If $f(x) =\sum_{n=0}^{\infty}(-1)^n a_n x^n$, then $\int_0^{\infty}\frac{f(x)}{1+x}dx = -\sum_{n=0}^{\infty}a'_n$ I've found using heuristic methods:
$$\int_0^{\infty}\frac{f(x)}{1+x}dx = -\sum_{n=0}^{\infty}a'_n$$
if
$$f(x) =\sum_{n=0}^{\infty}(-1)^n a_n x^n$$
and $a'_n \equiv \dfrac{d a_n}{dn}$.
This seems to work, although the summation may sometimes need to be redefined as $g(-1)$, where
$g(z) = \sum_{n=0}^{\infty}(-1)^n a'_n z^n$
with this analytically continued such that $g(-1)$ is well defined. 
The only exception seems to be cases where $f(x)$ has a pole on the unit circle, e.g. if $f(x) = \frac{1}{1+x}$, but in that case using $f(x) =  \frac{1}{a+x}$ and taking the limit $a\to 1$ at the end of the calculations gives the correct result.
Is there a rigorous version of this statement with rigorous conditions for $f(x)$?
 A: The following is not an answer, but a mere collection of some special cases, in which the proposed identity holds (if interpreted in a suitable way). It is intended to illustrate that despite the absolutely valid objections made by David C. Ullrich in the comments there is at least some truth in the suggested equation.
The functions $f \colon \mathbb{R}_{\geq 0} \to \mathbb{R}$ considered here have the following things in common:


*

*The integral $\int_0^\infty \frac{f(x)}{1+x} \, \mathrm{d} x$ converges.

*The function has a power series expansion $f(x) = \sum_{n=0}^\infty
   (-1)^n a(n) x^n$ about the origin with a non-zero (but not necessarily 
infinite) radius of convergence. 

*$a \colon \mathbb{N}_0 \to \mathbb{R}_{\geq 0}$ is a non-negative function. 
The expression $a(t)$ makes sense for all $t \geq 0$ , so there is a more or 
less natural continuation $a \colon \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 
   0}$ , which defines a differentiable function. 

*The series $\sum_{n=0}^\infty a'(n)$ converges.


Note that the third point is very vague and far from rigorous. Replacing $a(n)$ by $a(n) [1+\sin(2 \pi n)]$ obviously leads to the same power series but changes the series of the derivatives, which is exactly the problem mentioned in the comments. A proof of the identity must at first give a proper definition of a 'natural' continuation.
In the following examples the simplest choice turns out to be correct:


*

*For $0 < \alpha < 1$ consider $f(x) = \frac{1}{1+\alpha x}$ . The series 
expansion $$f(x) = \sum \limits_{n=0}^\infty (-1)^n \alpha^{n} x^n \, , \, 0 
\leq x < \frac{1}{\alpha} \, , $$ suggests the definition $a(t) = \alpha^t$ 
. We have $$ \int \limits_0^\infty \frac{f(x)}{1+x} \, \mathrm{d} x = 
\lim_{R \to \infty} \frac{1}{1-\alpha} \int \limits_0^R \left[\frac{1}{1+x}- 
\frac{\alpha}{1+\alpha x}\right] \, \mathrm{d} x = \frac{- \ln(\alpha)}{1- 
\alpha}$$ and $$ - \sum 
\limits_{n=0}^\infty a'(n) = - \sum \limits_{n=0}^\infty \ln(\alpha) 
\alpha^n = \frac{- \ln(\alpha)}{1-\alpha} \, . $$ The identity also holds in 
the limit $\alpha \nearrow 1$ and even for $\alpha > 1$, provided that the 
sum is regularised as suggested in the question. 

*For $f(x) = \frac{\ln(1+x)}{x}$ we have $$ f(x) = \sum \limits_{n=0}^\infty 
(-1)^n \frac{1}{n+1} x^n \, , \, 0 \leq x \leq 1 \, . $$ The definition 
$a(t) = \frac{1}{t+1}$ seems obvious. Indeed, $$ \int \limits_0^\infty 
\frac{f(x)}{1+x} \, \mathrm{d} x = \int \limits_0^\infty \frac{y} 
{\mathrm{e}^y - 1} \, \mathrm{d} y = \Gamma (2) \zeta (2) = \frac{\pi^2}{6} 
$$ and $$ - \sum \limits_{n=0}^\infty a'(n) =  \sum \limits_{n=0}^\infty 
\frac{1}{(n+1)^2} = \zeta (2) = \frac{\pi^2}{6}$$ coincide.

*If $$ f(x) = \mathrm{e}^{-x} = \sum \limits_{n=0}^\infty (-1)^n \frac{1}{n!} 
x^n \, , $$ the choice $a(t) = \frac{1}{\Gamma(1+t)}$ is natural. The value 
of the integral (known as the Gompertz constant) can be expressed in terms 
of the exponential integral $\operatorname{Ei}$: $$ \int \limits_0^\infty 
\frac{f(x)}{1+x} \, \mathrm{d} x = -\mathrm{e} \int \limits_{-\infty}^{-1} 
\frac{\mathrm{e}^y}{y} \, \mathrm{d} y = - \mathrm{e} \operatorname{Ei} 
(-1) \approx 0.59635 \, .$$ Using the relation $\psi (n+1) = H_n - \gamma$ 
between the digamma function, the harmonic numbers and the Euler-Mascheroni 
constant as well as the series representation of the exponential integral, 
the same value is obtained for the series: $$ - \sum \limits_{n=0}^\infty 
a'(n) = \sum \limits_{n=0}^\infty \frac{\psi(n+1)}{n!} =  - \mathrm{e} 
\operatorname{Ei} (-1) \, . $$


I have not found a way to formulate the continuation procedure in a rigorous manner, let alone a proof for the general case, but I hope that these results will encourage someone else to give it a try.
