let be the 2 functions defined by an integral
a) $$ F(x)= \int_{0}^{x}g(t) dt $$
b) $$ H(x)= \int_{0}^{\infty} dt \frac{g(t)}{1+tx} $$
assuming that i only can evaluate $ F(x)$ and $ H(x)$ by numerical methods , how could i expand this functions into a Laurent series ? $ \sum_{n=-\infty}^{\infty} a(n)z^{n} $
my initial guess would approximate each integral by a sum of K terms and then expand each term in the sum into a Laurent series but it is correct ? (my reasoning)
If $g$ is analytic in a neighbourhood of $0$ and has Maclaurin series $g(z) = \sum_{n=0}^\infty c_n z^n$, then $F$ is also analytic and its Laurent series about $0$ is a Maclaurin series $F(z) = \sum_{n=0}^\infty c_n z^{n+1}/(n+1)$. Now you need to evaluate the $c_n$. For this, you can use the generalized Cauchy formula
$$ c_n = \dfrac{1}{2\pi i} \oint_C \dfrac{g(z)}{z^{n+1}}\ dz$$
where $C$ is a simple closed contour with $0$ inside. You could, for example, take $C$ to be the circle of radius $r$ centred at $0$. Then $$ c_n = \dfrac{1}{2\pi r^n} \int_0^{2\pi} g(r e^{i\theta}) e^{-n i \theta}\; d\theta$$ which can be numerically approximated as $$ c_n \approx \dfrac{1}{N r^n} \sum_{k=0}^{N-1} g(r e^{2\pi i k/N})\; e^{-2\pi i k n/N} $$
EDIT: Similarly, if $g(z)$ is analytic for $|z| > R$ and has Laurent series $\sum_{n=-\infty}^\infty c_n z^n$ there with $c_{-1} = 0$, then $F$ is also analytic for $|z| > R$ and has Laurent series $\sum_{n \ne -1} c_n z^{n+1}/(n+1)$ there. Again the coefficients can be evaluated by contour integrals around circles, and these can be approximated by sums.
As for $H$, I don't think it has a Laurent series in general. The problem is that when $x$ is negative, the integral has a pole at $t = -1/x$. This may manifest itself in logarithmic terms.
For example, with $g(t) = 1/(1+t)$ I get $H(x) = \dfrac{\ln(x)}{x-1}$, which has a branch point at $\infty$ and therefore no Laurent series.
