# Determine if the following converges or diverges

1. Nov 5, 2007

### lovelylila

Determine if the following converges or diverges as x approaches infinity by either evalutation, the direct comparison test, or the limit comparison test: (It's a Calculus II, AP Calculus BC level of problem)

the integral of (lnx/(square root of (x^2-1))), from 1 to infinity.

* I do not know how to evaluate the integral analytically, so I tried to use either the direct comparison test or limit comparison test. I can't seem to find another function that will "sandwich" that function (and thus prove convergency) or one that will prove it's divergency. I've tried 1/x, 1/(x^2), etc and I'm stuck. Any help on a function to use would be very much appreciated- I'm frustrated beyond belief!

Direct Comparison Test:
0< f(x)< g(x) proves that f(x) converges if g(x) also converges
f(x)> g(x)---proves that f(x) diverges if g(x) diverges
Limit Comparison Test:
if the limit as x approaches infinity of f(x)/g(x) is a finite, non-zero number, then f(x) has the same behavior of convergence as g(x)

2. Nov 5, 2007

### EnumaElish

0 < Ln x < x-1 for all x > 1. Does this help?

3. Nov 5, 2007

### Midy1420

look at ln(x)/Sqrt(x^2-1) its only improper at infinity so if you look at what the function looks like when x --> infinity we get ln(x) /x . From this we can compare it to anything that diverges and is smaller than that...easiest example 1/x. ln(x)/Sqrt(x^2-1) > 1/x for all x > 1

4. Nov 5, 2007

### lovelylila

thank you, but don't you have to choose a function that is greater/less than for all numbers from 1 to infinity? I don't understand how you can say for x>1, because one itself is the lower limit of the integral and shouldn't it thus be included? or in this type of problem is it to be assumed that the f(x)> g(x) for x>1, not including one?

5. Nov 5, 2007

### EnumaElish

That's an easy fix; 0 < Ln x < Sqrt(x-1) for all x > 1. [See edit.]

[Although: "Integral of 1/Sqrt[1+x] does not converge on {1,\[Infinity]}" may be a problem.]

"x > 1" should be replaced with "x > 2.5 (approximately)."

I do not think either of these examples is a solution.

Last edited: Nov 5, 2007
6. Nov 5, 2007

### soccergal13

thank you all very much for your help , but does anyone know what a solution could be? what you're all suggesting makes sense, but nothing works 100%....

7. Nov 5, 2007

### EnumaElish

On a 2nd thought, 1/x works, but is a little involved.

a. For 1 < x < 2.5, 0 < ln(x)/Sqrt(x^2-1) < 1/x.

b. For x > 2.5, ln(x)/Sqrt(x^2-1) > 1/x.

Fact "a" is inconvenient because it seemingly precludes a direct comparison with 1/x (our hope for a divergence result). So here is the involved part. Below, I will use the expressions "F diverges" and "F = $\infty$" interchangeably.

$$\int_1^\infty f(x) dx = \int_1^{2.5} f(x) dx + \int_{2.5}^\infty f(x) dx.$$

In shorthand, I(1/x) = I1(1/x) + I2(1/x) for f(x) = 1/x.

1/x evaluates to a finite (positive) result over [1, 2.5]. Therefore I1(1/x) is finite (and positive). Let I1(1/x) = J > 0.

Then, I(1/x) = J + I2(1/x). Since I(1/x) diverges, so does I2(1/x): $\infty$ = J + I2(1/x) implies I2(1/x) = $\infty$ - J = $\infty$. Call this Result 1.

From fact "b," I2(ln(x)/Sqrt(x^2-1)) > I2(1/x) = $\infty$ (from Result 1). Therefore I2(ln(x)/Sqrt(x^2-1)) diverges. Call this Result 2.

Now...

We know we can write I(ln(x)/Sqrt(x^2-1)) = I1(ln(x)/Sqrt(x^2-1)) + I2(ln(x)/Sqrt(x^2-1)). Also, I1(ln(x)/Sqrt(x^2-1)) = K > 0 (positive & finite).

Can you take it from here?

Last edited: Nov 5, 2007
8. Nov 5, 2007

### EnumaElish

Forum rules prohibit giving out answers "100%." What are your thoughts on the subject?

9. Nov 6, 2007

### soccergal13

thanks again for all your help, i think i understand it now.

i wasn't asking for a "100%" answer, i was asking for a hint to a solution that worked 100% for this problem, there is a difference. i'll watch my wording next time.

10. Nov 6, 2007

### EnumaElish

Good, nice; am glad to help.

11. Nov 6, 2007

### Midy1420

there are numerous number of solution to these kind of problems. you asked to show whether it diverges or converges...it diverges if it is larger than something than diverges then it too diverges (remember the converse is not true) 1/(100x) could be an answer that can work all the times but so can 1/(1004x)

Last edited: Nov 6, 2007
12. Nov 6, 2007

### EnumaElish

Midy1420 is right; that's a much simpler solution.