- #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: