I'm looking for a way to prove that a function is periodic (or not periodic) by definition.
For example, I tried to prove that $\frac{1}{1+x^2}$ is not periodic by comparing it to $\frac{1}{1+(x+p)^2}$ ($p\in R$) and see if I can get a constant, and as expected I didn't.
Yet, with actual periodic function it didn't work.
How can I do it the right way?
$$f(x)=\left(\frac{1}{1+x^2}\right)$$ $$f(x+T)=\left(\frac{1}{1+(x+T)^2}\right)$$ $$f(x)=f(x+T)$$ $$\left(\frac{1}{1+x^2}\right)=\left(\frac{1}{1+(x+T)^2}\right)$$ $$x^2=(x+T)^2$$ $$x^2=x^2+T^2+2xT$$ $$T(T+2x)=0$$ Therefore, $$T=0;T=-2x$$
So, the values of $T$ that we've obtained aren't independent of $x$, neither are they positive real(s) greater than zero. Hence, the given function isn't periodic.
This is the method to determine if a function is periodic or not, "by definition".
If you weren't able to obtain a suitable value for T for a function you know is periodic, then you've simply made some sort of calci error.
Thus, please revert with the periodic function whose periodicity you weren't able to prove.