Contour Integral e(iqz)/z^4: Is it Infinite?

[pivoxa15]

Homework Statement

Find the contour integral of e(iqz)/z^4

The Attempt at a Solution

Is it infinite as the pole at z=0 is too high?

 

[Dick]

Around what contour? Don't you know some theorems that might apply here?

 

[HallsofIvy]

I second Dick. To have a "contour integral" you have to have a contour! For one thing, if the contour does not pass through or have 0 in its interior, the countour integral is 0 because the function is analytic inside the contour.

If the contour integral does include 0 in its interior, you probably want to look at the "residue" at z= 0. No, the integral is not (necessarily) 0 because "the pole at z=0 is too high". The pole is only of order 4! Write out the Taylor's series for e^iqz and divide by z⁴. That will give you the Laurent series for your function and that makes it easy to find the residue.

 

[pivoxa15]

I originally wanted to compute the integral of e(iqz)/z^4 from -infinity to infinity.

So I'll have a contour from that skips the origin but the small arc contribution near the origin could be too much because of the pole at z=0? So the integral of e(iqz)/z^4 from -infinity to infinity is infinite?

 

[Dick]

Actually, the contour around the origin is not a problem. It will pick up half the residue at the origin which is perfectly finite. The problem is the rest of the contour on the real axis. As the arc gets smaller and smaller it's that part that diverges. I don't think the integral can be defined.

 

[Avodyne]

If you define the integral by deforming the contour around the pole, then the answer is not infinite, but the value depends on how you choose to go around the pole (above or below). Once you've deformed the contour, you can add an arc at infinity (in either the upper or lower half plane, depending on the sign of q), and then you have a closed contour integral that can be evaluated by computing the residue, which is finite.

 

[Dick]

If you define the integral by deforming the contour around the pole, then the answer is not infinite, but the value depends on how you choose to go around the pole (above or below). Once you've deformed the contour, you can add an arc at infinity (in either the upper or lower half plane, depending on the sign of q), and then you have a closed contour integral that can be evaluated by computing the residue, which is finite.

What about the cos(qx)/x^4 divergence along the real axis?

 

[pivoxa15]

Actually, the contour around the origin is not a problem. It will pick up half the residue at the origin which is perfectly finite. The problem is the rest of the contour on the real axis. As the arc gets smaller and smaller it's that part that diverges. I don't think the integral can be defined.

Its the small or epsilon arc near the origin that creates the problem. So it is the contour near the origin that is the problem.

HOwever if it was 1/z not 1/z^4 then it would be okay.

 

[Avodyne]

There is no divergence if you go around the pole (in either direction). The apparent divergence along the real axis is canceled by the one from the little arc. It's easy to see this explicitly for q=0. We have

$$\int_{-\infty}^{-\varepsilon}{dz\over z^4} = {1\over 3\varepsilon^3}$$

$$\int^{\infty}_{\varepsilon}{dz\over z^4} = {1\over 3\varepsilon^3}$$

Now for the little arc, let $z = \varepsilon e^{i\theta}$, $dz = i\varepsilon e^{i\theta}d\theta$. To go above the contour, we integrate $\theta$ from $\pi$ to 0; to go below, we integrate from $-\pi$ to 0. We get

$$\int_{\rm arc}{dz\over z^4} = \int_{\pm\pi}^0{ i\varepsilon e^{i\theta}d\theta\over (\varepsilon e^{i\theta})^4} = {i\over\varepsilon^3}\int_{\pm\pi}^0{e^{-3i\theta}d\theta = -{1\over3\varepsilon^3}e^{-3 i\theta}\Big|^0_{\pm\pi} = -{1\over3\varepsilon^3}\bigl(1-(-1)\bigr) = -{2\over3\varepsilon^3}.$$

The sum of the three terms (left half of the real axis, right half, little arc) is zero.

