Integrating through singularities

In summary, the conversation discusses the concept of dealing with integrals around singularities, specifically the difference in treatment between integrals with removable and non-removable singularities. The example of \int_{-\infty}^{\infty}\frac{sin(x)}{x} dx is used to illustrate the idea of integrating through a removable singularity, while \int_{-\infty}^{\infty}\frac{1}{x^3} dx is left undefined due to its non-removable singularity. The concept of principle value integrals is also mentioned, where the integral is taken by leaving out a small neighborhood of the singularity and then shrinking it to zero. The conversation ends with a discussion about the
  • #1
dsr39
14
0
I am a little bit confused about dealing with integrals around singularities because my professor seems to treat some situations more rigorously than others.

We talked this integral and said

[tex] \int_{-\infty}^{\infty}\frac{1}{x^3} dx = undefined [/tex]

This seems a little bit unintuitive to me because clearly we have an odd function, so no matter what happens as you approach 0, you should get equal cancellations from the left and right sides... but since you cannot say EXACTLY what happens at x = 0, I can see calling this indeterminate.


But then when we came across this integral

[tex] \int_{-\infty}^{\infty}\frac{sin(x)}{x} dx [/tex]

and integrated right through the removable singularity without any problems. It seems hypocritical to deal with this one by just appealing to intuition and saying that the part near x=0 doesn't "blow up" so it has a negligible contribution when we don't in fact know what it does AT x=0.

Is there a good reason why in this case we can just integrate straight through the singularity even though we have an undefined value at x = 0?
 
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  • #2
I don't know if this answer is satisfactory to you, but

[tex]\lim_{x \to 0} \frac{\sin x}{x} = 1[/tex]
while
[tex]\lim_{x \to 0} \frac{1}{x^3} = \infty[/tex]

So I don't agree with your argument that
we don't in fact know what it does AT x=0.
You even said it yourself: it is a removable singularity, while in the first integral it is not. The word removable already indicates that it's not a real "hard" singularity.

In fact, your idea of integrating 1/x^3 while leaving out some small neighbourhood of the origin and then shrinking that to zero is very much possible, it is called principle value. In this case, you would find
[tex]\mathcal P \int_{-\infty}^{\infty}\frac{1}{x^3} dx = \lim_{\epsilon \downarrow 0} \left[
\int_{-\infty}^{-\epsilon}\frac{1}{x^3} dx + \int_{\epsilon}^{\infty}\frac{1}{x^3} dx
\right] = \lim_{\epsilon \downarrow 0} 0 = 0.[/tex]
 
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  • #3
To add to what CompuChip said about the principal value integral: note that the resulting value you get depends on how you take the limit of the integral bounds. Because the limit was taken symmetrically in the example he gave the result was zero. However, I could just as well do

[tex]\lim_{\epsilon \rightarrow 0} \left[ \int_{-\infty}^{-2\epsilon}\frac{1}{x^3} dx + \int_{\epsilon}^{\infty}\frac{1}{x^3} dx\right] = \lim_{\epsilon \rightarrow 0} -\frac{1}{2}\left[\frac{1}{4\epsilon^2} - \frac{1}{\epsilon^2}\right] \neq 0[/tex].

This is why the integral of 1/x^3 is typically left as undefined: in general you DON'T get equal contributions to the integral from both sides! Taking the limit in different ways can give different results! The principal value integral does turn out to be useful in physics, but as it doesn't give the only possible result of an integral with a singularity, such integrals are typically regarded as undefined.
 
  • #4
I have done principle parts, but even though sin(x)/x converges to 1 in the limit that x approaches 0, if we wish to integrate directly through the point x=0 how can we ignore the fact that it is undefined at 0. It is a removable discontinuity, but it is still a discontinuity.

Isn't integrating through the removable discontinuity as if it weren't there, essentially taking a principle part?
 
  • #5
dsr39 said:
Isn't integrating through the removable discontinuity as if it weren't there, essentially taking a principle part?

No, because the two limits

[tex]\lim_{a \rightarrow 0^+} \int_{a}^{\infty} \frac{\sin x}{x} \; dx[/tex]

and

[tex]\lim_{b \rightarrow 0^-} \int_{-\infty}^{b} \frac{\sin x}{x} \; dx[/tex]

converge independently.
 
  • #6
Instead of thinking of the integral as a procedure (going along x in a certain direction and adding up rectangles), it may be useful to think of the integral as the (signed) measure of the 2D region between 0 and the curve.
 
  • #7
These replies are all helpful. Thanks for the feedback.
 
  • #8
Log(|x|) also "blows up" at x=0, but the integral [tex]\int_{-1}^1 Log{|x|}dx[/tex] is well defined. Anything less singular than 1/x is integrable.
 

What is the concept of integrating through singularities?

Integrating through singularities is a mathematical process used to evaluate the integral of a function that contains singularities, or points where the function is undefined.

Why is it important to integrate through singularities?

Integrating through singularities is important because it allows us to accurately evaluate the integral of a function, even if the function has points where it is undefined. Without integrating through singularities, we would not be able to obtain a complete understanding of the function's behavior.

What types of functions typically have singularities?

Singularities can occur in a variety of functions, but they are most commonly found in rational functions, trigonometric functions, and logarithmic functions.

How do you integrate through a singularity?

There are several techniques that can be used to integrate through a singularity, such as using a change of variables, using Cauchy's residue theorem, or using a Taylor series expansion. The specific technique used will depend on the function and the type of singularity present.

Are there any limitations to integrating through singularities?

Yes, there are some limitations to integrating through singularities, such as when the singularity is an essential singularity or when the function has an infinite discontinuity at the singularity. In these cases, the integral may not be well-defined or may require more advanced techniques to evaluate.

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