# Complex contour integral, which one do I use?

1. Jul 29, 2011

### hunt_mat

I want to perform the following integral which comes from inverting a Laplace transform:
$$\lim_{R\rightarrow\infty}\int_{\sigma -Ri}^{\sigma +Ri}\frac{e^{sx}}{\sqrt{s^{2}-a^{2}}}ds$$
Would it be some kind of keyhole contour?

Mat

2. Jul 30, 2011

### Charles49

You need a semi-circle where the diameter is on the right (from your perspective) of the poles $$\pm a$$. That's easier to work with.

3. Jul 30, 2011

### hunt_mat

I am aware of the Bromwhich contour for the Laplace transform. Do I need to cut to disks out at $\pm a$ to make a double keyhole if you like?

Also if a is complex do I need to do the same thing? What about a slightly function?
$$\lim_{R\rightarrow\infty}\int_{\sigma -Ri}^{\sigma +Ri}\frac{e^{sx}}{(s-b)\sqrt{s^{2}-a^{2}}}ds$$
where b is real. Do I need the keyhole contour again?

4. Aug 1, 2011

### hunt_mat

This is the contour that you would usually use if there are any singularities on the real line but what is you singularities move? I have a value which moves from being completely real (hence the need for a keyhole contour shown to one being complex and I can just take a semi-circle.

Suggestions?

Mat

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5. Aug 2, 2011

6. Aug 2, 2011

### hunt_mat

I don't have this book. I am not going to buy this book either, our library does not have this book.

Care to describe the contour?

7. Aug 2, 2011

### Charles49

It doesn't matter if the pole is real or complex. You just move the location of the keyhole to circumvent the value. If there are more poles, you simple include more keyholes.

I included the link because the parts you are interested in is free to view in Google Books.

8. Aug 3, 2011

### hunt_mat

It's just showing me the cover of the book and I can't see inside the actual book.

Surely if $a$ is complex then then I don't need all this fuss about a keyhole contour, I can just use a semi-circle.

9. Aug 3, 2011

### hunt_mat

I have done another contour which I think it should be, is this more like it?

#### Attached Files:

• ###### contour3.jpg
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10. Aug 4, 2011

### jackmell

I believe direct contour integration is going to be tough. How about a different approach? Solve an easy one, get the answer (via Mathematica), then justify that answer. Then generalize it to arbitrary values of the parameter. Ok, let me try without helping you too much. Say:

$$\mathcal{L}^{-1}\left\{\frac{1}{\sqrt{s^2-1}}\right\}=\text{BesselJ[0,ix]}$$

Now, we know:

$$\text{BesselJ[0,ix]}=\frac{1}{2\pi i}\oint{e^{ix/2(t-1/t)}1/t dt}$$

So, if you're interested, can you come up with a suitable change of variable (and subsequent change of contour) and justify saying:

$$\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{e^{sx}}{\sqrt{s^2-1}}ds=\frac{1}{2\pi i}\oint{e^{ix/2(t-1/t)}1/t dt}=\text{BesselJ[0,ix]}$$

I've not (completely) proven this and would personally find it a challenge to do so. Maybe though, you may wish to pursue it.

Last edited: Aug 4, 2011
11. Aug 4, 2011

### hunt_mat

I think the best approach here would be to use the keyhole contour, the integral in question is somewhat more complicated than the initial one I posted as I wanted to get a feel for the square root in the denominator.

12. Aug 4, 2011

### jackmell

If you wish to integrate it directly, then I think you need to use a dumb-bell contour around the branch points $\pm a$ enclosed inside a larger closed contour which includes the Bromwich path. These two contours then traverse an entirely analytic function so they are equal to one another and I think the dumb-bell contour is not too difficult to analyze. The reason I suggest that is because the Riemann surface for n'th roots of n'th degree polynomials are analytic outside the hull of polynomial zeros. I barely understand it myself for that matter. Tell you what, when I do that for the simple case I used above, I obtain:

$$\int_{1}^{-1}\frac{e^{r x}+e^{-{rx}}}{\sqrt{r^2-1}}dr$$

and that integral can be written in terms of $\text{BesselJ[0,ix]}$

however, I've not analyzed the remaining parts of the contour to justifiy equating that answer as the inverse transform.

