I am using integration by part concept in my proofs in which I will use it recursively infinite times.
In this regard, I am wondering, what will be the last term of the resulting series which will have infinite order derivative of the first function.
Can anybody give me abstract mathematical form or the resulting infinite sum?
If your integral is of the form $\int_I u v$ (let's say on an segment I with both $C^\infty$ functions
$\int_I uv = [u \cdot \int v]_I - \int_I (u' \int v)$
If you iterate, you'll get
$\int_I uv = [u \cdot \int V]_I - [u' \cdot \int^{\circ 2} v]_I + \cdots + (-1)^{n} [u^{(n)} \int^{\circ n+1} v]_I + (-1)^{n+1} \int_I (u^{(n+1)} \int^{\circ n+1} v)$
With $\int^{k} v$ a $k$-th primitive of $v$ and if $I = [a,b]$, $[f]_I = f(b) - f(a)$.
Now if $\int_I (u^{(n+1)} \int^{\circ n+1} v) \to 0$ when $n \to \infty$, you will get $$\int_I uv = \sum_{n=0}^\infty (-1)^n [u^{(n)} \int^{\circ n+1} v]_I $$
I don't know if that's what you were looking for, but without further explanation It seems impossible to do more.