What does the integral along a path of x+i diverge to?

In summary: There are other improper integrals that do have the same result, depending on the domain you're working in.2: Basically, what you're doing is trying to integrate over a path that goes from -infty to +infty, but the limits don't change after you substitute y=x-\frac{i}{2\alpha}(k_0-k). This is an improper integral because you're trying to integrate something that has a limit that is not itself a limit.In summary, the integral limits diverge to infinity and -infinity after the substitution of y=x-\frac{i}{2\alpha}(k_0-k).
  • #1
Whazupp
8
0
[tex]\lim_{x\to\infty}(x+i)=?[/tex]

What does this diverge to? Intuitively i would have thought (realinfinity + i).

EDIT:

Which brings the other question:

[tex]\int^\infty_0}(x+i)dx=?[/tex]

How should i interpret this integral considering x is a real number? And what does it diverge to?
 
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  • #2
It depends on what points you have placed at infinity.

For example, if you were working in [itex]\bar{\mathbb{R}}^2[/itex], where [itex]\bar{\mathbb{R}}[/itex] is the extended real numbers, then every number is of the form a + bi, where a and b are any extended real number. (even an infinite one!) And then, you could show

[tex]
\lim_{x \rightarrow +\infty} x + i = +\infty + i
[/tex]

(even if x is not restricted to the reals!)


Actually, if you are using any extension of the complexes such that + is continuous at [itex](\infty, i)[/itex] (whatever [itex]\infty[/itex] happens to mean in this context), then it would be true that your limit converges to [itex]\infty + i[/itex]. But it might happen (as it does on the complex projective line and the real projective plane) that [itex]\infty + i = \infty[/itex]. (for the appropriate meaning of [itex]\infty[/itex])
 
  • #3
Great answer :). That did actually clarify some things.

Now here's the real context of the problem. I'm trying to simplify the following integral. I'm still confused as to which of those systems i should use for this one:

[tex]\int^\infty_{-\infty}\left(e^{-\alpha(x-\frac{i}{2\alpha}(k_0-k))^2\right)dx}[/tex]

I can simplify it by substituting [tex]y=x-\frac{i}{2\alpha}(k_0-k)[/tex]. However in the model solutions for this problem, the integral limits do not change after the substitution, even though I'm subtracting a complex number from the real number x. Any reason why the limits should still be INFINITY and -INFINITY after the substititution? (and not [tex]y=\infty-\frac{i}{2\alpha}(k_0-k)[/tex] and [tex]y=-\infty-\frac{i}{2\alpha}(k_0-k)[/tex]
 
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  • #4
Integrals1 don't happen between limits: they happen along paths.

In the simple world of the reals, any path is (essentially) determined by its endpoints. So, we can get away with defining a definite integral in such a fashion.

But for what you're doing, you really ought to be specifying the path. Your original integral really ought to be specified as being integrated over the path

x = t ([itex]t \in \mathbb{R}[/itex])​

and your second integral as being over the path

y = t - (i / 2a)(k0 - k) ([itex]t \in \mathbb{R}[/itex]).​

However, I have seen people use notation similar to what you want to use as shorthand for denoting such vertical or horizontal lines.


The other thing to note is that (I think), this is an improper integral, which is defined in terms of a limit of things with bounded domain. On each of those bounded things, you don't have any of the problems you've been having!



1: Well, this type of integral, anyways.
 

FAQ: What does the integral along a path of x+i diverge to?

What is a complex limit?

A complex limit is a mathematical concept that describes the behavior of a function as its input values approach a complex number. It is similar to a regular limit, but instead of dealing with real numbers, it deals with complex numbers, which have a real and imaginary component.

How do I determine the limit of a complex function?

To determine the limit of a complex function, you need to evaluate the function as the input values approach the complex number in question. This can be done by plugging in values that get closer and closer to the complex number, and observing the output values. If the function approaches a single value, that is the limit. If the function approaches different values depending on the direction of approach, the limit does not exist.

What is the difference between a complex limit and a regular limit?

The main difference between a complex limit and a regular limit is the type of numbers involved. A regular limit deals with real numbers, while a complex limit deals with complex numbers. This means that the behavior of a function near a complex number can be more complex and varied compared to a function near a real number.

Can a complex limit be infinite?

Yes, a complex limit can be infinite. Just like with regular limits, a complex limit can approach positive or negative infinity. This can happen when the function approaches a complex number in such a way that the output values become larger and larger, without approaching a specific value.

Why are complex limits important in mathematics?

Complex limits are important because they help us understand the behavior of functions in more complicated scenarios. They allow us to analyze functions that involve complex numbers, which are often used in various branches of mathematics, such as calculus and physics. Complex limits also have practical applications in fields like engineering and economics.

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