Last edited: Aug 4, 2011
13. Aug 8, 2011

### hunt_mat

I had a go at doing the integral myself and it reduces to an integral on the real line which I can't compute, can anyone lend a hand or spot where I have made a mistake?

#### Attached Files:

• ###### Contour_integral.pdf
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14. Aug 9, 2011

### jackmell

. . . that's not the correct contour to use although it may turn out to give you the correct answer or maybe the neg of the answer because it's computationally similar to what I said earlier. However, using it to solve this problem in my opinion clearly indicates a lack of understanding of the underlying Riemann geometry. No offense ok. It's just if you don't understand that geometry for this one, then you'll have problems solving others like it that likely will not be so forgiving if you use the wrong contour.

Last edited: Aug 9, 2011
15. Aug 9, 2011

### hunt_mat

My Riemannian geometry isn't bad (I have an MSc in it from Oxford university in geometry) but I don't see how it pertains to this problem.

Let us examine the reason why the contour may be a good one to use. First off, for the inverse Laplace transform, the Bromwich contour, so whatever contour id used, it must be a deformation of this in some way. There is a square root involved, so there must be a branch cut of some kind, so a branch cut is taken from -a to +a and the contour is deformed around the two branch points, so that is what I did.

As I homework helper, I am always polite and helpful. I am never condescending to any of the people I help out. I would ask you to do the same.

16. Aug 9, 2011

### jackmell

I did not feel I was being condenscending. Actually I thought I was being quite polite. Sorry if I came across that way. Also, I should have said, "Riemann surface" and not Riemann geometry. The two legs you have going from -infty to -b are completely unnecessary. The function is analytic there so you can just cross-over the negative imaginary axis with one big circular contour (which includes the Bromwich path) and equate that contour it to the remaining "dumb-bell" contour going around the two branch-points unless the pole at a is "between" the two contours in which case you'd have to adjust the calculation to include that pole. I stand by what I said: in order to do well integrating over multifunctions, I believe it's essential to have a good understanding of the underlying Riemann surfaces that are being integrated over and understanding why you can delete the two "excipient" legs from -infty to -b is one way to exhibit such an understanding.

Also, unless you know what the answer is, I suggest you numerically integrate over the Bromwich path for some value of x and compare that with the inverse transform expression you obtain. If these agree, then I think it's a very good indication your answer is correct.

Last edited: Aug 9, 2011
17. Aug 9, 2011

### Charles49

I am assuming a and b are real. Also is a>b or is b>a?

18. Aug 9, 2011

### Charles49

I couldn't evaluate the integral you posted using Mathematica. I suggest that the inverse you want is the convolution of $$BesselJ[0, i a t]$$ and $$e^{b t}$$ where BesselJ is the Bessel function of the first kind. The convolution is defined as $$f*g=\int_0^t f(y)g(t-y)dy.$$ This follows from the convolution theorem which says that the product of functions in the s domain equals to the Laplace transform of the convolution of the inverse Laplace transform of the individual factors.

Last edited: Aug 9, 2011
19. Aug 9, 2011

### jackmell

Does that answer agree with the numerical result:

Code (Text):
In[16]:=
a = 2;
b = 1;
x = 1;
NIntegrate[(1/(2*Pi*I))*((I*Exp[s x])/
((s - a)*Sqrt[s^2 - b^2])) /. s -> 3 + I*y,
{y, -250, 250}]

Out[19]= 3.34859 + 4.35036*10^-15 I

20. Aug 9, 2011

### Charles49

No, it doesn't. How did you get s=3+iy?