# 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?

Last edited:

Homework Helper
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.

Homework Helper
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.

Homework Helper
Yup, a textbook example.

estro
Thanks a lot!