# Improper Integrals

1. Sep 27, 2008

### demersal

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

$$\int$$ $$\frac{dx}{x\sqrt{x^2-4}}$$ from 2 to infinity

2. Relevant equations

Trigonometric substitution, improper integrals

3. The attempt at a solution

$$\int$$ $$\frac{dx}{x\sqrt{x^2-4}}$$ from 2 to infinity

= $$\underbrace{lim}_{t->inf}$$ $$\int$$ $$\frac{dx}{x\sqrt{x^2-4}}$$ from 2 to t

x = 2 sec $$\theta$$
dx = 2 sec $$\theta$$ tan $$\theta$$
(x^2-4)^(1/2) = 2 tan $$\theta$$

= $$\underbrace{lim}_{t->inf}$$ $$\int$$ $$\frac{2sec\theta tan\theta d\theta}{2sec\theta2tan\theta}$$ from 2 to t

= $$\underbrace{lim}_{t->inf}$$ $$\int$$ $$\frac{d\theta}{2}$$ from 2 to t

= $$\underbrace{lim}_{t->inf} \theta/2$$

= $$\underbrace{lim}_{t->inf}$$ $$(1/2)(arc tan(\frac{2}{\sqrt{x^(2)-4)}}$$ and plugging in the limits of the integrand ...

= $$\underbrace{lim}_{t->inf}$$ $$(1/2)(arc tan(\frac{2}{\sqrt{x^(2)-4)}} - ((1/2)(arc tan(\frac{2}{\sqrt{(2)^(2)-4)}})$$

What I'm wondering is why the answer to the integral I got isn't negative (I checked on an online integral calculator and it has -.5arctan ...) because I just don't see where that would come into play! Also, what do I do now that I plugged that 2 in and got an undefined number? Should I have split this up into the product of two integrals? How would I do that and was all this work done in vain? (I hope not!!)

2. Sep 27, 2008

Staff Emeritus
Somehow your TeX was mangled, making your message hard to figure out. However...

Your answer has to be positive, since 1/x is positive from 2 to infinity, and sqrt(anything) is defined to be positive. So your integrand is always positive, and adding up a lot of positive things will be positive.

3. Sep 27, 2008

### Dick

If x=2*sec(theta), then x=2 corresponds to theta=0 and x->infinity to theta->pi/2. Just use the theta limits to evaluate the integral instead of going all the way back to x and having to use a limiting process.

4. Sep 27, 2008

### demersal

If I do it that way, I get the answer as pi/4 and that is the end of it. I checked the answer online and it said the answer to the larger limit was 0, not pi/4. Am I neglecting something by just plugging in pi/2 for t?

5. Sep 27, 2008

### Dick

You didn't solve for theta correctly. theta=arctan(sqrt(x^2-4)/2). You have the argument of the arctan upside down. That's where the extra minus is coming from.

6. Sep 27, 2008

### demersal

I completely understand now, thank you! Just one last question ... how does t -> infinity correspond to theta -> pi/2? I don't see how I could just plug infinity into sqrt(x^2-4) or something

7. Sep 27, 2008

### Dick

You have to take the limit, not 'plug in'. As x->infinity, sqrt(x^2-4)/2->infinity. If you look at the graph of arctan, you'll see it approaches pi/2 as the argument goes to infinity.

8. Sep 27, 2008

### demersal

Oh, ok, I misinterpreted your original statement. Thank you for your help!