When complex integration depends on the method of evaluation

[karlzr]

Homework Statement

It is about an example from Essential Mathematical methods for physicists by Weber & Arfken, which describes a scattering process:
$$I(\sigma)=\int^{+\infty}_{-\infty}\frac{x \sin x dx}{x^2-\sigma^2}$$2. The attempt at a solution
The straightforward way is to contruct a contour and one can find the result is $\pi \cos\sigma$. Then the author said that since this is not an outgoing scattering wave, we should try a different technique. He moved both singular points off the real axis by letting $\sigma→\sigma+i\gamma$, and obtained $\pi e^{-i\sigma}$ or $\pi e^{i\sigma}$.
My question is, how can one integration has two results? Both methods make sense to me, but lead to totally different values. This problem has bothered me a lot when discussing Green function in quantum mechanics.

 

[mfb]

I think the integral is not well-defined in mathematics (unless σ=0 or with an imaginary part). You have a pole, and if you try to split up your integral, those parts will diverge. In physics, it can give meaningful results if you integrate in some specific way to avoid those issues, but then the result can depend on the integration method.

 

[Ray Vickson]

Homework Statement

It is about an example from Essential Mathematical methods for physicists by Weber & Arfken, which describes a scattering process:
$$I(\sigma)=\int^{+\infty}_{-\infty}\frac{x \sin x dx}{x^2-\sigma^2}$$2. The attempt at a solution
The straightforward way is to contruct a contour and one can find the result is $\pi \cos\sigma$. Then the author said that since this is not an outgoing scattering wave, we should try a different technique. He moved both singular points off the real axis by letting $\sigma→\sigma+i\gamma$, and obtained $\pi e^{-i\sigma}$ or $\pi e^{i\sigma}$.
My question is, how can one integration has two results? Both methods make sense to me, but lead to totally different values. This problem has bothered me a lot when discussing Green function in quantum mechanics.

You need to worry about how you define such an integral. Let's look at the integral over [0,∞), which we can regard as having f(x) in the numerator and (x^2-s^2) in the denominator; here, I write s instead of σ because it is easier. Assume s>0. The integral
$$J = \int_0^{\infty} \frac{f(x)}{x^2-s^2} \, dx$$ is improper, so let us try to define it through a limiting procedure. First, though, note that
$$J = \int_0^{\infty} \frac{f(x)-f(s)}{x^2-s^2} \, dx +<br /> \int_0^{\infty} \frac{f(s)}{x^2-s^2} \, dx.$$
The first integral is OK, so we need worry only about the second one; that is, we need to worry about the value of
$$K = \int_0^{\infty} \frac{1}{x^2-s^2} \, dx.$$
This integral is improper, so must be defined in terms of some type of limiting procedure. Let us try the following:
$$K = \lim_{a,b \rightarrow 0+} J_{ab}, \text{ where } J_{ab} = <br /> \int_0^{s-a} \frac{1}{x^2-s^2}\, dx <br /> + \int_{s+b}^{\infty} \frac{1}{x^2-s^2}\, dx.$$
We have
$$J_{ab} = \frac{1}{2s} \ln \left(\frac{2s+b}{2s-a}\right) + \frac{1}{2s} \ln (a/b).$$
If we let a and b go to zero independently, J_ab need not have a limit, but if we let a and b go to zero in a fixed ratio k = a/b, the limit is
$$\frac{1}{2s} \ln (k).$$ You can see from this that the value of k matters a lot.

The most common form of such improper integrals is the "symmetric" version (called the principle value), which takes a = b and gives
$$J = \lim_{a \rightarrow 0+} \int_0^{s-a} \frac{1}{x^2-s^2} \, dx <br /> + \int_{s+a}^{\infty} \frac{1}{x^2-s^2} \, dx = 0.$$
However, the undeniable fact is that how you define it can affect its value.

As to your question about getting the two possible values π*exp(± i σ), well I would worry even more about getting a complex value for a real integral; the integrand becomes real in the limit as γ → 0, but the integral remains stubbornly complex, with imaginary part equal to ± π sin(σ).

RGV

 

[karlzr]

I think the integral is not well-defined in mathematics (unless σ=0 or with an imaginary part). You have a pole, and if you try to split up your integral, those parts will diverge. In physics, it can give meaningful results if you integrate in some specific way to avoid those issues, but then the result can depend on the integration method.

This kind of integration is quite common in physics and is an important part in every math physics textbook. In general, we need to find out the principal value of this integration. That is where ambiguity originates from, at least I think it is. Then how to deal with such integrations? I mean, it is possible that sometimes it is not so clear as to which method we should take to do the integration.

 

[karlzr]

You need to worry about how you define such an integral. Let's look at the integral over [0,∞), which we can regard as having f(x) in the numerator and (x^2-s^2) in the denominator; here, I write s instead of σ because it is easier. Assume s>0. The integral
$$J = \int_0^{\infty} \frac{f(x)}{x^2-s^2} \, dx$$ is improper, so let us try to define it through a limiting procedure. First, though, note that
$$J = \int_0^{\infty} \frac{f(x)-f(s)}{x^2-s^2} \, dx +<br /> \int_0^{\infty} \frac{f(s)}{x^2-s^2} \, dx.$$
The first integral is OK, so we need worry only about the second one; that is, we need to worry about the value of
$$K = \int_0^{\infty} \frac{1}{x^2-s^2} \, dx.$$
This integral is improper, so must be defined in terms of some type of limiting procedure. Let us try the following:
$$K = \lim_{a,b \rightarrow 0+} J_{ab}, \text{ where } J_{ab} = <br /> \int_0^{s-a} \frac{1}{x^2-s^2}\, dx <br /> + \int_{s+b}^{\infty} \frac{1}{x^2-s^2}\, dx.$$
We have
$$J_{ab} = \frac{1}{2s} \ln \left(\frac{2s+b}{2s-a}\right) + \frac{1}{2s} \ln (a/b).$$
If we let a and b go to zero independently, J_ab need not have a limit, but if we let a and b go to zero in a fixed ration k = a/b, the limit is
$$\frac{1}{2s} \ln (k).$$ You can see from this that the value of k matters a lot.

The most common form of such improper integrals is the "symmetric" version (called the principle value), which takes a = b and gives
$$J = \lim_{a \rightarrow 0+} \int_0^{s-a} \frac{1}{x^2-s^2} \, dx <br /> + \int_{s+a}^{\infty} \frac{1}{x^2-s^2} \, dx = 0.$$
However, the undeniable fact is that how you define it can affect its value.

As to your question about getting the two possible values π*exp(± i σ), well I would worry even more about getting a complex value for a real integral; the integrand becomes real in the limit as γ → 0, but the integral remains stubbornly complex, with imaginary part equal to ± π sin(σ).

RGV

Yeah, it is weird that the second method which let's $\sigma→\sigma+i\gamma$ seems reasonable but leads to a complex result. So is this ambiguity common in complex integration or it just happens under some condition?

 

[mfb]

This kind of integration is quite common in physics

I know, and quantum field theory lectures have them everywhere. If it works, it can be fine. But you can get that ambiguity or completely meaningless results, if you do it "wrong" (= not in the precise way the specific calculation requires).