Complex inversion formula (branch cut issue)

[jrp131191]

Hi, first time trying to to use the complex inversion formula and I'm confusing myself.

I'm trying to find L^-1{ ln(1 + (1/s))}

I can see that there is an essential pole at s=0 (is this even right? The power series has an infinite amount of z^n's in the denominator..) and branch points at s=0 and s=-1.

So.. I've created a contour which excludes the branch cut from x=0 to x=-1

But then if I integrated around this contour, there are no singularities in my contour so by cauchys theorem won't my integral be equal to 0?...

 

[jackmell]

Are you looking to show this:

\mathcal{L}^{-1}\left\{\log(1+1/s)\right\}=\frac{1}{2\pi i}\mathop\int\limits_{\sigma-i\infty}^{\sigma+i\infty}e^{st}\log(1+1/s)ds=\frac{1-e^{-t}}{t}

If so, then I think the points 0 and -1 are branch-points for log.

And after looking at it for a little while, seems you need two contours: The outer one which includes the inversion contour, and an inner "dumb-bell" contour. Equate the two, bingo-bango, done. Can you see how to do this?

 

[orangesun]

Hi Jack Mell,

I was just wondering if you could lead on with maybe a few more steps in evaluating this?

Thankyou :)

 

[jackmell]

Hi Jack Mell,

I was just wondering if you could lead on with maybe a few more steps in evaluating this?

Thankyou :)

Alright, you got the two contour thing down right? First need to realize that the function \log(u) is completely analytic around a small circle going around the point u=1 and you have \log(1+1/s) and with |s|>>1, the expression 1+1/s is a small circle going around the point s=1 right? That means for a large contour going around your function, it's analytic (no branch cuts) and in fact for any contour that's larger than a cicrcle going around the two branch-points. So if we have two such contours, since they are in an analytic region of the function, integrals over them are equal. Now make a dog-bone contour (dumb-bell contour) around the two branch-points. That's one contour. Now make that larger contour which has as one leg, the inversion contour, and close it with a large circular arc. Now, regardless of what \sigma is, say even 1000, as that contour grows to infinity, the arc angle extended by the circular arc of that contour tends to go from \pi/2 around to 3\pi/2.

So now you got everything:

\mathop\oint\limits_{\text{big contour}} f(s)ds=\mathop\oint\limits_{\text{dog-bone}} f(s)ds

I'll do the big one:

\mathop\oint\limits_{\text{big contour}}f(s)ds=\mathop\int_{\sigma-i\infty}^{\sigma+i\infty} f(s)ds+\int_{\pi/2}^{3\pi/2} f(s)ds

The first integral on the right is the inversion integral. Can you make the substitution s=Re^{it} in the second one and show that as R\to\infty, the integral is zero?

 

[jrp131191]

Alright, you got the two contour thing down right? First need to realize that the function \log(u) is completely analytic around a small circle going around the point u=1 and you have \log(1+1/s) and with |s|>>1, the expression 1+1/s is a small circle going around the point s=1 right? That means for a large contour going around your function, it's analytic (no branch cuts) and in fact for any contour that's larger than a cicrcle going around the two branch-points. So if we have two such contours, since they are in an analytic region of the function, integrals over them are equal. Now make a dog-bone contour (dumb-bell contour) around the two branch-points. That's one contour. Now make that larger contour which has as one leg, the inversion contour, and close it with a large circular arc. Now, regardless of what \sigma is, say even 1000, as that contour grows to infinity, the arc angle extended by the circular arc of that contour tends to go from \pi/2 around to 3\pi/2.

So now you got everything:

\mathop\oint\limits_{\text{big contour}} f(s)ds=\mathop\oint\limits_{\text{dog-bone}} f(s)ds

I'll do the big one:

\mathop\oint\limits_{\text{big contour}}f(s)ds=\mathop\int_{\sigma-i\infty}^{\sigma+i\infty} f(s)ds+\int_{\pi/2}^{3\pi/2} f(s)ds

The first integral on the right is the inversion integral. Can you make the substitution s=Re^{it} in the second one and show that as R\to\infty, the integral is zero?

Yup thanks, I've done it by looking over at my lecturers notes but I am very so confused as to something he had done. I was fine showing that the arc segments of the contour equal zero but when it came to the straight segments in the dog-bone I had to refer to his notes (He actually did the exact same question..)

Anyway he wrote ln((s+1)/s) = ln(s+1)-ln(s) which is fair enough, but he then set the parametrization for ln(s+1) as s=(1-x)e^0i and for ln(s) as s=xe^-pi, could you explain to me what he has done here?

I actually have no idea whatsoever, does it have to do with the branch cuts? I have never seen anybody split a function up like to traverse a contour.. I mean, I'm absolutely stumped as to the intention behind doing this.

 

[jackmell]

Anyway he wrote ln((s+1)/s) = ln(s+1)-ln(s) which is fair enough, but he then set the parametrization for ln(s+1) as s=(1-x)e^0i and for ln(s) as s=xe^-pi, could you explain to me what he has done here?

I actually have no idea whatsoever, does it have to do with the branch cuts? I have never seen anybody split a function up like to traverse a contour.. I mean, I'm absolutely stumped as to the intention behind doing this.

Don't know why he did it that way.

. . . branching is a formidable undertaking in Complex Analysis. Our task is to determine what the value of \log(1+1/s) is along the upper cut and what it is along the lower cut (the branch-cut between -1 and zero). But really, we're using the limiting case where there is no upper or lower, because they are exactly over one another, i.e., along the real axis (in the limiting case that we actually use). First consider the principal branch of \log(u) for values of u slightly above and below the real axis. Say:
z=-1/2+0.01i
s=-1/2-0.01i
Now compute what the quantity 1+1/z and 1+1/s is. You'll find the first to be a number close to the negative real axis in the third quadrant and the second close to the axis in the second quadrant. That is, the argument are close to -pi and pi respectively. So that in the limit as we approach the real axis:

\text{upper contour: }\log(1+1/z)=\ln|1+1/z|-\pi i
\text{lower contour: }\log(1+1/s)=\ln|1+1/s|+\pi i

but at the limiting case, z=re^{\pi i}=-r on the upper contour and s=re^{-\pi i}=-r on the lower contour so that we can write for the value of this function on both legs as:

\text{upper contour: }\log(1+1/z)=\ln(1-1/r)-\pi i
\text{lower contour: }\log(1+1/s)=\ln(1-1/r)+\pi i

and when you do the integrations, the log(1-1/r) terms cancel but the pi i terms add to 2pi i.