Dirichlet's Convergence Test - Improper Integrals

[estro]

Hello,

I have question about using Dirichlet's Convergence Test which states:
1. if f(x) is monotonic decreasing and $$\lim_{x\rightarrow \infty} f(x)=0$$
2. $$G(x)=\int_a^x g(t)dt$$ is bounded.
Then $$\int_a^\infty f(x)g(x)dx$$ is convergent.

But what about the following situation:
f(x)=1/x
g(x)=cosxsinx

Can I say that $$\int_a^x costsintdt=\int_{sina}^{sinx} tdt=t^2/2=sin^2x/2-sin^2a/2$$
for every x G(x) is bounded and by the Dirichlet's Convergence Test the integral is convergent?

 

[lanedance]

shouldn't you consider the integral
$$\int_a^{\infty} f(t) g(t) dt = \int_a^{\infty} t.cost.sint.dt$$

then you can say that is convergent

$$G(x) \int^x costsintdt=\int^x g(t)dt$$
is bounded but oscillating function

 

[estro]

It is just general question, can I do what I did in my first post?
f(x) is monotonic and deceasing with limit of 0.
and $$G(x)=\int_a^x g(x)dx$$ is bounded for every x.
The only thing that makes me unsure is that when x goes to infinity I simply can't integrate $$\int_a^x g(x)dx$$ but I can for every x.

 

[Gib Z]

You don't need to be able to integrate the infinite case, you just need to show it is bounded. That's one reason why this test is useful - often the integral you need to be bounded is just bounded, but oscillating and not convergent, like if $f(x) = 1/x$ and $g(x) = \sin x$ , or in the example you did. The conditions on f(x) (monotone decreasing, tending to 0) are strong enough to ensure that even though $\int^x_a g(t) dt$ may oscillate too much for it to converge, as long as it's bounded, then the extra factor of f dampens it enough for $\int^x_a f(t) g(t) dt$ to converge.

 

[estro]

Gib Z, thanks for your response.
So basically what I've done in my first post is a legitimate use of the Dirichlet test, right?
As $$\int_a^x sintcostdt\$$ is bounded for every x.

 

[Gib Z]

Yup, a textbook example.

 

[estro]

Thanks a lot!