Integrals of Complex Functions

Click For Summary

Homework Help Overview

The discussion revolves around the integration of complex functions, specifically focusing on the function defined as ##f(t) = u_1(t) + i v(t)## over a real interval. The original poster attempts to prove that multiplying an integral by a complex constant can be interchanged with the integral of the function itself.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the validity of passing the imaginary unit ##i## through the integral sign, questioning whether this can be done similarly to real-valued scalars. There is also mention of the relationship between integrals and limits of summations, suggesting a deeper exploration of foundational concepts.

Discussion Status

Participants are actively engaging with the problem, raising questions about the assumptions involved in the proof and referencing the textbook for guidance. Some suggest that the proof may be straightforward, while others express uncertainty about the necessary theorems or definitions that may apply.

Contextual Notes

There is a noted lack of information regarding Riemann sums in the textbook being used, which may be relevant to the understanding of the integral's properties. Participants are encouraged to refer to specific sections of the textbook for potential insights.

Bashyboy
Messages
1,419
Reaction score
5

Homework Statement


Suppose we have the function ##f : I \rightarrow \mathbb{C}##, where ##I## is some interval of ##\mathbb{R}## the functions can be written as ##f(t) = u_1(t) + i v(t)##. Furthermore, suppose this function is integral over the interval ##a \le t \le b##, which can be found by computing

##\int_a^b f(t) = \int_a^b u(t) + i \int_a^b v(t)##

Let ##c## be some arbitrary complex constant. Proof that ## c \int_a^b f(t) = \int_a^b c f(t)##

2. Homework Equations

The Attempt at a Solution



## c \int_a^b f(t) = c \left[ \int_a^b u(t) + i \int_a^b v(t) \right] ##

## = c \int_a^b u(t) + ci \int_a^b v(t)##

Let ##c = a + bi##,

##c \int_a^b f(t) = (a+bi) \int_a^b u(t) + (a+bi) i \int_a^b v(t)##

##= a \int_a^b u(t) + bi \int_a^b u(t) + ai \int_a^b v(t) - b \int_a^b v(t)##

Here is where I had some trouble. The integrals of ##u(t)## and ##v(t)### are integrals of real-valued functions, and so I know that I can pass real-valued scalars through the integral sign; however, I do not know if I can pass ##i## through. I tried various manipulations, but all were positively unhelpful.
 
Physics news on Phys.org
Passing i through is what you are asked to prove (for the case c = i), so it becomes a bit circular then, doesn't it ?

On the other hand, would you have a problem "passing i through" for a sum ? And an integral is the limit of a summation.
 
BvU said:
Passing i through is what you are asked to prove (for the case c = i), so it becomes a bit circular then, doesn't it ?

Yes, that it what I figured the trouble to be.

BvU said:
On the other hand, would you have a problem "passing i through" for a sum ? And an integral is the limit of a summation.

Hmmm, I am not sure. We have not yet viewed the integral as the limit of a sum yet, nor has the textbook represented it as such. There is obviously something elementary that I am missing.

Here is the textbook that we are using: http://www.jiblm.org/downloads/jiblmjournal/V090515/V090515.pdf

We are presently on chapter four, which begins on page 30. Perhaps you might see some theorem I am suppose to use. I will continue to scour through the text myself.
 
Wow, there really isn't anything about riemann sums or anything...
Anyways, there's a couple useful things (ones a definition and the other's a theorem) on the next couple pages. Try using those.
 
Since, in the notes, it says "the proof of this theorem is very straightforward" I wouldn't worry about "passing through" the factor i.

You could split in real part and imaginary part to verify.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
21
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K