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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: $int_2^{infty}left({frac{log(x+n)-log(x)}{log(x)}}right)dx = int_2^{infty}left({frac{logleft(1+frac{n}{x}right)}{log(x)}}right)dx = int_2^{infty}log_xleft(1+frac{n}{x}right)dx$. 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!

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,

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

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
$$
lim_{tto 0}frac{ln(1+t)}{t}=1
$$
we get
$$
lim_{xto+infty}biggl(frac{ln(1+n/x)}{ln x}biggr)biggm/frac{n}{xln x}=1
$$
Now, by the comparison theorem (in the form of limit) we find that
$$
int_2^{+infty}Bigl(frac{ln(x+n)}{ln x}-1Bigr),dxquadtext{diverges}
$$
since
$$
int_2^{R}frac{n}{xln x},dx=bigl[nln(ln x)bigr]_2^{R}to+inftyquadtext{as $Rto+infty$},
$$
i.e. $int_2^{+infty} n/(xln x),dx$ diverges.