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The integral $$I(k) = \int_{-\infty}^\infty \frac{s e^{isr}}{(s-k)(s+k)} ds \tag{1}$$ where $k$ is real and the integration is for real $s$, is not really well-defined. This is precisely because the integrand has singularities on the integration domain. However consider if $k$ is a complex number $k = k_r + ik_i$ with $k_i >0$. Then the integrand is ...

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Integration operates on functions, correct? No. Integration dates back to ca. 1670. The notion of a function gradually evolved and didn't get put into its modern form until ca. 1830. Let's look at your expression $$\int_{p_1}^{p_2} dp .$$ This is Leibniz's notation, and what he means by it is a sum of infinitely many terms $dp$. Each of these terms ...

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What does it mean to integrate $\frac{d\mathbf p}{dt}dt$? First, and in scalar form, recall from elementary calculus that $$\int_{x_1}^{x_2} dx = x_2 - x_1$$ Second, recall that $$f(x + dx) = f(x) + f'(x)dx$$ where $$f'(x) = \frac{df(x)}{dx}$$ Denoting the differential of $f$ as $$df = f(x + dx) - f(x)$$ we have $$df = f'(x)dx$$ Since ...

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While in some contexts, $$\frac{d\textbf{p}}{dt}dt = d\textbf{p}$$ is correct and is mathematically rigorous, there is a straightforward way to derive impulse as the change in momentum. Consider a function $f(x)$ where $f:\mathbb{R}^m \to \mathbb{R}^n$ and $f \in C^1$. Suppose its derivative is $g = f'(x)$. We can consider integrating $g(x)$ over some ...

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