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I was thinking about functions $f(x)$ such that $\lim_{x \to \infty}{\frac{f(x+n)}{f(x)}=1}$, where $n \geq 0$. I was looking for some special properties of them, and decided to first focus on $\log(x)$. My first natural question was on the rate of divergence of $f(x+n)$ vs $f(x)$, and I'm now trying to solve the following integral:

$$\int_2^{\infty}\left({\frac{\log(x+n)}{\log(x)}-1}\right)dx$$

So far, I've tried the following simplifications: $\int_2^{\infty}\left({\frac{\log(x+n)-\log(x)}{\log(x)}}\right)dx = \int_2^{\infty}\left({\frac{\log\left(1+\frac{n}{x}\right)}{\log(x)}}\right)dx = \int_2^{\infty}\log_x\left(1+\frac{n}{x}\right)dx$. I examined each form and did various substitutions, none of which led to any obvious findings. Does anyone know a better way to approach this integral?

Would also appreciate any ideas about the class of functions I'm thinking about more generally. Thanks!

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The integral does not converge for $n>0$. Note that:

$$\frac{\ln(x+n)}{\ln(x)}-1>\frac n{(x+n)\ln(x)}>\frac n{(x+n)\ln(x+n)}$$

Which can be proven by multiplying both sides by $\ln(x)$ to see that

$$\ln(x+n)-\ln(x)=\int_x^{x+n}\frac1t~\mathrm dt>\int_x^{x+n}\frac1{x+n}~\mathrm dt=\frac n{x+n}$$

From here it is very easy to show that it diverges,

$$\begin{align}\int_2^\infty\frac{\ln(x+n)}{\ln(x)}-1~\mathrm dx&>\int_2^\infty\frac n{(x+n)\ln(x+n)}~\mathrm dx\\&=\int_{2+n}^\infty\frac n{x\ln(x)}~\mathrm dx\\&=\int_{\ln(2+n)}^\infty\frac nx~\mathrm dx\\&=+\infty\end{align}$$

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We have $$ \frac{\ln(x+n)}{\ln x}-1=\frac{\ln x+\ln(1+n/x)}{\ln x}-1=\frac{\ln(1+n/x)}{\ln x}>0 $$ Since $$ \lim_{t\to 0}\frac{\ln(1+t)}{t}=1 $$ we get $$ \lim_{x\to+\infty}\biggl(\frac{\ln(1+n/x)}{\ln x}\biggr)\biggm/\frac{n}{x\ln x}=1 $$ Now, by the comparison theorem (in the form of limit) we find that $$ \int_2^{+\infty}\Bigl(\frac{\ln(x+n)}{\ln x}-1\Bigr)\,dx\quad\text{diverges} $$ since $$ \int_2^{R}\frac{n}{x\ln x}\,dx=\bigl[n\ln(\ln x)\bigr]_2^{R}\to+\infty\quad\text{as $R\to+\infty$}, $$ i.e. $\int_2^{+\infty} n/(x\ln x)\,dx$ diverges.

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