If $f$ be a periodic function with period $k$ and $f(-x)=-f(x)$ in $\bigg[-\frac{k}{2}\;,\frac{k}{2}\bigg]$. Then prove that $\displaystyle \int^{x}_{a}f(t)dt$ is a periodic function with period $k$
Solution i tried
Let $\displaystyle F(x)=\int^{x}_{a}f(t)dt$. Then $\displaystyle F(x+k)=\int^{x+k}_{a}f(t)dt$
$\displaystyle F(x+k)=\int^{x}_{a}f(t)dt+\int^{x+k}_{x}f(t)dt$
$\displaystyle F(x+k)-F(x)=\int^{x+k}_{x}f(t)dt$
I want some help How to prove $F(x)$ is periodic function. Help me plaese
Hint: $\int_x^{x+k}f(t)dx=\int_{{-k}\over 2}^{k\over 2}f(t)dt=0$. Since $f$ is $k$-periodic, (make the change of variables $u=t-x-{k\over 2}$).
Write $u=-x$, $\int_0^{k\over 2}f(t)dt=-\int_0^{{-k}\over 2}f(-u)du=\int_0^{{-k}\over 2}f(u)du=-\int_{{-k}\over 2}^0f(u)du$.
Hint
Because $f$ is supposed to be periodic $$\int^{x+k}_{x}f(t)dt = \int^{\frac{k}{2}}_{-\frac{k}{2}}f(t)dt$$
And as $f$ is supposed to be odd on $[-\frac{k}{2},\frac{k}{2}]$, the RHS of above equality vanishes. Hence $F(x+k)-F(x)=0$.
Hint: let $u = t-x-\frac{k}{2}$ in your final integral and use the fact that $f$ is odd in $[-k/2,k/2]$.
$$\displaystyle F(x+k)-F(x)=\int^{x+k}_{x}f(t)dt$$
You're almost there. Note that
$$\displaystyle \int^{\frac{k}{2}}_{-\frac{k}{2}}f(t)dt = 0$$ because the function is odd. Also, since its periodic with period $k$, that means any shifted version of the above integral as follows
$$\displaystyle \int^{\frac{k}{2} + x_0}_{-\frac{k}{2} + x_0}f(t)dt = 0$$
is also 0. Choose $x_0=x +\frac{k}{2}$, then
$$\int^{x+k}_{x}f(t)dt=0=F(x+k) -F(x)$$