How to calculate primitive functions on maximal intervals for periodic functions?

[Lotto]

Homework Statement

My task is to find all primitive functions to $f(x)= \frac{\sin^2 x}{1+\sin^2 x}$ on maximal intervals.

Relevant Equations

For intervals $\left(-\frac{\pi}{2}+k\pi, \frac{\pi}{2}+k\pi \right)$ it is quite easy: $F(x)=x-\frac{1}{\sqrt 2} \arctan{\left(\sqrt 2 \tan x\right)}+C_k$. ($C_k$ corresponds to a particular $k$)

Since $f(x)$ is continuous in $\mathbb R$, it has a primitive function in $\mathbb R$ as well, so we have to define $F(x)$ also for points $\frac{\pi}{2}+k\pi$.

$\lim_{x \to \left(\frac{\pi}{2}+k\pi \right)^-} F(x) =\frac{\pi}{2}+k\pi -\frac{\pi}{2\sqrt 2}+C_k$
$\lim_{x \to \left(\frac{\pi}{2}+k\pi \right)^+} F(x) =\frac{\pi}{2}+k\pi +\frac{\pi}{2\sqrt 2}+C_{k+1}$

And since it is continuous, we can write that $C_{k+1}= -\frac{\pi}{\sqrt 2}+C_k$. But now I don't know how to continue. Am I close to the solution? How to write the whole definition for $F(x)$?

 

[WWGD]

Er, no , be careful. Not every a.e. continuous nor continuous function has a well-defined antiderivative. Absolute Continuity is the minimal condition. The derivative of the Cantor Function is a classic example. The derivative of any a.e. 0 not identically 0 function will integrate to 0, but its antiderivative isn't a constant.

 

[PeroK]

Homework Statement: My task is to find all primitive functions to $f(x)= \frac{\sin^2 x}{1+\sin^2 x}$ on maximal intervals.
Relevant Equations: For intervals $\left(-\frac{\pi}{2}+k\pi, \frac{\pi}{2}+k\pi \right)$ it is quite easy: $F(x)=x-\frac{1}{\sqrt 2} \arctan{\left(\sqrt 2 \tan x\right)}+C_k$. ($C_k$ corresponds to a particular $k$)

Since $f(x)$ is continuous in $\mathbb R$, it has a primitive function in $\mathbb R$ as well, so we have to define $F(x)$ also for points $\frac{\pi}{2}+k\pi$.

$\lim_{x \to \left(\frac{\pi}{2}+k\pi \right)^-} F(x) =\frac{\pi}{2}+k\pi -\frac{\pi}{2\sqrt 2}+C_k$
$\lim_{x \to \left(\frac{\pi}{2}+k\pi \right)^+} F(x) =\frac{\pi}{2}+k\pi +\frac{\pi}{2\sqrt 2}+C_{k+1}$

And since it is continuous, we can write that $C_{k+1}= -\frac{\pi}{\sqrt 2}+C_k$. But now I don't know how to continue. Am I close to the solution? How to write the whole definition for $F(x)$?

As $x \to \frac \pi 2$, we have $\arctan(\sqrt 2 \tan x) \to \frac \pi 2$. So, we can define $F(\frac \pi 2) = \frac \pi 2 - \frac \pi {2\sqrt 2}$ and we have a continous primitive function.

 

[Lotto]

As $x \to \frac \pi 2$, we have $\arctan(\sqrt 2 \tan x) \to \frac \pi 2$. So, we can define $F(\frac \pi 2) = \frac \pi 2 - \frac \pi {2\sqrt 2}$ and we have a continous primitive function.

So, could we write the solution in this form?

For $\left(-\frac{\pi}{2}+k\pi, \frac{\pi}{2}+k\pi\right)$: $F(x) = x-\frac{1}{\sqrt 2} \arctan{\left(\sqrt 2 \tan x\right)}+C_k-k \frac{\pi}{\sqrt 2}$
For $\frac{\pi}{2}+k\pi$: $F(x) = \frac{\pi}{2}+k\pi-\frac{1}{\sqrt 2}\frac{\pi}{2}+C_k - k \frac{\pi}{\sqrt 2}$

We subtract $k \frac{\pi}{\sqrt 2}$ because when we go from one interval with $k$ to another one with $k+1$ (when considering the unmodified function), the value at point $\frac{\pi}{2}+k\pi$ is greater by $k \frac{\pi}{\sqrt 2}$, so if we want the function to be continuous, we always have to shift it on given interval by $k \frac{\pi}{\sqrt 2}$ downwards.

I imagine it this way because the graph of the unmodified function looks like this

 

[pasmith]

For a continuous function with period $T$ (such as $\sin^2 x/(1 + \sin^2 x)$), its antiderivative on $[0, \infty)$ can be written as $$\begin{split} F(t) &= \int_0^t f(s)\,ds + C \\ &= \sum_{n=0}^{\lfloor t/T \rfloor - 1} \int_{nT}^{(n+1)T} f(s)\,ds + \int_{\lfloor t/T \rfloor T}^t f(s)\,ds + C \\ &= \left\lfloor \frac{t}{T} \right\rfloor \int_0^T f(s)\,ds +\int_0^{t - \lfloor t/T \rfloor T}f(s)\,ds + C \end{split}$$ so that the only integral you actually need to calculate is $\int_0^t f(s)\,ds$ for $t \in [0,T]$. This can be extended to negative $t$, but one must be mindful of how $\lfloor \cdot \rfloor$ works for negative numbers.