Integral Calculus - Spot the Error

[MermaidWonders]

The big blue circle has been put there by my math prof to denote the location of the error in the following solution. Why is this an error? I'm lost. :(

 

[Ackbach]

$$\int_1^{\infty}\frac1x \, dx$$
does not converge. If you were to try to get an antiderivative, the only candidate is $\ln|x|$, so formally, you'd have
$$\int_1^{\infty}\frac1x \, dx = \ln|x|\big|_{1}^{\infty} =\ln(\infty)-\ln(1)=\ln(\infty)=\infty.$$
So it doesn't converge.

 

[MermaidWonders]

Ah, so is it like we're not supposed to do the "by comparison" for integrals, especially when we don't know whether the integral will be equal to infinity or not? I mean, why didn't she just circle the "converges as well" part?

 

[tkhunny]

Ah, so is it like we're not supposed to do the "by comparison" for integrals, especially when we don't know whether the integral will be equal to infinity or not? I mean, why didn't she just circle the "converges as well" part?

Comparison is fine, but it must be a VALID comparison. The idea would be capsulized as "closer to zero".

Bad Example: It is also true that for $$x > 1$$, we have $$-e^{x} < \dfrac{1}{x^{2}}$$, but that is not helpful.

In other words, your comparison must be on the same side of the x-axis.

... the only candidate is ln|x|...

This is a little excessive. For $$x \ge 1$$, we have $$\ln(x)$$ as an additional candidate.

 

[MermaidWonders]

Can you explain what you mean by the fact that the "comparison must be on the same side of the x-axis"?

 

[HOI]

You error is that the "comparison test" applies only to positive functions. You cannot just say that \(\displaystyle -\frac{1}{x}< \frac{1}{x^2}\). Sometimes you can use the absolute value but it is NOT TRUE that \(\displaystyle \left|-\frac{1}{x}\right|= \frac{1}{x}< \frac{1}{x^2}\).

 

[tkhunny]

You error is that the "comparison test" applies only to positive functions.

Well, BOTH positive, sure, but also BOTH negative would do.

 

[HOI]

Well, BOTH positive, sure, but also BOTH negative would do.

But "the other way around". If f(x) and g(x) are both positive with f(x)< g(x) and $\int g(x) dx$ converges then $\int f(x) dx$ converges.

If f(x) and g(x) are both negative and f(x)< g(x) (so that |g(x)|< |f(x)| and $\int f(x)dx$ converges then $\int g(x) dx$ converges.