MHB Anti-derivatives of the periodic functions

cbarker1
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Dear Everyone,

I do not know how to begin with the following problem:Suppose that $f$ is $2\pi$-periodic and let $a$ be a fixed real number. Define $F(x)=\int_{a}^{x} f(t)dt$, for all $x$ .
Show that $F$ is $2\pi$-periodic if and only if $\int_{0}^{2\pi}f(t)dt=0$.
Thanks,
Cbarker1
 
Last edited:
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Cbarker1 said:
...if and only if $\int_{0}^{2\pi}f(t)dt$.
if and only if [math]\int_0^{2 \pi}f(t)~dt[/math] is what?

-Dan
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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