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From Arfken's Mathematical Methods for Physicists (7th ed.)...

The remainder, $R_n$, is given by the n-fold integral $$R_n=\int^{x}_{a}dx_n\int^{x_n}_{a}dx_{n-1}\ldots\int^{x_2}_{a}dx_1f^{(n)}(x_1)$$ This remainder may be put into a perhaps more practical form by using the mean value theorem of integral calculus: $$\int^{x}_{a}g(x)dx=(x-a)g(\xi)$$ with $a\leq\xi\leq x$.By integrating n times we get the Lagrangian form of the remainder: $$R_n={(x-a)^n\over n!}f^{(n)}(\xi)$$

I know of the MVT of integral calculus and Taylor's theorem but I cannot understand this proof. Did Arken took $\xi$ as a constant? I mean, if we apply MVT to the first intrgral...$$\int^{x_2}_{a}dx_1f^{(n)}(x_1)=(x_2-a)f^{(n)}(\xi)$$ with $a\leq \xi=\xi(x_2)\leq x_2$. If we integrate this one... $$\int^{x_3}_{a}dx_2 (x_2-a)f^{(n)}(\xi)={(x_3-a)^2\over2!}f^{(n)}(\xi(x_3))-\int^{x_3}_{a}dx_2{(x_2-a)^2\over2!}\xi'(x_2)f^{(n+1)}(\xi(x_2))$$ Definitely something gone wrong. Is Arfken's proof valid? Then how can I show it?

generic properties
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  • Perhaps this answer to a related question will help. In short: Apply the mean value argument to the n-fold integral first to pull out $f^{(n)}(\xi)$, then evaluate the n-fold integral with constant integrand. – Bob Pego May 13 '14 at 03:15

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