Drawing the contour with a branch cut

  • #1
I am trying to determine the contour required in solving part b. The branch points (poles) are at s=0 and s= -a and in between these two values, there is a branch cut.

I know that the branch cut cannot be included in the contour so does this mean the poles also cannot be in the contour? Would it be the one shown in the second image?

Thank you.

EDIT: Got branch points and poles mixed up. Reattempting now.
EDIT2: Hmm, are the branch points the same as poles in this case? Clarification would be appreciated.

Screenshot 2017-12-06 22.27.53.png
export.png
 

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  • #2
stevendaryl
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I am trying to determine the contour required in solving part b. The branch points (poles) are at s=0 and s= -a and in between these two values, there is a branch cut.

I know that the branch cut cannot be included in the contour so does this mean the poles also cannot be in the contour? Would it be the one shown in the second image?

Thank you.

EDIT: Got branch points and poles mixed up. Reattempting now.
EDIT2: Hmm, are the branch points the same as poles in this case? Clarification would be appreciated.

View attachment 216246 View attachment 216247
The idea behind contour integrals is that you have some contour that you want to integrate over, [itex]\mathcal{C}_0[/itex], which is typically not a closed curved. Then you add a bunch of other contours: [itex]\mathcal{C}_1, \mathcal{C}_2, ...[/itex] such that
  • If you connect them, then the add up to a closed curve, [itex]\mathcal{C} = \mathcal{C}_0 + \mathcal{C}_1 + ...[/itex]
  • The closed curve [itex]\mathcal{C}[/itex] encloses no singularities (so the integral over it is zero).
  • The new contours [itex]\mathcal{C}_1, \mathcal{C}_2, ...[/itex] are easy (or at least possible) to do.
Then you can compute the original integral by:

[itex]\int_{\mathcal{C}_0} ds + \int_{\mathcal{C}_1} ds + \int_{\mathcal{C}_2} + ... = \int_{\mathcal{C}} ds = 0[/itex]

In your case, you have [itex]\mathcal{C}_0[/itex] is the vertical line on the right side of your drawing, and [itex]\mathcal{C}_1[/itex] is that funny shape with two bumps. The problem with your drawing is that it's not clear how you should compute the integral around the funny-shaped contour.
 
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EDIT2: Hmm, are the branch points the same as poles in this case? Clarification would be appreciated.

View attachment 216246 View attachment 216247
Branch points refer to singular points of multivalued functions, poles are singular points of single-valued functions. Also, the contour for an inverse Laplace transform is just the Bromwich contour: a straight line from ##\gamma-i\infty## to ##\gamma+i\infty## such that ##\gamma## is to the right of all the singular points of the integrand so say ##\gamma=1## for example.

Did you try to just numerically integrate it to get a ball-park of what's going on? Note in the definition below, we need dz=i along that path so I canceled i in the denominator.

Code:
g[(x_)?NumericQ, (a_)?NumericQ] :=
 (1/(2*Pi))*NIntegrate[Exp[x*z]/Sqrt[z*(z + 1)] /. z -> 1 + I*y, {y, -a, a}];
And after you do that and get a sense of what happens as x gets large, can you explain why it tends to that value? What about just computing what the actual inverse transform is and then taking the limit of that as x goes to infinity?
 

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