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Thread: [SOLVED] Convergence/divergence of an integral

  1. #1
    Moo
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    [SOLVED] Convergence/divergence of an integral

    Hi again !

    This time, I'm really struggling with a problem...


    Text

    Let $\displaystyle C_0(\mathbb{R}^+)$ be the set of functions defined over $\displaystyle \mathbb{R}^+$, continuous, with real values.

    $\displaystyle f \in C_0(\mathbb{R}^+)$. Let F be the antiderivative of f that annulates in 0.

    Let E be the subset of functions f in $\displaystyle C_0(\mathbb{R}^+)$ such that $\displaystyle I(f)=\int_0^\infty \frac{F(t)}{(1+t)^2} \ dt$ converges.

    Previous questions...
    1/ Determine the positive functions f in E such that $\displaystyle I(f)=0$.

    It's ok, f=0.

    2/ Let f be a positive function of $\displaystyle C_0(\mathbb{R}^+)$

    Show : $\displaystyle \int_0^\infty \frac{f(t)}{1+t} \ dt \text{ converges } \Longleftrightarrow f \in E$

    It's ok for this.


    Problem :
    Find an example for f (necessarily not of constant sign) in E and such that $\displaystyle \int_0^\infty \frac{f(t)}{1+t} \ dt$ diverges.




    Thanks for your help, we are really struggling with that

    (and I have no assignment, I'm asking several questions these days because I have an exam tomorrow )
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Moo View Post
    Text

    Let $\displaystyle C_0(\mathbb{R}^+)$ be the set of functions defined over $\displaystyle \mathbb{R}^+$, continuous, with real values.

    $\displaystyle f \in C_0(\mathbb{R}^+)$. Let F be the antiderivative of f that annulates in 0.

    Let E be the subset of functions f in $\displaystyle C_0(\mathbb{R}^+)$ such that $\displaystyle I(f)=\int_0^\infty \frac{F(t)}{(1+t)^2} \ dt$ converges.


    Problem :
    Find an example for f (necessarily not of constant sign) in E and such that $\displaystyle \int_0^\infty \frac{f(t)}{1+t} \ dt$ diverges.
    this is a good question! an example is this: $\displaystyle f(t)=\sin t + (t+1)\cos t.$ then $\displaystyle F(t)=(t+1)\sin t.$ to see why this function

    satisfies the condition, we first prove that $\displaystyle J=\int_0^{\infty} \frac{\sin t}{t+1} \ dt$ is convergent. this is easy to prove, because using integration

    by parts we have $\displaystyle J=1 - \int_0^{\infty} \frac{\cos t}{(t+1)^2} \ dt,$ and $\displaystyle \int_0^{\infty} \frac{\cos t}{(t+1)^2} \ dt$ is (absolutely) convergent because $\displaystyle \left|\frac{\cos t}{(t+1)^2} \right| \leq \frac{1}{(t+1)^2}.$

    thus $\displaystyle \int_0^{\infty} \frac{F(t)}{(t+1)^2} \ dt = J$ is convergent and $\displaystyle \int_0^{\infty} \frac{f(t)}{t+1} \ dt = J + \int_0^{\infty} \cos t \ dt,$ which is clearly divergent. Q.E.D.
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  3. #3
    Moo
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    Thanks a bunch !

    You're great
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