Fie \( f:[0,\infty)\to\mathbb{R} \) o functie continua astfel incat
\( \lim_{x\to\infty}\left(f(x)+\int_0^xf(t)dt\right) \) exista.
Aratati ca \( \lim_{x\to\infty}f(x)=0 \).
Functie continua pentru care exista limita de integrala
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Functie continua pentru care exista limita de integrala
Last edited by Cezar Lupu on Sat Mar 08, 2008 10:41 pm, edited 1 time in total.
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
\( F(x)=\int_{0}^{x}f(t)dt \)
\( F^{\prime}(x)+F(x)=g(x) \) tends to \( L\in R \)
Solving the differential equation
\( F(x)=e^{-x}\int_{0}^{x}e^{t}g(t)dt+Ae^{-x} \)
with A fixed real, we get
\( f(x)=F^{\prime}(x)=-\frac{\int_{0}^{x}e^{t}g(t)dt}{e^{x}}+g(x)-Ae^{-x} \)
Now if \( g \) tends to \( L \), then \( \frac{\int_{0}^{x}e^{t}g(t)dt}{e^{x}} \) tends also to \( L \).
Finally \( f \) tends to \( 0 \).
\( F^{\prime}(x)+F(x)=g(x) \) tends to \( L\in R \)
Solving the differential equation
\( F(x)=e^{-x}\int_{0}^{x}e^{t}g(t)dt+Ae^{-x} \)
with A fixed real, we get
\( f(x)=F^{\prime}(x)=-\frac{\int_{0}^{x}e^{t}g(t)dt}{e^{x}}+g(x)-Ae^{-x} \)
Now if \( g \) tends to \( L \), then \( \frac{\int_{0}^{x}e^{t}g(t)dt}{e^{x}} \) tends also to \( L \).
Finally \( f \) tends to \( 0 \).