Proof of 3 cases of convergence of the Dirichlet Integral(Please verify)

In summary: This isn't right. Again, you're using the p-test but the limits are inappropriate and the p-test is not applicable to this series. Also, the statement "We know that eventhough n becomes larger the entire sum decreases" is not true for the series in question.
  • #1
Hummingbird25
86
0

Homework Statement



Case (1)

Given the Dirichlet Integral

[tex]I = \int_ {0}^{\infty }\frac{sin(x)}{x} dx[/tex]

Prove that this is convergent.

Case(2)

Given the Dirichlet Integral

[tex]I = \int_ {0}^{\infty }\frac{|sin(x)|}{x} dx[/tex]

prove that it is divergent.

case(3)

Given the series

[tex]I = \int_ {0}^{n \pi }\frac{sin(x)}{x} dx[/tex]

prove that it converges

The Attempt at a Solution



Proof (1)

Acording to the the definition of the improper integral

The above integral can be written as

[tex]I = \int_ {0}^{\infty }\frac{sin(x)}{x} dx = \int_ {0}^{1 }\frac{sin(x)}{x} dx + \int_ {1}^{\infty }\frac{sin(x)}{x} dx[/tex]

The first integral is clearly convergent by the p-test and test by comparison.

The next integral

[tex]\int_ {1}^{\infty }\frac{sin(x)}{x} dx[/tex] needs to be analyzed futher.

By the use partial integral

[tex]I = \int_ {0}^{t }\frac{sin(x)}{x} dx = [ \frac{-cos(x)}{x}]_{1}^{t} - \int_ {1}^{t }\frac{cos(x)}{x^2} dx[/tex]

Where

[tex]\mathop{\limits} \lim_{t \to \infty} [ \frac{-cos(x)}{x}]_{1}^{t} = cos(1)[/tex]

Observing the remaining integral

[tex]\mathop{\limits} \lim_{t \to \infty} \int_{1}^{t} \frac{cos(x)}{x^2} dx = \int_{1}^{\infty} \frac{cos(x)}{x^2} dx[/tex]

Therefore

[tex]\int_ {1}^{\infty }\frac{sin(x)}{x} dx = cos(1) - \int_{1}^{\infty} \frac{cos(x)}{x^2} dx[/tex]

since

[tex]\frac{|cos(x)|}{x^2} \leq \frac{1}{x^2}[/tex] is convergent by the comparison test.

then

[tex]\int_{1}^{\infty} \frac{|cos(x)|}{x^2} dx[/tex] is convergent.

and thusly

[tex]\int_{1}^{\infty} \frac{cos(x)}{x^2} dx[/tex] is convergent.

Therefore

[tex]\int_{1}^{\infty} \frac{sin(x)}{x} dx = cos(1) - \int_{1}^{\infty} \frac{cos(x)}{x^2} dx[/tex] is convergent.

Thusly
[tex]\int_{0}^{\infty} \frac{sin(x)}{x} dx[/tex] is convergent.

q.e.d.

proof(2)

By the test of comparison, the integral is divergent since

[tex]\frac{|sin(x)|}{x} > \frac{1}{x}[/tex] and since the looking at this as a p-series where p = 1 also leads to divergens.

q.e.d.

proof (3)

By increasing n, then the limit of the series will always tend to zero, this shows that the series is bounded and continous, therefore it is also convergent!

q.e.d.

Sincerely Maria.

p.s. I hope there is someone who will review my proofs and conclude if they are okay and if there is something missing!
 
Last edited:
Physics news on Phys.org
  • #2
Hummingbird25 said:

[tex]I = \int_ {0}^{\infty }\frac{sin(x)}{x} dx = \int_ {0}^{1 }\frac{sin(x)}{x} dx + \int_ {1}^{\infty }\frac{sin(x)}{x} dx[/tex]

The first integral is clearly convergent by the p-test and test by comparison.

I'm not understanding how you can use test by comparison to determine that the first integral is convergent. Would you explain.

[tex]I = \int_ {0}^{t }\frac{sin(x)}{x} dx = [ \frac{-cos(x)}{x}]_{1}^{t} - \int_ {1}^{t }\frac{cos(x)}{x^2} dx[/tex]

Typo here. The bounds on the LHS should be from 1 to t.

Other than that, proof (1) looks OK. In proof (2), you incorrectly state that |sin x|/x > 1/x. In proof (3), couldn't you just use the result of proof (1)?
 
  • #3
e(ho0n3 said:
I'm not understanding how you can use test by comparison to determine that the first integral is convergent. Would you explain.

I treat the first integral as a p-series where p = 1 and from this conclude that that integral is convergent.

Typo here. The bounds on the LHS should be from 1 to t.

I will change that
In proof (2), you incorrectly state that |sin x|/x > 1/x.

Would say its wrong to use the comparison test here?

In proof (3), couldn't you just use the result of proof (1)?

You mean I shold say that RHS integral is divergent by the comparison test?

Sincerely
Maria
 
  • #4
Hummingbird25 said:
I treat the first integral as a p-series where p = 1 and from this conclude that that integral is convergent.

I still don't understand. The p-test works for series of the form 1/xp. I assume that you're using the fact that sin x / x < 1/x and you're letting p = 1. But then how are you applying the p-test? The limits aren't even appropriate. You can use the comparison test though, but if you do, the integral of 1/x from 0 to 1 diverges.

Would say its wrong to use the comparison test here?

No it isn't. It's just that you wrote |sin x| / x > 1/x when it should be |sin x| / x < 1/x. See comment above.

