$f$ is integrable on $[0,\infty)$, and $\int_0^{\infty} |f(y)|dy < \infty$.
Prove:
Then $ F(x)=\int_0^{\infty}\frac{f(y)}{x+y}dy $ is continuous on $(0,\infty)$ and differentiable, and have $\lim\limits_{x\to \infty} F(x)=0$.
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What's your question? – May 04 '13 at 18:46
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What's the question here? Prove the latter given the former? If so please add that with an edit – rajb245 May 04 '13 at 18:46
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Prove the latter statement. – Jimmy Wang May 04 '13 at 18:47
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$f$ integrable on $[0,\infty)$ has the same meaning that $\int_0^\infty |f(y)|dy<\infty$ and $<$ not $\leq$. – May 04 '13 at 18:50
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@SamiBenRomdhane They should be same. – Jimmy Wang May 04 '13 at 18:57
1 Answers
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A brief proof
- the function $x\mapsto \frac{f(y)}{x+y}$ is continuous on $(0,+\infty)$
- $\lim_{x\to\infty}\frac{f(y)}{x+y}=0$
- $\forall a>0$ and if $x\geq a$ we have $\frac{|f(y)|}{x+y}\leq\frac{|f(y)|}{a+y}\leq\frac{1}{a}|f(y)|\quad(1)$
then use dominated convergence theorem and you have an answer for 2 points.
- the function $x\mapsto\frac{f(y)}{x+y}$ is differentiable and $x\mapsto-\frac{f(y)}{(x+y)^2}$ its derivative then use the same idea as in $(1)$ and the dominated convergence theorem and you have an answer for the last point