It is well known that
$$\int \frac{\sin(x)}{x} \,dx$$
cannot be expressed in terms of elementary functions. However, if we repeatedly use integration by parts, we seem to be able to at least approximate the integral through the formula
$$\int f(x) \,dx \approx \sum_{n=1}^a \frac{(-1)^{n-1}\cdot f^{(n-1)}(x)\cdot x^n}{n!}$$
where $a \in \mathbb{N}$. When plugging this in to a graphing calculator, it converges, but very slowly. It also tends to converge more quickly for functions that tend to $0$ as $x \to \infty$. My guess is that
$$\int f(x) \,dx = \lim_{a\to\infty}\sum_{n=1}^a \frac{(-1)^{n-1}\cdot f^{(n-1)}(x)\cdot x^n}{n!}$$
at least on a certain interval, but I am uncertain where to look to learn more about these series. Any ideas? Thanks!
Answer
You've essentially rediscovered Taylor series.
Let $G(x)$ be an antiderivative of $f(x)$, so $f^{(k)}(x) = G^{(k+1)}(x)$ for $k \ge 0$. If $G$ is analytic in a neighbourhood of $0$, and $x$ is in that neighbourhood,
$$ G(0) = \sum_{k=0}^\infty G^{(k)}(x) \frac{(-x)^k}{k!} = G(x) + \sum_{k=1}^\infty f^{(k-1)}(x) \frac{(-x)^k}{k!}$$
i.e.
$$ \int_0^x f(t)\; dt = G(x)- G(0) = \sum_{k=1}^\infty f^{(k-1)}(x) \frac{(-1)^{k-1} x^k}{k!} $$
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