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

[Hummingbird25]

Homework Statement

Case (1)

Given the Dirichlet Integral

$$I = \int_ {0}^{\infty }\frac{sin(x)}{x} dx$$

Prove that this is convergent.

Case(2)

Given the Dirichlet Integral

$$I = \int_ {0}^{\infty }\frac{|sin(x)|}{x} dx$$

prove that it is divergent.

case(3)

Given the series

$$I = \int_ {0}^{n \pi }\frac{sin(x)}{x} dx$$

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

$$I = \int_ {0}^{\infty }\frac{sin(x)}{x} dx = \int_ {0}^{1 }\frac{sin(x)}{x} dx + \int_ {1}^{\infty }\frac{sin(x)}{x} dx$$

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

The next integral

$$\int_ {1}^{\infty }\frac{sin(x)}{x} dx$$ needs to be analyzed futher.

By the use partial integral

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

Where

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

Observing the remaining integral

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

Therefore

$$\int_ {1}^{\infty }\frac{sin(x)}{x} dx = cos(1) - \int_{1}^{\infty} \frac{cos(x)}{x^2} dx$$

since

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

then

$$\int_{1}^{\infty} \frac{|cos(x)|}{x^2} dx$$ is convergent.

and thusly

$$\int_{1}^{\infty} \frac{cos(x)}{x^2} dx$$ is convergent.

Therefore

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

Thusly
$$\int_{0}^{\infty} \frac{sin(x)}{x} dx$$ is convergent.

q.e.d.

proof(2)

By the test of comparison, the integral is divergent since

$$\frac{|sin(x)|}{x} > \frac{1}{x}$$ 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!

 

[e(ho0n3]

$$I = \int_ {0}^{\infty }\frac{sin(x)}{x} dx = \int_ {0}^{1 }\frac{sin(x)}{x} dx + \int_ {1}^{\infty }\frac{sin(x)}{x} dx$$

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.

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

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)?

 

[Hummingbird25]

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

 

[e(ho0n3]

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/x^p. 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).

 

[Hummingbird25]

I still don't understand. The p-test works for series of the form 1/x^p. 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):

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

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

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

q.e.d.

How does this look now?

Sincerely Maria.

 

[e(ho0n3]

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):

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

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

$$\mathop{\lim} \limit_{n \to infinity} I_n = 0$$ 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.

 

[e(ho0n3]

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.

 

[Hummingbird25]

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) $$\mathop{\lim} \limit_{n \to 0} \frac{sin(x)}{x} = 0$$ therefore the integral

$$\int_{0}^{1} \frac{sin(x)}{x}$$ 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.