# How Can You Solve an Integral with a 2nd Order Pole on the Real Axis?

• gysush
In summary, the conversation involves a complex analysis problem from Arfken, specifically solving the integral of (cos(b*x)-cos(a*x))/(x^2) over the entire real line and showing it is equal to Pi*(a-b). A related problem is discussed and a method involving complex contour integration is proposed, with the final answer being derived through integration by parts.
gysush
I'm working on a complex analysis problem from Arfken.

Integrate[ (cos(b*x)-cos(a*x))/(x^2), {x, -Infinity, Infinity}]

and show that it is equal to Pi*(a-b)

Attempt:

I first look at a related problem (or one that I think is related).

=> f(x) = sin(x)/x
=> f(z) = sin(z)/z

For this problem I can look at P[Integral(f(z)dz)]

then...sin(z)/z = [exp(i*z) - exp(-i*z)]/2*i*z = Integral 1 - Integral 2

where Integral 1 is the +i and Integral 2 is -i => this determines if we should close our contour in the UHP or LHP

Then Integral 1 = P[exp(i*z)/2*i*z] = Pi*i*Res(f(0))

Res(f(0)) is simple to calculate => 1/2*i

Likewise, Integral 2 = P[exp(-i*z)/2*i*z] = -Pi*i*Res(f(0))

Then Integral 1 - 2 = Pi*i/2*i - (-Pi*i/2*i) = pi

Upon inspection of my original integral I realize I have a 2nd order pole on the axis instead of a simple pole. Prof said the above method is only valid for simple poles. I then examine my notes/boas/arfken/leia for an example of 2nd order pole on real axis as in this problem...but have not been successful.

One example that I see is of form f(cos,sin)/(x^2 - a^2) which has two simple poles at a and -a

Another is of the form f(cos,sin)/(x^2 + a^2) which has two simple poles at +/- ia
For this integral, do not need principal value. We can consider, i.e. sin/(x^2 + a^2)...and from there calculate Integral[exp(iz)/z)...and set it equal to sum of residues etc...then separate out im/real parts and math them on LHS and RHS.

I'm kinda lost on the problem that I originally posted...

Try integrating by parts once to get an integrand with only a simple pole.

Thank you!

I'll give you a quick sketch. Perhaps you can go through it and fill in the details. Also, that's a removable singularity in the original problem. But how about considering the integral:

$$\int_C \frac{e^{iaz}-e^{ibz}}{z^2}dz$$

where C is the half-disc in the upper half-plane with an indentation around the now simple pole. We would then have:

$$\int_{-\infty}^0+\int_{\Gamma}+\int_{0}^{\infty}+\int_U=0$$ so that (after some justification):

$$\int_{-\infty}^{\infty}+\int_{\Gamma}=0$$

where the integral over U is the half-circle arc and gamma is the indentation at the origin and that is just $-i\pi r$

where:

$$r=\mathop\text{Res}_{z=0}\left\{\frac{e^{iaz}-e^{ibz}}{z^2}\right\}=i(a-b)$$

I'm assuming the sine terms drop out because they are odd but I've not gone through this carefully and the integral over the large semi-circle tends to zero but I've not checked that.

So assuming all that is ok, then we have:

$$\int_{-\infty}^{\infty}=-\pi(a-b)=\pi(b-a)$$

Note I'm getting a minus sign different from yours. You can look into that. I left out details that you can go through.

Yes, thank you as well. But the answer fell out almost trivially (since I already know Integral[sinx/x] once I did the integration by parts. Just had to make sure I took out the b or the a once I diff cos(ax)

## 1. What is a 2nd order pole on the Real axis?

A 2nd order pole on the Real axis is a type of singularity in a complex function that occurs when the denominator of the function equals zero twice. This results in a vertical asymptote on the Real axis, which can affect the behavior of the function.

## 2. How can a 2nd order pole on the Real axis be identified?

A 2nd order pole on the Real axis can be identified by setting the denominator of the complex function equal to zero and solving for the variable. If the solution yields two identical values, then the function has a 2nd order pole on the Real axis.

## 3. What is the significance of a 2nd order pole on the Real axis?

A 2nd order pole on the Real axis can affect the behavior of a complex function, especially in the vicinity of the pole. It can cause the function to have a vertical asymptote, making it undefined at certain points. This can also impact the convergence of series expansions of the function.

## 4. How does a 2nd order pole on the Real axis differ from a 1st order pole?

A 2nd order pole on the Real axis differs from a 1st order pole in that it has a higher degree of singularity. This means that the function will have a steeper slope at the pole, and its behavior will be more affected in the vicinity of the pole.

## 5. Can a 2nd order pole on the Real axis be avoided or eliminated?

In general, a 2nd order pole on the Real axis cannot be avoided or eliminated, as it is a natural singularity that arises from the function itself. However, in some cases, it may be possible to use techniques such as partial fraction decomposition to simplify the function and potentially eliminate the pole.

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