Let $f(x)$ be a continuous function on all real numbers such that $|f(t)| \le Ke^{at}$ for all $t \ge 0$ and for some constants $K>0$ and $a$ a real number. And let $f(x)$ be a periodic function for a period T, $f(x)=f(x+T)$ Prove that $\int_0^{\infty}f(t)e^{-st}dt = \frac{1}{1-e^{-sT}} \int_0^{T} f(t)e^{-st}dt$. I tried everything, using properties of periodic functions, but I'm stuck. Only thing I proved is that function function converges when $s>a$ (which was a first part to the problem). Any ideas?


Hint: Split the integral up into a sum over intervals of the form $[nT,(n+1)T]$. On each interval use periodicity of $f$ and properties of exp to translate it into an integral over $[0,T]$.

  • $\begingroup$ Thank you for your answer, it's a little bit more clear to me, but I'm still kind of stuck. If I understood correctly should I split that integral over infinite sum like from 0 to T, then T to 2T, then 2T to 3T.. and so on. What should I do next with it when I look at that interval? I changed f(t) to f(t+T). Should I use an substitution for z=t+T? Elaborate please if you can. Thank you! $\endgroup$ – Darko Dekan Sep 17 '16 at 14:20
  • $\begingroup$ Yes an infinite sum as you say. On each interval make a substitution $t=nT+x$ with $0\leq x\leq T$ and use $e^{a+b}=e^a e^b$. You will see a geometric series popping out. $\endgroup$ – H. H. Rugh Sep 17 '16 at 14:26
  • $\begingroup$ I finally solved it. I was actually thinking that somewhere should appear geometric series because of that constant on right side of equation. Thank you so much for your help! $\endgroup$ – Darko Dekan Sep 17 '16 at 15:22

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