- 11-Oct-2011 --- Various updates and typo corrections.
- 08-Oct-2011 --- Extra comments on problems 1, 4, and 5.
- 07-Oct-2011 --- Expanded comments on large R contour in problem 3.
- 06-Oct-2011 --- Original version.

**Contour integrals**- As in the example done in class,
choose the height of the rectangular contour so that the top
edge (where z = x + i times height) is such that the integrand
is a multiple of the original integrand. Should the height
depend on a, b, both, or neither?
The conditions on a and b are important to ensure that the
right and left edges vanish as you take
*R*to infinity; make sure you see how that happens. For the Mathematica evaluation, you want to tell it to assume that*b > Re(a) > 0*. You can do this with the`Assumptions`option to the`Integrate`command. In this case,`Assumptions -> {b > Re[a] > 0}`is what you want. (After`Assumptions`is a minus sign and a greater-than sign, which together make an arrow.) -
You could avoid the branch point here by changing variables
to
*y = x*, but what fun would that be? :) Instead, design your contour to avoid crossing a branch cut on the positive real axis and follow the steps as in the example done in class or similar ones in the texts. This problem is quite similar to Example 2.23 in Lea and you might use the solution there as a guide. Be careful to check for contributions from all the pieces (big circle, little circle, both sides of the branch cut) and remember that you'll get a different answer for the integration on top (θ = 0) and bottom (θ = 2π) of the cut. Also remember when calculating the residues that the branch cut was chosen so^{1/3}*0 < θ < 2π*, so determine the angles of the poles consistent with this range. Mathematica gives the result for this integral without any tricks.

- As in the example done in class,
choose the height of the rectangular contour so that the top
edge (where z = x + i times height) is such that the integrand
is a multiple of the original integrand. Should the height
depend on a, b, both, or neither?
The conditions on a and b are important to ensure that the
right and left edges vanish as you take
**Integral representation of step function**- Consider the two cases,
*t>0*and*t<0*, separately. In each case, choose a contour. Remember Jordan's lemma in deciding on what contour to pick. (Note that Jordan's lemma applies to closing the contour in the lower half-plane with a large semi-circle as well as the upper half-plane. Which half-plane you close in is determined by whether the exponential*e*has positive or negative^{ikz}*k*.) Where is the singularity (or singularities)? In Mathematica, you can get the*t<0*case with:

`Integrate[Exp[I k t]/(k - I eps), {k, -Infinity, Infinity}, Assumptions -> {t < 0, eps > 0}]`

and similarly with*t>0*. You can take the limit of ε to zero in your head, or use the Mathematica`Limit`function (`Limit[stuff,eps->0]`). - If you use the definition of a principal value integral
we originally used in class (the average of the integrals with
*k - iε*and*k + iε*), then this result follows quickly from your result for part (a) and a very similar result with the opposite sign for ε. But you can also follow the texts or the lecture notes on page 57 treating principal value integrals. To get Mathematica to do a principal value integral, add the option`PrincipalValue->True`to the`Integrate`function.

- Consider the two cases,
**Fresnel integrals**

What is an appropriate choice of*f(z)*that can get you both the sine and cosine integrals? An integral in the radial direction from the origin in the complex*z*plane can be parametrized as*z = r e*with a fixed θ and^{iθ}*r*going from 0 to the*R*. When considering the contribution from the curved part of the integration contour, what matters is the dependence on*R*. If there are parts that are pure phases (*e*for some real α) --- do these affect whether the contribution blows up or dies off as^{iα}*R*gets large? What about real exponentials? You can make a upper bound on the integral as it is done in the proof of Jordan's Lemma (Arfken pg. 467 or Lea pg. 140), leaving an integral you can do with the conclusion that the contribution vanishes as*R*goes to infinity. You might want to review the class discussion from page 56 in the notes.**Laguerre's equation**

Follow the examples in Arfken or Lea for the Frobenius method (see the reading) or pages 60-61 in the lecture notes. Regular at the origin means that it doesn't blow up at*x=0*. We try a solution that is a Taylor expansion times*x*with^{p}*p*to be determined (Arfken uses the variable*k*instead of*p*). Substitute the series, doing the derivatives term by term, and equate coefficients of the each term. The lowest power of*x*determines*p*(you should find*p=0*here), and then you get relations between coefficients of successively higher powers of*x*. (Note that it is only the equation with the lowest power of*x*that you solve for*p*. An equation for the next lowest power may determine whether*a*is zero or a free parameter.) You should be able to identify the general term, but it is easiest to see if you try the first few individually. To show that when α is an integer you get a polynomial, you must show that the equation for the coefficients of the series are zero after a finite number. You should be able to check your general result against the first few Laguerre polynomials, which you will find in quantum books (or just Google it)._{1}**Bessel equation**

Again, follow the examples in Arfken or Lea for the Frobenius (see the reading) or pages 60-61 in the lecture notes. The same comments as for Laguerre's equation apply here (in both cases you are expanding about 0). A difference here is that you will find two possible values of*p*, which you should consider separately. You should find that*a*and_{0}*a*are independent parameters (for one of the_{1}*p*values), which means you obtain two solutions. You should recognize the series that you get in the two cases (once you pull out an overall power of*x*). You can check it by noting that the solutions are Bessel functions of order 1/2, which you can find (for example) in Wikipedia under "Bessel function" (search for spherical bessel functions).**More contour integrals**- We'd like to convert this to an integral around the entire unit circle (and not just half of it); how do we do that and justify it? This is then one of the standard integral types we considered and you get to practice finding poles and calculating residues. (You can use Mathematica to find the poles.) Mathematica gives the answer for the full integral directly.
- What kind of contour works for this one? (Does a semi-circle
at large
*R*vanish? If not, what are the alternatives?) Is it useful to extend the integral to minus infinity? Why can you do this? If you have to evaluate an integral of*(cosh ax)*without the^{-1}*x^2*, you can just quote the result from Mathematica or a table of integrals. If you give this integral to Mathematica, it will give an answer but tell you it is only true for*a>0*. Is this a true limitation?

**Proving an identity with residues**

This is the same type integral as 6(a), and the suggestions there apply here as well. Remember the binomial expansion (you can look it up in the index in Arfken), which tells you the coefficient of each term in*(a + b)*(with integer^{m}*m*). Can you integrate term by term? Which terms give non-zero contributions? (If you think about each term as being a one-term Laurent expansion, the answer is immediate; you can also use the residue formula for m-order poles.) For the double factorial expression, note equation (8.33c) in Arfken, which you can quote.**Atomic collision integral**

Look at Arfken Example 7.1.4 for a similar integral. You may find it useful to split up*|p| > 1*into the two cases of positive and negative*p*(and with the*|p| < 1*condition as well). If you do this, you can apply the result of problem 2b) and get the result in a couple of lines in terms of theta functions.

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Last modified: 07:59 am, October 11, 2011.

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