^x \sin x \; dx = \frac{e^x}{2}(\sin x - \cos x)$$

So you see the "summation" implied by multiple IBP need not converge in the sense that the sum $\sum_{n \geq 0} \frac{x^n}{n!}$ converges. All that one needs for multiple IBP is some finite number of terms to work with.

Hope this helps!

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The wording in the question is a bit misleading but it's not actually saying that the repeated IBP implies the limit (the sentence does not end with the formula). –  Joel Cohen Mar 14 '12 at 19:30
    
given the "cyclic" nature of the integration in your post; i don't think the repeated IBP will be of any practical use !! i think a precise formulation of my question should mention that for IBP to be performed infinitely many times, none of the terms should be equal to $\pm \int f(x)g(x)dx$ . the limit is imposed over the formula, and is not implied by the formula . –  Mohammad Al Jamal Mar 14 '12 at 20:37
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