How to use similar ($\sim)$ functions correctly (if they change their signs) Are there any restrictions when replacing one function with an equivalent (a similar function)? I've read somewhere that when a function changes its sign in an interval, one cannot freely replace the function with an equivalent. For example, if I have the function $f(x)=\frac{\sin{x}}{x^3+1}$, can I say that $f(x)\sim\frac{\sin{x}}{x^3}$ for $x\in[1, \infty)$? This function is a part of a definite integral, which I'm trying to solve.
 A: The equivalence is at one point only, not an interval like $[1,+\infty)$ as you have written.
Here $x^3\sim x^3+1$ is only true at infinity so to have $f(x)\sim g(x)$ we have to check if their ratio goes to $1$.
$f(x)=\dfrac{\sin(x)}{x^3+1}$ and $g(x)=\dfrac{\sin(x)}{x^3}$ then the ratio cancels the $\sin$ anyway (so you see we don't care of its sign changes) and we get $\dfrac{f(x)}{g(x)}=\dfrac{x^3}{x^3+1}\to_\infty 1$
Therefore, the statement below is true
$$\dfrac{\sin(x)}{x^3+1}\sim_\infty\dfrac{\sin(x)}{x^3}$$
Where you have to be careful, is when performing additions. Multiplying equivalents poses no issues, exponentiating you should start to be careful, but adding equivalents is a no-no.
E.g. $f(x)=x^3+x$ and $g(x)=-x^3+7$ then $f(x)\sim x^3$ and $g(x)\sim -x^3$ but we don't have $f(x)+g(x)\sim x^3-x^3=0$ since in fact $f(x)+g(x)=x+7\to\infty$.
This is the reason it does not work inside an integral. An integral is basically a sum of values of $f(x)$ and you cannot rely on an equivalent that changes sign else you would add equivalents which is wrong.
If $f\sim g$ then $\int^{\infty} f$ and $\int^\infty g$ have the same behaviour (convergence, divergence) only when $f,g$ are of constant sign, i.e. we are interested only in absolute convergence, same as required with discrete series.
$a_n\sim b_n$ then $\sum a_n$ and $\sum b_n$ have same behaviour only when $a_n,b_n$ have constant sign.
A classical example of constant sign necessity is given by $\sum \dfrac{(-1)^n}{\sqrt{n}+(-1)^n}$ which is divergent while the a priori equivalent $\sum \dfrac{(-1)^n}{\sqrt{n}}$ is convergent via the alternated series criteria ($\sqrt{n}\nearrow$).
I have no example in mind for integrals, but the reason is pretty similar.
