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Differentiating $\displaystyle f(x)=\int_{0}^{x}\int_{1}^{x-s}\frac{g(t)}{t}\space dt\space ds$, I obtained $\displaystyle \int_{1}^{0}\frac{g(t)}{t}\space dt + \int_{0}^{x} \frac{\partial}{\partial x}\int_{1}^{x-s}\frac{g(t)}{t}\space dt\space ds$, with $\displaystyle \int_0^{x}\frac{\partial}{\partial x}\int_{1}^{x-s}\frac{g(t)}{t}\space dt\space ds = \int_{0}^{x}\frac{g(x-s)}{x-s}\space ds$, so that $\displaystyle f'(x)=\int_{0}^{x}\frac{g(x-s)}{x-s}\space ds - \int_{0}^{1}\frac{g(t)}{t}\space dt $.

Is there a property that the functions $g:\mathbb{R}\to\mathbb{R}$ must necessarily satisfy (or have in common) such that $f$ is $C^1$ or even $C^{\infty}$, apart from being polynomials (in order to cancel the problematic $\frac{1}{x-s}$ and $\frac{1}{t}$ terms in the given integrands)? Through trial and error, I've worked out that $g(x)=\sin(x),\ln(x+1),e^x-1$ all work and that $\displaystyle \lim_{x \to 0} \frac{g(x)}{x}$ exists.

I appreciate the insight.

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  • $\begingroup$ First insight: "Walls of text" are hard to read and follow. Break your posts up into short paragraphs. $\endgroup$ Oct 23, 2022 at 16:21

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For $C^1$, you already have the obvious condition: $\frac{g(x)}x$ must be integrable.

If $g$ is integrable away from $0$, then $\frac{g(x)}x$ will be as well, so the only issue is the behavior at $0$. But for this, you just need $g(x) = xh(x)$ for some integrable function $h$. Integrations are always continuous, so that part is handled. If you know that $g$ is $C^1$, then it is sufficient that $g(0) = 0$.

For $C^\infty$, you need $g$ to be $C^\infty$ and $g(0) = 0$.

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