 

[Dick]

There is no divergence if you go around the pole (in either direction). The apparent divergence along the real axis is canceled by the one from the little arc. It's easy to see this explicitly for q=0. We have

$$\int_{-\infty}^{-\varepsilon}{dz\over z^4} = {1\over 3\varepsilon^3}$$

$$\int^{\infty}_{\varepsilon}{dz\over z^4} = {1\over 3\varepsilon^3}$$

Now for the little arc, let $z = \varepsilon e^{i\theta}$, $dz = i\varepsilon e^{i\theta}d\theta$. To go above the contour, we integrate $\theta$ from $\pi$ to 0; to go below, we integrate from $-\pi$ to 0. We get

$$\int_{\rm arc}{dz\over z^4} = \int_{\pm\pi}^0{ i\varepsilon e^{i\theta}d\theta\over (\varepsilon e^{i\theta})^4} = {i\over\varepsilon^3}\int_{\pm\pi}^0{e^{-3i\theta}d\theta = -{1\over3\varepsilon^3}e^{-3 i\theta}\Big|^0_{\pm\pi} = -{1\over3\varepsilon^3}\bigl(1-(-1)\bigr) = -{2\over3\varepsilon^3}.$$

The sum of the three terms (left half of the real axis, right half, little arc) is zero.

Good explanation!

 

[pivoxa15]

There is no divergence if you go around the pole (in either direction). The apparent divergence along the real axis is canceled by the one from the little arc. It's easy to see this explicitly for q=0. We have

$$\int_{-\infty}^{-\varepsilon}{dz\over z^4} = {1\over 3\varepsilon^3}$$

$$\int^{\infty}_{\varepsilon}{dz\over z^4} = {1\over 3\varepsilon^3}$$

Now for the little arc, let $z = \varepsilon e^{i\theta}$, $dz = i\varepsilon e^{i\theta}d\theta$. To go above the contour, we integrate $\theta$ from $\pi$ to 0; to go below, we integrate from $-\pi$ to 0. We get

$$\int_{\rm arc}{dz\over z^4} = \int_{\pm\pi}^0{ i\varepsilon e^{i\theta}d\theta\over (\varepsilon e^{i\theta})^4} = {i\over\varepsilon^3}\int_{\pm\pi}^0{e^{-3i\theta}d\theta = -{1\over3\varepsilon^3}e^{-3 i\theta}\Big|^0_{\pm\pi} = -{1\over3\varepsilon^3}\bigl(1-(-1)\bigr) = -{2\over3\varepsilon^3}.$$

The sum of the three terms (left half of the real axis, right half, little arc) is zero.

What happens if q isn't 0?

You haven't achieved anything with this explanation. There are no poles inside so the whole integral must be 0 as you have shown. THe goal is to calculate the integral of e^iqx/x^4 from -infinity to infinity.

 

[Avodyne]

The Laurent expansion
$${e^{iqz}\over z^4} = {1\over z^4} + {iq\over z^3} - {q^2\over 2z^2} - {iq^3\over 6z} + \ldots$$
shows that the residue of the pole at $z=0$ is $-iq^3/6$.

If $q>0$, you can add an arc at infinity in the upper half plane. If the contour on the real axis is deformed to go above the pole, the integral is zero. If the contour is deformed to go below the pole, the integral is $(+2\pi i)(-iq^3/6)=\pi q^3/3$.

If $q<0$, you can add an arc at infinity in the lower half plane. If the contour is on the real axis is deformed to go below the pole, the integral is zero. If the contour is deformed to go above the pole, the integral is $(-2\pi i)(-iq^3/6)=-\pi q^3/3$.

 

[pivoxa15]

But I am worried that the deformed integral near the origin will blow to infinity due to 1/z^4.
So the residue value dosen't matter in any case.

 

[Avodyne]

But I am worried that the deformed integral near the origin will blow to infinity due to 1/z^4.

I already showed explicity that this doesn't happen. Nonzero q won't affect this explanation, because at small epsilon, e^(iqz) is always very close to one. And in any case, the contour evaluation that gives either 0 or $\pi |q|^3/3$ (depending on how the contour is deformed) is rigorously valid.

On the other hand, if you use a principal-value prescription at the pole (which is equivalent to dropping the contribution from the small arc), then the integral is indeed infinite.

 

[pivoxa15]

FOr the small arc we have z=Ee^(i(theta)) , E=epsilon, small

z^4 in the denominator means E^4 which as E->0 will go to 0 fast unless there is something but in the numerator which there isn't. So the small arc will diverge to infinity in the limit.

 

[Avodyne]

See post #9.