关于递归无限次应用分部积分法所得级数的末项及无穷和形式的技术问询

Hey there! Let's break this down step by step since you're working with recursive infinite integration by parts—super useful for proofs, by the way.

First, let's start with the standard integration by parts formula to set the stage. For two differentiable functions ( u(x) ) and ( v(x) ), we know:

( \int u(x) v'(x) dx = u(x) v(x) - \int u'(x) v(x) dx )

If we apply this recursively, let's define some shorthand to keep things clean:

• Let ( u^{(n)}(x) ) denote the nth derivative of ( u(x) ) (so ( u^{(0)}(x) = u(x) ), ( u^{(1)}(x) = u'(x) ), etc.)
• Let ( v_{n}(x) ) denote the nth antiderivative of ( v'(x) ) (so ( v_1(x) = v(x) ), ( v_2(x) = \int v_1(x) dx ), ( v_3(x) = \int v_2(x) dx ), and so on)

After k recursive applications, we get the finite expansion:

( \int u(x) v'(x) dx = \sum_{n=0}^{k-1} (-1)^n u^{(n)}(x) v_{n+1}(x) + (-1)^k \int u^{(k)}(x) v_k(x) dx )

Now, when we talk about extending this to an infinite number of recursions, the key thing here is convergence. For the infinite series to equal the original integral, the remainder term ( (-1)^k \int u^{(k)}(x) v_k(x) dx ) must approach 0 as ( k \to \infty ). This is non-negotiable—if the remainder doesn't vanish, the infinite sum won't converge to your original integral.

Assuming that remainder term goes to zero, the abstract infinite sum form is:

( \int u(x) v'(x) dx = \sum_{n=0}^{\infty} (-1)^n u^{(n)}(x) v_{n+1}(x) )

As for the "last term" with the infinite-order derivative of ( u ): in a convergent infinite series, there isn't a traditional "last term" in the way finite series have one. Instead, the remainder term (which carries the kth derivative of ( u )) shrinks to zero as k grows without bound. If this condition holds, we don't end up with a leftover term involving ( u^{(\infty)}(x) )—the infinite sum itself fully represents the integral.

A quick example: this works great for functions where derivatives decay rapidly (like ( u(x) = e^{-x} )) or polynomials (though polynomials will terminate the series after finite steps since their higher derivatives hit zero). If the derivatives of ( u ) don't decay fast enough, or the antiderivatives of ( v' ) grow too quickly, the remainder term won't vanish, and the infinite series won't be valid for representing the integral.

备注：内容来源于stack exchange，提问作者Engr. Moiz Ahmad