Let $f(x)$ be a real-valued function of real variable $x$, whose anti-derivative is difficult to obtain. Suppose we wish to compute the definite integral $$I=\int_a^\infty f(x)\text{d}x,$$ where $a$ is finite.
Assuming all standard attempts to evaluate the integral have failed, is the following method plausible and/or known?:
Compute the power series of $f(x)$ at $\infty$ (so that we obtain a power series in terms of powers of $1/x$). Integrate the the resulting power series, possibly (?) over an interval, say $[0,y]$ or $[1,y]$, so as to avoid constants of integration, and assuming convergence, let $y\longrightarrow 0$. This should (?) give the limit of the anti-derivative of $f(x)$ at $\infty$.
Compute the lower limit in a similar way. Subtract the two limits. The result should be the real number $I$.
While I don't really understand the strategy which you describe in the last two paragraphs of the question, the general method of integrating the Taylor series will work provided that
The assumption of analyticity is important, for otherwise you could take a positive smooth bump function supported in $(a,\infty)$ and the Taylor series at $\infty$ would have all zeros for coefficients - which would give an integral of zero by integration of series, but the integral is certainly not zero since the function is nonnegative and nonzero. Also, the assumption that $f=O(1/|z|^2)$ at $\infty$ is important, as demonstrated by various elementary examples like $1/z$ (which is not integrable on $(a,\infty)$).
For the given hypotheses, the proof would go as follows: First, $\int_a^\infty |f(x)|dx<\infty$ by hypothesis 3 and continuity on $[a,\infty)$.
If $f(x)=\sum_{j=2}^\infty c_j(1/x)^j$ is the Taylor series then by hypothesis 2, $\sum_{j=2}^\infty |c_j|(1/a)^j<\infty$ so \begin{align} \sum_{j=2}^\infty\int_a^\infty |c_j|(1/x)^jdx&= \sum_{j=2}^\infty |c_j|a^{1-j}/(j-1)\\ &=a\sum_{j=2}^\infty |c_j|(1/a)^j/(j-1)\\ &\leq a\sum_{j=2}^\infty |c_j|(1/a)^j\\ &<\infty \end{align} therefore $[x\mapsto \sum_{j=2}^\infty |c_j(1/x)^j|]\in L^1(a,\infty)$ by monotone convergence. Thus \begin{align} \int_a^\infty f(x)dx&=\int_a^\infty\lim_{n\to \infty}\sum_{j=2}^nc_j(1/x)^jdx \\ &=\lim_{n\to \infty}\int_a^\infty \sum_{j=2}^nc_j(1/x)^jdx \\ &=\lim_{n\to \infty}\sum_{j=2}^nc_j\int_a^\infty (1/x)^jdx \\ &=\sum_{j=2}^\infty c_ja^{1-j}/(j-1). \end{align} The first equality is the pointwise convergence of the series to the function $f$ which is valid by hypothesis 2, the second equality is by dominated convergence, which is justified by the monotone convergence argument given above.
Example: $$ f(x)=\frac{1}{1+x^2}=\frac{1}{x^2}\frac{1}{1-(-1/x^2)}=\frac{1}{x^2}\sum_{k=0}^\infty (-1)^k(1/x^2)^k=\sum_{k=0}^\infty(-1)^k(1/x^{2k+2}) $$ the third equality being valid only if $x>1$, in which case $$ \int_a^\infty f(x)dx=\sum_{k=0}^\infty(-1)^ka^{-1-2k}/(1+2k) $$ provided $a>1$. If $a<1$ then the series does not even converge because, despite the real analyticity of $1/(1+x^2)$ on the entire real line, the expansion at $\infty$ can't be continued to the unit disk since there are singularities at $\pm i$ on the unit circle. This is why hypothesis 2 is crucial.
To check this, observe that $1/(1+x^2)$ has a nice antiderivative $\arctan(x)$... so you can compare the series answer to the usual answer $$ \int_a^\infty 1/(1+x^2)dx=\pi/2-\arctan(a). $$ Since $a\mapsto \pi/2-\arctan(a)$ and $a\mapsto \sum_{k=0}^\infty(-1)^ka^{-1-2k}/(1+2k)$ have the same derivative on $(1,\infty)$ and the same limit as $a\rightarrow \infty$, these answers are the same.