Tough Calc Problem I Saw Recently

Calculus Level 3

Does $\displaystyle \int_1^{\infty} \frac{1}{x\log^n(x)} \ dx$ converges for all numbers $n$ greater than 1?

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To prove that you haven't guessed, please post a solution down below.

Yes No

2 solutions

Chew-Seong Cheong
Jan 27, 2020

Consider the case $n=0$ . Then $\displaystyle I_0 = \int_0^\infty \frac 1x dx = \lim_{x \to \infty} \ln x - \ln 1$ , which does not converge. No , the integral does not converge.

I assumed n was an integer at least 1. Allowing n=0 removes this problem from category of 'tough calc problem'.

Richard Desper - 1 year, 4 months ago

It clearly says any real number $n$ . I just like to find the easiest way to solve a problem.

Chew-Seong Cheong - 1 year, 4 months ago

Sorry to all of you. I forgot to state it in the problem, but n must be 1 or higher. I will update the problem. I apologize.

Aashish Cheruvu - 1 year, 3 months ago

A Former Brilliant Member
Jan 26, 2020

The value of the indefinite integral is $\dfrac{\ln^{1-n} x}{1-n}+C$ , where $C$ is the integration constant. So the definite integral is not convergent for all values of $n$

To clarify: this converges for n > 1, right? What happens when n=1, where this formula does not apply?

Richard Desper - 1 year, 4 months ago

Apparently if n =1, the antiderivative is ln ln x. So this problem is actually pretty simple.

Richard Desper - 1 year, 4 months ago

For $n>1$ , the value of the integral at the lower limit becomes infinite. So in this case also, the integral diverges.

A Former Brilliant Member - 1 year, 4 months ago

This does not converge for n>1.

Aashish Cheruvu - 1 year, 3 months ago

Tough Calc Problem I Saw Recently

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