Do I have to use residue theory here?

[Nusc]

Homework Statement

$$\begin{subequations} \begin{eqnarray} s + \kappa + g^{2} \int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'} &=& s + \kappa + g^{2} \int_{-\infty}^{\infty} d \Delta'\; \frac{1}{(s^{2}+\omega^{2})}\, \frac{1}{(s+\gamma+i\Delta')} \\ \nonumber &=& s + \kappa + g^{2} \int_{-\infty}^{\infty} d \Delta'\; \frac{1}{(s+i\omega)(s-i\omega)}\, \frac{1}{(s+\gamma+i\Delta')} \\ \nonumber \end{eqnarray} \end{subequations}$$

Homework Equations

The Attempt at a Solution

Do I have to use residue theory here?

 

[CompuChip]

If there are no hidden dependences, try calculating
$$\frac{d}{d\Delta'} \left( \frac{\log(s + \gamma + i \Delta')}{(s + i\omega)(s - i\omega)} \right).$$

If there are (like $\omega$ depends on $\Delta'$) probably calculus of residues may be the way to go (if you can safely turn it into a contour integral, that is).

 

[Nusc]

Here is the correct integral then:

$$\begin{subequations} \begin{eqnarray} s + \kappa + g^{2} \int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'} &=& s + \kappa + g^{2} \int_{-\infty}^{\infty} d \Delta'\; \frac{1}{\pi}\frac{\omega}{(\Delta'^{2}+\omega^{2})}\, \frac{1}{(s+\gamma+i\Delta')} \\ \nonumber &=& s + \kappa + g^{2} \int_{-\infty}^{\infty} d \Delta'\; \frac{\omega}{(\Delta'+i\omega)(\Delta'-i\omega)}\, \frac{1}{(s+\gamma+i\Delta')} \end{eqnarray} \end{subequations}$$

Which method of contour integration do I use again? it's been a while since I've done this.

the poles are at $$\pm i\omega, -i(s+\gamma)$$,now what?

 

[CompuChip]

So basically, you close the contour by a semi-circle of infinite radius, either in the upper or in the lower half plane (which of the two it is, should be dictated by the requirement that the extra arc you add does not contribute).

Then the result will be $2\pi i$ (or minus that, for clockwise integration) multiplied by the residue of the function, which is basically
$$\sum_{p} \left( f(z) (z - p) \right)_{z = p}$$
where the sum runs over all the (single) poles p and f(z) is the integrand.

However, this one looks tricky, so I suggest you check that it is allowed (or maybe you need a different contour).

 

[Nusc]

What are the conditions for which this is allowed? What other contour could I use?

 

[CompuChip]

I'm actually a physicist so usually I am a bit sloppy with these things :)
But IIRC you can parametrize the arc like
$$\gamma_R: t \mapsto R e^{i t}$$
for t between 0 and pi, for example. Then you can (in principle) do the integration:
$$\oint f(z) \, dz = \int_{-R}^{R} f(x) \, dx + \int_{\gamma_R} f(z) = \int_{-R}^{R} f(x) \, dx + \int_0^\pi f(\gamma(t)) \gamma'(t) \, dt$$
If you send R to infinity (make the semi-arc larger) then the first term approaches what you want to calculate, namely
$$\int_{-\infty}^\infty f(x) \, dx,$$
therefore the second part should vanish in that limit.

Anyway, that's the formal part. In this case, your problem seems to be a bit more tricky than one would think at first sight, because such a semi-arc does not vanish. One can still use contour integration then, but one needs to evaluate
$$\int_0^\pi dt \, \frac{i R e^{it}}{(R e^{it} + i \omega)(R e^{it} - i \omega)(i R e^{it} + s + \gamma}$$
which is probably more complicated than your original integral. I checked it with Mathematica and it does work out, but you really need to add the contributions from the closed contour as well as the additional semi-arc.

I don't really see what the best way to solve the integral would be at the moment, I will think about it for a bit.

 

[Nusc]

Iirc?

 

[LowlyPion]

If I recall correctly

 

[Dick]

Let me write the integration variable as z, ok? (Your Delta'). The integrand is ~ 1/z^3. The contribution for the semiarc at radius R goes to zero as R->infinity. So use the usual contour, along the x-axis for -R to R and close with a semiarc in either half plane. As long as you keep track of the signs, you'll get the same thing. You'll want to do it once assuming s+gamma is positive and again with it negative. They are two different cases.

 

[Nusc]

I'm actually a physicist so usually I am a bit sloppy with these things :)
But IIRC you can parametrize the arc like
$$\gamma_R: t \mapsto R e^{i t}$$
for t between 0 and pi, for example. Then you can (in principle) do the integration:
$$\oint f(z) \, dz = \int_{-R}^{R} f(x) \, dx + \int_{\gamma_R} f(z) = \int_{-R}^{R} f(x) \, dx + \int_0^\pi f(\gamma(t)) \gamma'(t) \, dt$$
If you send R to infinity (make the semi-arc larger) then the first term approaches what you want to calculate, namely
$$\int_{-\infty}^\infty f(x) \, dx,$$
therefore the second part should vanish in that limit.

