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

1. Feb 27, 2008

### Hummingbird25

1. The problem statement, all variables and given/known data

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

3. 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 remaing 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!

Last edited: Feb 27, 2008
2. Feb 27, 2008

### e(ho0n3

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

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. Feb 27, 2008

### Hummingbird25

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

I will change that
Would say its wrong to use the comparison test here?

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

Sincerely
Maria

4. Feb 27, 2008

### e(ho0n3

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.

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.

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. Feb 27, 2008

### Hummingbird25

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 refered 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.

Last edited: Feb 27, 2008
6. Feb 27, 2008

### e(ho0n3

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

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

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. Feb 27, 2008

### 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.

8. Feb 27, 2008

### Hummingbird25

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.

Last edited: Feb 27, 2008