Does -ln(-∞) Equal ln(∞) in Complex Analysis?

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Homework Help Overview

The discussion revolves around the expression -ln(-∞) and its relationship to ln(∞) within the context of complex analysis and logarithmic functions. Participants explore the implications of taking the natural logarithm of negative and infinite values.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants question the validity of taking the logarithm of negative infinity and whether it can be equated to other logarithmic expressions. There is exploration of the definitions and implications of logarithmic functions in both real and complex contexts.

Discussion Status

The discussion is active, with participants providing insights and clarifications about the nature of logarithms of negative numbers and infinity. Some guidance has been offered regarding the use of complex numbers, but there is no consensus on the interpretation of ln(-∞).

Contextual Notes

There is mention of constraints related to the use of real versus complex numbers in logarithmic functions, and some participants express uncertainty about the definitions involved.

DrCrowbar
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For instance, say I have

-ln(-∞)​

Does the negative sign on the natural log cancel with the negative sign on the infinity?

Is this true?
-ln(-∞) = ln(∞)​

Thank you

-Drc
 
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Hi DrCrowbar! :smile:

There is no such thing as the ln of a negative number. :wink:

(unless we're allowed complex numbers, in which case eg ln(-1) = πi)
 
Hi Tim!

Ah... clumsy me. I knew that. I really did. Well, used to... :wink:

So if you have ln(-∞), is that essentially ∞ or -∞?

I'm finding the solution for a calculus III improper integrals question. I guess it doesn't matter, though, because either way it diverges (a possible solution).

Thanks again

-Drc
 
DrCrowbar said:
Hi Tim!

Ah... clumsy me. I knew that. I really did. Well, used to... :wink:

So if you have ln(-∞), is that essentially ∞ or -∞?
No, it's simply not defined, as tiny-tim said. I'm assuming you're working with real numbers.
tiny-tim said:
Hi DrCrowbar! :smile:

There is no such thing as the ln of a negative number. :wink:

(unless we're allowed complex numbers, in which case eg ln(-1) = πi)
 
DrCrowbar said:
So if you have ln(-∞), is that essentially ∞ or -∞?

No, that's ∞i :wink:
 
Ah, ok.

So ln(-#) is the same as -ln(#)... That makes sense, actually. I had forgotten what the graph of the natural log function looks like.

Thanks guys. It's the first time I've asked a math question online and actually received a correct answer!

-Drc
 
DrCrowbar said:
So ln(-#) is the same as -ln(#)

no it isn't!

ln(-#) is an imaginary number (something times i)

if we're only allowed to use real numbers, then ln(-#) doesn't exist!
 
Oh, ok.

So you can have a negative natural log function (-ln|cscx+cotx| for instance) but you cannot have the natural log of a negative number unless you involve imaginary numbers.

I haven't really seen much of imaginary numbers, but I hear they're used a bit in D.E.
 
Last edited:
yes, they'll be useful later

for now, forget about them

the graph of ln(x) is like the graph of √x …

it simply has no value for x < 0
 
  • #10
tiny-tim said:
No, that's ∞i :wink:

What?? In what context is ln(-\infty)=\infty i? What does infty i even mean?
 
  • #11
oops! :redface:

i should have said ln(-∞) = πi + ∞ :rolleyes:
 

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