Anyway, that's the formal part. In this case, your problem seems to be a bit more tricky than one would think at first sight, because such a semi-arc does not vanish. One can still use contour integration then, but one needs to evaluate
$$\int_0^\pi dt \, \frac{i R e^{it}}{(R e^{it} + i \omega)(R e^{it} - i \omega)(i R e^{it} + s + \gamma}$$
which is probably more complicated than your original integral. I checked it with Mathematica and it does work out, but you really need to add the contributions from the closed contour as well as the additional semi-arc.

I don't really see what the best way to solve the integral would be at the moment, I will think about it for a bit.

Why doesn't the semi-arc vanish? Dick said it does.

 

[Nusc]

It should be valid by the estimation lemma right?

 

[Dick]

Why doesn't the semi-arc vanish? Dick said it does.

It does vanish. The length of the arc is proportional to R and the function goes like 1/R^3. This is a perfectly simple exercise in adding up the pole contributions.

 

[Nusc]

Do I still have to parametrize the function?

 

[CompuChip]

I checked it with Mathematica and it does work out, but you really need to add the contributions from the closed contour as well as the additional semi-arc.

So it turns out that I made a very stupid calculation error (even the best make a little mistake sometimes ) for which I deeply apologize.
The intuitive argument that if you'd parametrize the arc by $$\Delta = R e^{it}$$ then you would get an integrand which goes like $$R / R^3$$ (the R on top comes from the derivative you have to add in $\int_\gamma f = \int f(\gamma(t)) \gamma'(t) dt$) and as R goes to infinity this vanishes (and quite rapidly too). So the problem in all its simplicity was just confusing to me, and I probably confused you a bit too.

Thanks to Dick for waking me

 

[Dick]

It should be valid by the estimation lemma right?

No, just draw your contour, figure out what poles are inside the contour, sum the residues and be comfortable with the notion that in sending the radius of the arc to infinity, it's contribution will vanish.

 

[Nusc]

Only $$+i\omega$$ should be in the contour provided that you have $$s+\gamma <0$$

The residues would then be $$2\pi i f(i\omega)$$

How do I find $$f(i\omega)$$
?

 

[Dick]

How do you usually find a residue of a function like 1/((z-a)*(z-b)*(z-c)) at say z=a? This is one of the easiest residue theorem cases of all, they are all simple poles.

 

[Nusc]

$$\begin{subequations} \begin{eqnarray} \lim \limits_{\Delta' \to i\gamma_{n}} \frac{\gamma_{n}}{(\Delta'+i\gamma_{n})}\, \frac{1}{(s+\gamma+i\Delta')} &=& \frac{1}{2i}\frac{1}{s+\gamma-\gamma_{n}} \\ \lim \limits_{\Delta' \to -i\gamma_{n}} \frac{\gamma_{n}}{(\Delta'-i\gamma_{n})}\, \frac{1}{(s+\gamma+i\Delta')} &=& \frac{-1}{2i}\frac{1}{s+\gamma+\gamma_{n}} \\ \lim \limits_{\Delta' \to -i(s+\gamma)} \frac{\gamma_{n}}{(\Delta'+i\gamma_{n})(\Delta'-i\gamma_{n})} &=& \frac{\gamma_{n}}{(-s-\gamma+\gamma_{n})(s+\gamma+\gamma_{n})} \end{eqnarray} \end{subequations} \begin{subequations} \begin{eqnarray} \oint_{\gamma_{R}(t)} f(z) \, dz &=& 2\pi i[\frac{1}{2i}\frac{1}{s+\gamma-\gamma_{n}} + \frac{-1}{2i}\frac{1}{s+\gamma+\gamma_{n}} + \frac{\gamma_{n}}{(-s-\gamma+\gamma_{n})(s+\gamma+\gamma_{n})}] \\ &=&2\pi i[-\frac{(1+i)\gamma_{n}}{(s+\gamma-\gamma_{n})(s+\gamma+\gamma_{n})}] \end{eqnarray} \end{subequations}$$

Is that right way?

$$\begin{subequations} \begin{eqnarray} \int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'} &=& 2i[-\frac{(1+i)\gamma_{n}}{(s+\gamma-\gamma_{n})(s+\gamma+\gamma_{n})}] \end{eqnarray} \end{subequations}$$

I guess that's assuming $$s+\gamma >0$$

 

[Dick]

Not sure. I can't really tell what you are doing. Going back to the expression in post 3, the interesting part of your integral is 1/((z+iw)*(z-iw)*(s+gamma+iz)). To get the residue at z=iw you just multiply that by (z-iw) and take the limit as z->iw. It shouldn't be that complicated.

 

[Nusc]

That's what I did for all three poles or do I do it for the one above the x-axis? Then I just sum them all multiplied by 2pi i. Right?

 

[Dick]

That's what I did for all three poles or do I do it for the one above the x-axis? Then I just sum them all multiplied by 2pi i. Right?

You only sum the residues over the poles that are enclosed by your contour.

 

[Nusc]

So all the cases are as follows:

The top region with

s+gamma<0

then again with s+gamma > 0 both with clockwise orientation.

Do I need to do below the x-axis as well?

 

[Dick]

You don't NEED to do the integral closing below the x-axis as well. You should get the same answer as closing above once you allow for the reversed sign coming from the contour being in the opposite direction. Doing it closing in the lower half-plane and checking you get the same thing as closing in the upper would be good practice though. And not very hard.