[SOLVED] Convergence/divergence of an integral

**Moo** · May 26th 2008, 12:06 PM

Hi again !

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

Text

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

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

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

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

It's ok, f=0.

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

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

It's ok for this.

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

)

**NonCommAlg** · May 26th 2008, 05:02 PM

Originally Posted by Moo

Text

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

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

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

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

this is a good question! an example is this: $f(t)=\sin t + (t+1)\cos t.$ then $F(t)=(t+1)\sin t.$ to see why this function

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

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

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

**Moo** · May 27th 2008, 09:02 AM

Thanks a bunch !

You're great

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