Math 455                         Homework              Saff&Snider
Sec 6.3
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#3.  Since the integrand is an even function, we can integrate from minus
     infinity to infinity and then divide by 2.  Now consider the function

     f(z) =(z^2 + 1) / (z^4 + 1).

     This function has four simple poles, two in the upper half plane and
     two in the L.H.P.  Therefore, we can use the contour gamma = Re^(it)
     for this problem.  We have

     int_{gamma} f(z) dz = 2pi i sum {Residues of f in the U.H.P.}

     Also,

     int_{gamma} f dz = int_{-R}^{R} f dz + int_{semicircle} f dz.

     First, prove that

     limit_{R -> infinity} int_{upper semicircle} f(z) dz = 0.

     Then, calculate residues of f in the U.H.P.  There are two of them
     and they both are simple poles.  So, use either one of the two
     techniques we covered in class.  Then,  you will have he answer

     int_{0}^{infinity} f(x) dx = 1/2 int_{-infinity}^{infinity} f dx

     = 1/2 limit_{R->infinity} int_{-R}^{R} f(x) dx

     (1/2) (2pi i) sum (residues of f in the U.H.P.)
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#10. For this problem use the contour that is suggested in your book.
     Consider  f(z) = e^(-z^2).  This function is entire.  Hence its
     integral over the specified contour is zero.  First prove that

     int_{R}^{R+iL} f(z) dz -> 0 as R -> infinity.

     Then, prove that

     int_{0}^{L} f(z) dz = i int_{0}^{L} e^(t^2) dt.

     Now we have

     int_{specified contour} f(z) dz = 0 = integral over the four pieces
     of the specified contour. Hence,

     int_{0}^{R} f(x) dx + i int_{0}^{L} f_{z} dz 

     = int_{0}^{R} e^{(t + iL)^2} dt + (an integral that goes to 0)

     Finally, if you let R go to infinity and take the real and imaginary
     part of the complex number above, you get the desired result.
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#15c. Here we need to modify the equation from problem 14.  Note that
      the function f(z)=1/z^2 has a pole of order two at zero.  Thus,
      since cot(pi z) has a simple pole at z=0, the function 
      g(z)=pi f(z) cot(pi z) has a pole of order three at z=0. Now,
      let gamma be the same contour as in problem 14.  Then, as in 
      problem 14 the integral of g over gamma is zero.  Hence, by the
      residue thm we have

      0 = sum_{k=1}^{infinity} Res(g;k) 

	+ sum_{k=-1}^{-infinity} Res(g;k)

	+ Res(g;0)

      Now, as before as long as k is not zero  Res(g;k) = f(k).  And since
      f(-k) = f(k) we have

      sum_{k=1}^{infinity} f(k)  =  - (1/2) Res(g;0)

      Hence, what is left is to compute the Res(g;0).  g has a pole of
      order 3 at zero.  Therefore, 

      Res(g;0) = (1/2) lim_{z->0} (z^3 g)".

      So, compute the second derivative of z^3 g and take the limit as
      z -> 0.  You will get -pi/3, which will provide the desired result.

      ADDITIONAL COMMENTS FOR THIS PROBLEM:

      1. If k is a non-zero integer, then Res(g;k) can be obtained from the
      formula  P(k) / Q'(k). This is due to the fact that g(z) has a simple
      pole at these integer points. If you make this calculation you will
      find that  Res(g;k) = f(k) = 1 / (k^2).

      2. Now the calculations for Res(g;0). We have

      z^3 g(z) = \pi z cot(\pi z), which gives 

      (z^3 g)" = \pi {2 \pi csc^2 (\pi z)}{(\pi z / sin(\pi z))cos(\pi z) -1}

      If you let z -> 0 above, you will find that the quantity in the first
      set of braces goes to infinity, and the quantity in the second set of
      braces goes to 0. Hence, to take the limit we must use one of the 
      standard techniques of calculus. Namely, convert the problem to the
      form when you get 0/0, then apply the L'Hopitals rule to get the limit.

      3. As an alternative method, you can write down the Laurent expansion
      for g(z) explicitly and extract the residue at zero. If you have
      trouble with the L'Hopital's rule, you should use this technique. It
      works equally well.