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I'm wondering if a classification of analytic functions, $f\,$ (it may be that $C^1$ is enough, but I'm not taking any chances, if you have a reason why I only need to consider a larger class of functions, I would enjoy that as well) with the property $F(x):=\int f(x) dx = \mathcal{O}(xf(x))$ as $x\to\infty$ AND $xf(x)=\mathcal{O}(F(x))$ where $\mathcal{O}$ is big-O notation. I'm not sure if it's standard or not, but I'll denote this condition by

$$\mathcal{O}(F(x))=\mathcal{O}(xf(x))\qquad (*)$$

I'm motivated by the naïve notion of integration from the mistake many first semester students in calculus make when trying to take anti-derivatives and keep on thinking the same thing over and over: the power rule.

It is straightforward to check that $f(x)=x^n\quad n\ne -1$ and $f(x)=\log^n x\quad n\ge 0$ satisfy the condition $(*)$, the first just by checking and the second by induction on n and integration by parts. It's also easy to see that this is not the case for $x^ne^x,\; n\in\mathbb{Z}$ again by integration by parts.

A preliminary investigation yields some interesting first starts:

1. A valid refomulation of the problem is ${F\over f}$ has a slant asymptote, i.e. ${F\over f}=kx+\mathcal{o}(1)$, and if you know that the derivative of this little o function is also little o of 1, and say $k=1$ then you can rephrase this as $1-{d\over dx}(\log f(x))\cdot {F\over f}(x)=1+\mathcal{o}(1)$

2. If the function is increasing we can get half of this inequality since $F(x)\le xf(x)$ since $F(x)=\int_a^xf(x)$, but the other half fails.

3. A convex $\mathcal{o}(1)$ in (1) might imply that the derivative is also $\mathcal{o}(1)$

I'm wondering if a classification of analytic functions, $f\,$ (it may be that $C^1$ is enough, but I'm not taking any chances, if you have a reason why I only need to consider a larger class of functions, I would enjoy that as well) with the property $F(x):=\int f(x) dx = \mathcal{O}(xf(x))$ as $x\to\infty$ AND $xf(x)=\mathcal{O}(F(x))$ where $\mathcal{O}$ is big-O notation. I'm not sure if it's standard or not, but I'll denote this condition by

$$\mathcal{O}(F(x))=\mathcal{O}(xf(x))\qquad (*)$$

I'm motivated by the naïve notion of integration from the mistake many first semester students in calculus make when trying to take anti-derivatives and keep on thinking the same thing over and over: the power rule.

It is straightforward to check that $f(x)=x^n\quad n\ne -1$ and $f(x)=\log^n x\quad n\ge 0$ satisfy the condition $(*)$, the first just by checking and the second by induction on $n$ and integration by parts. It's also easy to see that this is not the case for $x^ne^x,\; n\in\mathbb{Z}$ again by integration by parts.

A preliminary investigation yields some interesting first starts:

1. A valid refomulation of the problem is ${F\over f}$ has a slant asymptote, i.e. ${F\over f}=kx+\mathcal{o}(1)$, and if you know that the derivative of this little o function is also little o of 1, and say $k=1$ then you can rephrase this as $1-{d\over dx}(\log f(x))\cdot {F\over f}(x)=1+\mathcal{o}(1)$

2. If the function is increasing we can get half of this inequality since $F(x)\le xf(x)$ since $F(x)=\int_a^xf(x)$, but the other half fails.

3. A convex $\mathcal{o}(1)$ in (1) might imply that the derivative is also $\mathcal{o}(1)$

I'm wondering if a classification of analytic functions, $f\,$ (it may be that $C^1$ is enough, but I'm not taking any chances, if you have a reason why I only need to consider a larger class of functions, I would enjoy that as well) with the property $F(x):=\int f(x) dx = \mathcal{O}(xf(x))$ as $x\to\infty$ AND $xf(x)=\mathcal{O}(F(x))$ where $\mathcal{O}$ is big-O notation. I'm not sure if it's standard or not, but I'll denote this condition by

$$\mathcal{O}(F(x))=\mathcal{O}(xf(x))\qquad (*)$$

I'm motivated by the naïve notion of integration from the mistake many first semester students in calculus make when trying to take anti-derivatives and keep on thinking the same thing over and over: the power rule.

It is straightforward to check that $f(x)=x^n\quad n\ne 1$ and $f(x)=\log^n x\quad n\ge 0$ satisfy the condition $(*)$, the first just by checking and the second by induction on n and integration by parts. It's also easy to see that this is not the case for $x^ne^x,\; n\in\mathbb{Z}$ again by integration by parts.

A preliminary investigation yields some interesting first starts:

1. A valid refomulation of the problem is ${F\over f}$ has a slant asymptote, i.e. ${F\over f}=kx+\mathcal{o}(1)$, and if you know that the derivative of this little o function is also little o of 1, and say $k=1$ then you can rephrase this as $1-{d\over dx}(\log f(x))\cdot {F\over f}(x)=1+\mathcal{o}(1)$

2. If the function is increasing we can get half of this inequality since $F(x)\le xf(x)$ since $F(x)=\int_a^xf(x)$, but the other half fails.

3. A convex $\mathcal{o}(1)$ in (1) might imply that the derivative is also $\mathcal{o}(1)$

I'm wondering if a classification of analytic functions, $f\,$ (it may be that $C^1$ is enough, but I'm not taking any chances, if you have a reason why I only need to consider a larger class of functions, I would enjoy that as well) with the property $F(x):=\int f(x) dx = \mathcal{O}(xf(x))$ as $x\to\infty$ AND $xf(x)=\mathcal{O}(F(x))$ where $\mathcal{O}$ is big-O notation. I'm not sure if it's standard or not, but I'll denote this condition by

$$\mathcal{O}(F(x))=\mathcal{O}(xf(x))\qquad (*)$$

I'm motivated by the naïve notion of integration from the mistake many first semester students in calculus make when trying to take anti-derivatives and keep on thinking the same thing over and over: the power rule.

It is straightforward to check that $f(x)=x^n\quad n\ne -1$ and $f(x)=\log^n x\quad n\ge 0$ satisfy the condition $(*)$, the first just by checking and the second by induction on n and integration by parts. It's also easy to see that this is not the case for $x^ne^x,\; n\in\mathbb{Z}$ again by integration by parts.

A preliminary investigation yields some interesting first starts:

1. A valid refomulation of the problem is ${F\over f}$ has a slant asymptote, i.e. ${F\over f}=kx+\mathcal{o}(1)$, and if you know that the derivative of this little o function is also little o of 1, and say $k=1$ then you can rephrase this as $1-{d\over dx}(\log f(x))\cdot {F\over f}(x)=1+\mathcal{o}(1)$

2. If the function is increasing we can get half of this inequality since $F(x)\le xf(x)$ since $F(x)=\int_a^xf(x)$, but the other half fails.

3. A convex $\mathcal{o}(1)$ in (1) might imply that the derivative is also $\mathcal{o}(1)$