[SOLVED] Convergence/divergence of an integral

• May 26th 2008, 12:06 PM
Moo
[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 :))
• May 26th 2008, 05:02 PM
NonCommAlg
Quote:

Originally Posted by Moo
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.
• May 27th 2008, 09:02 AM
Moo
Thanks a bunch !

You're great (Bow)