# Thread: Help with proving that Improper Integral is Divergent

1. ## Help with proving that Improper Integral is Divergent

The problem is attached in this post.

Lim t -> ∞ ∫ dx/xlnx from 1 to t

u-substitution:

u=lnx
du=1/x dx

Lim t -> ∞ ∫ 1/u du

Lim t -> ∞ ln u

Lim t -> ∞ ln(lnx) from 1 to t

Lim t -> ∞ ln(lnt) - ln(0)

= ∞ - ∞ = 0 (This is incorrect since the answer is that the integral is divergent).

2. ## Re: Help with proving that Improper Integral is Divergent

Originally Posted by student93
The problem is attached in this post.

Lim t -> ∞ ∫ dx/xlnx from 1 to t

u-substitution:

u=lnx
du=1/x dx

Lim t -> ∞ ∫ 1/u du

Lim t -> ∞ ln u

Lim t -> ∞ ln(lnx) from 1 to t

Lim t -> ∞ ln(lnt) - ln(0)

= ∞ - ∞ = 0 (This is incorrect since the answer is that the integral is divergent).

you're fine up until trying to evaluate the definite integral. You can't do algebra with infinities and get usable answers. What you have to do is use the integral test with the harmonic series. Since you know the harmonic series diverges it must be that the integral of 1/x diverges as well.

3. ## Re: Help with proving that Improper Integral is Divergent

Do you mean the Direct Comparison Test? Also could you show how would do an integral test for this question?

4. ## Re: Help with proving that Improper Integral is Divergent

No I mean the Integral test.

Just read that link on what the test is. Given that the harmonic series $\displaystyle{\sum_{k=1}^{\infty}} \dfrac{1}{k}$ diverges it should be obvious how to apply the integral test to this problem.

5. ## Re: Help with proving that Improper Integral is Divergent

You don't need to know about the integral test for convergence of a series for this problem. Since the integral is improper at both endpoints, the convergence needs to be analyzed separately (you should have a special definition for this case), and it needs to converge on both sides to be called convergent. So just take some a, 1<a<\infty, and

$\int_1^\infty \frac{1}{x\ln{x}}\,dx = \int_1^a \frac{1}{x\ln{x}}\,dx + \int_a^\infty \frac{1}{x\ln{x}}\,dx$

and you correctly calculated that both integrals diverge. So the original integral diverges.

- Hollywood

6. ## Re: Help with proving that Improper Integral is Divergent

you can take the lim as b approaches infinity with b being your upper bound of integration then solve the indefinite integral then take the limit of F(b) - F(1)

or you can use the limit of the integral a smaller function and if it diverges then the larger function will also diverge, this is the comparison test.

it does diverge