Solving Improper Integrals: \int \frac{dx}{x\sqrt{x^2-4}}

[demersal]

Homework Statement

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

Homework Equations

Trigonometric substitution, improper integrals

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!)

 

[Vanadium 50]

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.

 

[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.

 

[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?

 

[Dick]

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?

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.

 

[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

 

[Dick]

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

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.

 

[demersal]

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