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Dirichlet's Convergence Test - Improper Integrals

  • Thread starter estro
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  • #1
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Hello,

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

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

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

  • #2
lanedance
Homework Helper
3,304
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shouldn't you consider the integral
[tex]\int_a^{\infty} f(t) g(t) dt = \int_a^{\infty} t.cost.sint.dt[/tex]

then you can say that is convergent

[tex]G(x) \int^x costsintdt=\int^x g(t)dt [/tex]
is bounded but oscillating function
 
  • #3
241
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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 [tex]G(x)=\int_a^x g(x)dx[/tex] is bounded for every x.
The only thing that makes me unsure is that when x goes to infinity I simply can't integrate [tex]\int_a^x g(x)dx [/tex] but I can for every x.
 
  • #4
Gib Z
Homework Helper
3,346
5
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 [itex] f(x) = 1/x [/itex] and [itex] g(x) = \sin x [/itex] , or in the example you did. The conditions on f(x) (monotone decreasing, tending to 0) are strong enough to ensure that even though [itex] \int^x_a g(t) dt [/itex] may oscillate too much for it to converge, as long as it's bounded, then the extra factor of f dampens it enough for [itex] \int^x_a f(t) g(t) dt [/itex] to converge.
 
  • #5
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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 [tex]\int_a^x sintcostdt\[/tex] is bounded for every x.
 
  • #6
Gib Z
Homework Helper
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Yup, a textbook example.
 
  • #7
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Thanks a lot!
 

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