Infinite series by integration by parts

[disregardthat]

Hi, I wonder if this hypothesis is true:

Let $$f_n$$ be an arbitrarily chosen n'th anti-derivative of the function $$f_0$$. Similarly, let $$g_n$$ be the n'th derivative of the function $$g_0$$.

Now, $$\int^b_a f_0 g_0 \rm{d}x=[f_1g_0]^b_a-\int^b_a f_1g_1 \rm{d}x=[f_1g_0-f_2g_1+...]^b_a+(-1)^n \int^b_a f_{n+1}g_n \rm{d}x$$.

hypothesis:

If $$\lim_{n \to \infty} f_{n+1}g_n =0$$ for all continuous intervals of and never diverges. Then

$$\int^b_a f_0 g_0 \rm{d}x = [\sum^{\infty}_{n=0} (-1)^n f_{n+1}g_n]^b_a$$

This seems intuitively correct, but I wonder how to prove it.

 

[Jame]

I'm thinking Mean Value Theorem. The integral equals $$\epsilon(b-a)(f_{n+1}g_n)|_{x=c}$$ for some c in the interval [a, b] and $$0 \leq \epsilon \leq 1$$ which approaches zero.

 

[disregardthat]

this was double post, sorry

 

[disregardthat]

Excellent, thank you.

We have by induction that

$$\int^b_a f_0g_0 \rm{d}x = [\sum^{n-1}_{k=0} (-1)^kf_{k+1}g_{k}]^b_a+(-1)^n\int^b_af_{n+1}g_n \rm{d}x=[\sum^{n-1}_{k=0} (-1)^kf_{k+1}g_{k}]^b_a+(b-a)(f_{n+1} \circ g_n)(t)$$

For some $$t \in [a,b]$$, and any non-negative integer n.

However, $$\lim_{n \to \infty} (f_{n+1} \circ g_{n} )(t)=0$$ is given, so

$$\int^b_a f_0g_0 \rm{d}x = \lim_{n \to \infty} \int^b_a f_0g_0 \rm{d}x= \lim_{n \to \infty} [\sum^{n-1}_{k=0} (-1)^kf_{k+1}g_{k}]^b_a+(-1)^n\int^b_af_{n+1}g_n \rm{d}x=\lim_{n \to \infty}[\sum^{n-1}_{k=0} (-1)^kf_{k+1}g_{k}]^b_a+(b-a)(f_{n+1} \circ g_n)(t)=[\sum^{\infty}_{k=0} (-1)^kf_{k+1}g_{k}]^b_a$$

But this is not so obvious if $$\lim_{n \to \infty} (f_{n+1} \circ g_{n} )(t)$$ not always equal 0 i.e. is finite for discrete values of x, or if either of the limits are infinite. Can someone help me there?

Perhaps if the limits are infinite, we can let n tend towards infinity at a rate which make $$f_{n+1}g_n$$ dominate a limit, say b i.e. so $$\lim_{b,n \to \infty} b \cdot f_{n+1}g_n = 0$$ Can we choose it to be like that?