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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!

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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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