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Improper Integrals

  • #1

Homework Statement


Suppose 0<a<1.
1) Show that
0<Integral(0 to a)1/(sqrt(1-x^2)<=Integral(0 to a)1/(sqrt(1-x)<=2

2) Show that I(a)=Integral(0 to a)1/(sqrt(1-x^2) is increasing and bounded by 2.

3) Deduce that Integral(0 to 1)1/(sqrt(1-x^2) exists and has an improper integral.


Homework Equations


Not sure that there are any relevant equations that are too useful.


The Attempt at a Solution


I can prove with simple algebra that Integral(0 to a)1/(sqrt(1-x^2)<=Integral(0 to a)1/(sqrt(1-x). But I'm not sure how to show that it is all less then 2. I don't have an atttempt for parts 2 and 3 because they heavily rely on step 1.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


Suppose 0<a<1.
1) Show that
0<Integral(0 to a)1/(sqrt(1-x^2)<=Integral(0 to a)1/(sqrt(1-x)<=2

2) Show that I(a)=Integral(0 to a)1/(sqrt(1-x^2) is increasing and bounded by 2.

3) Deduce that Integral(0 to 1)1/(sqrt(1-x^2) exists and has an improper integral.


Homework Equations


Not sure that there are any relevant equations that are too useful.


The Attempt at a Solution


I can prove with simple algebra that Integral(0 to a)1/(sqrt(1-x^2)<=Integral(0 to a)1/(sqrt(1-x). But I'm not sure how to show that it is all less then 2. I don't have an atttempt for parts 2 and 3 because they heavily rely on step 1.
Just evaluate last integral by integrating it.

Hint for part (b): If

[tex]I(a) = \int_0^a \frac{1}{\sqrt{1-x^2}}\ dx[/tex]

what is I'(a)?
 
  • #3
By the Fundamental Theorem of Calculus, if
[tex]I(a) = \int_0^a \frac{1}{\sqrt{1-x^2}}\ dx[/tex], then
I'(a)=1/(sqrt{1-x^2}.
Then, since 0<x<1, we see that sqrt(1-x^2)>0.
Thus, I is increasing since I'(a) is positive.
I'm still not sure how to show that I is bounded by 2...
 
  • #4
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5,192
What's the value of I(a)? The integral is an easy trig substitution.
 
  • #5
LCKurtz
Science Advisor
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By the Fundamental Theorem of Calculus, if
[tex]I(a) = \int_0^a \frac{1}{\sqrt{1-x^2}}\ dx[/tex], then
I'(a)=1/(sqrt{1-x^2}.
I'(a) has no x in it.

Then, since 0<x<1, we see that sqrt(1-x^2)>0.
Thus, I is increasing since I'(a) is positive.
I'm still not sure how to show that I is bounded by 2...
You have shown your integral is less or equal to

[tex]\int_0^1 \frac 1 {\sqrt{(1-x)}}\, dx[/tex]

Just integrate that. It's easy.
 
  • #6
So the integral is equal to inverse sine of x. So evaluated from 0 to 1 we have that it is equal to pi/2 <2?
 

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