# Do I have to use residue theory here?

1. Jan 24, 2009

### Nusc

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

Do I have to use residue theory here?

2. Jan 24, 2009

### CompuChip

Re: Integral

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).

3. Jan 24, 2009

### Nusc

Re: Integral

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?

Last edited: Jan 24, 2009
4. Jan 24, 2009

### CompuChip

Re: Integral

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).

5. Jan 24, 2009

### Nusc

Re: Integral

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

6. Jan 25, 2009

### CompuChip

Re: Integral

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.

7. Jan 25, 2009

### Nusc

Re: Integral

Iirc?

8. Jan 25, 2009

### LowlyPion

Re: Integral

If I recall correctly

9. Jan 25, 2009

### Dick

Re: Integral

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.

10. Jan 25, 2009

### Nusc

Re: Integral

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

11. Jan 25, 2009

### Nusc

Re: Integral

It should be valid by the estimation lemma right?

12. Jan 25, 2009

### Dick

Re: Integral

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.

13. Jan 26, 2009

### Nusc

Re: Integral

Do I still have to parametrize the function?

14. Jan 26, 2009

### CompuChip

Re: Integral

So it turns out that I made a very stupid calculation error (even the best make a little mistake sometimes :tongue:) 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

15. Jan 26, 2009

### Dick

Re: Integral

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.

16. Jan 26, 2009

### Nusc

Re: Integral

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 $$f(i\omega)$$
?

17. Jan 26, 2009

### Dick

Re: Integral

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.

18. Jan 26, 2009

### Nusc

Re: Integral

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

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 $$s+\gamma >0$$

Last edited: Jan 26, 2009
19. Jan 26, 2009

### Dick

Re: Integral

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.

20. Jan 26, 2009

### Nusc

Re: Integral

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?