You mean I shold say that RHS integral is divergent by the comparison test?

What I'm saying is that if you know that (1) is true, then certainly (3) is true if n goes to infinity because it reduces to (1).
 
  • #5
e(ho0n3 said:
I still don't understand. The p-test works for series of the form 1/xp. I assume that you're using the fact that sin x / x < 1/x and you're letting p = 1. But then how are you applying the p-test? The limits aren't even appropriate. You can use the comparison test though, but if you do, the integral of 1/x from 0 to 1 diverges.

How would you recommend that I change then to make it more understandable? "No it isn't. It's just that you wrote |sin x| / x > 1/x when it should be |sin x| / x < 1/x. See comment above."

Proof(2):
So basic I say. By the result in (1), then the integral diverges by the comparison test
|sin x| / x < 1/x

"What I'm saying is that if you know that (1) is true, then certainly (3) is true if n goes to infinity because it reduces to (1)."

To give it more independence from the above (1) and (2) could I use the fact that the integral since it referred as series then the integral

Proof(3):

[tex]I_n = \int_{0}^{n \pi} \frac{sin(x)}{x} dx[/tex] has the corresponding series

[tex]I_n = \sum_{j = 0}^{n \pi} \frac{sin(j)}{j}[/tex] converges. if and only [tex]n \neq 0[/tex] and tends to infinity. We know that eventhough n becomes larger the entire sum decreases and since

[tex]\mathop{\lim} \limit_{n \to infinity} I_n = 0[/tex] then the series I_n converges. By the series convergens theorem.

q.e.d.

How does this look now?

Sincerely Maria.
 
Last edited:
  • #6
Hummingbird25 said:
How would you recommend that I change then to make it more understandable?

Easy: provide the details. Show me, in detail, why the first integral converges.

So basic I say. By the result in (1), then the integral diverges by the comparison test |sin x| / x < 1/x

No. (1) does not imply (2).

To give it more independence from the above (1) and (2) could I use the fact that the integral since it referred as series then the integral

Proof(3):

[tex]I_n = \int_{0}^{n \pi} \frac{sin(x)}{x} dx[/tex] has the corresponding series

[tex]I_n = \sum_{j = 0}^{n \pi} \frac{sin(j)}{j}[/tex] converges. if and only [tex]n \neq 0[/tex] and tends to infinity. We know that eventhough n becomes larger the entire sum decreases and since

[tex]\mathop{\lim} \limit_{n \to infinity} I_n = 0[/tex] then the series I_n converges. By the series convergens theorem.

q.e.d.

How does this look now?

This looks fine. But, it is easier and clearer, in my opinion, that (3) reduces to (1) as n goes to infinity and thus converges.
 
  • #7
In your proof of (3): You seem to be applying the integral-test backwards, i.e. an integral converges if its corresponding series converges. I don't know if this is true. I'm reasoning it isn't because the integral is a larger sum than its corresponding series.
 
  • #8
e(ho0n3 said:
In your proof of (3): You seem to be applying the integral-test backwards, i.e. an integral converges if its corresponding series converges. I don't know if this is true. I'm reasoning it isn't because the integral is a larger sum than its corresponding series.

Hi

Regarding Proof(1) [tex]\mathop{\lim} \limit_{n \to 0} \frac{sin(x)}{x} = 0[/tex] therefore the integral

[tex]\int_{0}^{1} \frac{sin(x)}{x}[/tex] converges.

Regarding Proof(2)

How would you recommend aproaching it? If not by the comparison test?

Regarding Proof(3)

Oh :-( I really thought I had it. How would you approach if the proof of this is suppose to to be independent of (1) ?

Cheers
Maria

p.s. I hope there is somebody who maybe would help me correct this within the next couple of hours, because I need present this tomorrow and therefore natural should be correct. Thanks in Advance and God bless.
 
Last edited:

Related to Proof of 3 cases of convergence of the Dirichlet Integral(Please verify)

1. What is the Dirichlet integral?

The Dirichlet integral is an improper integral that is used to evaluate the convergence or divergence of certain types of infinite series. It is named after mathematician Peter Gustav Lejeune Dirichlet.

2. What are the three cases of convergence for the Dirichlet integral?

The three cases of convergence for the Dirichlet integral are when the function being integrated is a bounded and monotonic function, when the function is a bounded and piecewise continuous function, and when the function is a bounded and continuously differentiable function.

3. How do you verify the convergence of the Dirichlet integral?

To verify the convergence of the Dirichlet integral, you must evaluate the integral using one of the three convergence cases. If the integral converges, it means that the infinite series being represented by the integral also converges. If the integral diverges, the series also diverges.

4. Why is it important to verify the convergence of the Dirichlet integral?

It is important to verify the convergence of the Dirichlet integral because it allows us to determine whether an infinite series is convergent or divergent. This is crucial in many mathematical and scientific applications, as it helps us understand the behavior of functions and make accurate predictions.

5. Can the Dirichlet integral be used to evaluate all types of infinite series?

No, the Dirichlet integral can only be used to evaluate certain types of infinite series. It is specifically designed to evaluate series that have alternating signs and that either decrease or increase monotonically. Other types of series may require different methods of evaluation.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
721
  • Calculus and Beyond Homework Help
Replies
13
Views
533
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
807
  • Calculus and Beyond Homework Help
Replies
23
Views
999
  • Calculus and Beyond Homework Help
Replies
5
Views
737
  • Calculus and Beyond Homework Help
2
Replies
47
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
379
  • Calculus and Beyond Homework Help
Replies
2
Views
878
Back
Top