Math 455                    Homework Solution               Saff&Snider
Section 6.4
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#5.  For this problem you want to consider

     int_{-infinity}^{infinity} e^(3ix) x/(x^4+4) dx.

     Then, what we will need is the imaginary part of the complex number
     above.  So, let f(z) = z/(z^4+4) and compute the contour integral

     int_{gamma} e^(3iz) f(z) dz,

     where gamma = R e^(it), 0 \leq t \leq pi.

     First note that f(z) has four simple poles (the fourth roots of -4),
     two in the UHP and two in the LHP.  Hence,

     int_{gamma} e^(3iz) f(z) dz = 2i pi sum (residues of f in the UHP).

     Now, first prove (using Jardan's Lemma) that the integral over the 
     upper semicircle goes to zero as R->infinity.  Hence, we get that
     as R -> infinity

     int_{-infinity}^{infinity} e^(3ix) f(x) dx = 2i pi sum(residues of
     f(z) in the UHP).

     Finally, compute the residues using any of the two techniques we have
     covered (there are only two simple poles in the UHP) and take the
     imaginary part of the resulting complex number.
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#9.  Since cos(2x) = [e^(2ix)+e^(-2ix)]/2, we can write

     I = (I_1 + I_2) / 2,

     where I is the integral that we want to compute, and

     I_1 :=  p.v. int_{reals} e^(2ix) / (x - 3i) dx,

     I_2 :=  p.v. int_{reals} e^(-2ix) / (x-3i) dx.

     Consider, f(z) = e^(2iz) / (z-3i).  Then,

     I_1 = lim_{R->infinity} [int_{-R}^{R} f(z) + int_{gamma1} f(z)]
     where gamma1=R e^{it}, t \in [0, pi].  By the residue thm
     I_1 = 2i pi Res(f;3i) = 2i pi e^(-6), and by Jordan's Thm the
     integral over gamma1 goes to zero as R->infinity.

     I_2 = lim_{R->infinity} [int_{-R}^{R} f + int_{gamma2} f]
     where gamma2 = R e^{it}, t \in [pi, 2pi].  By Cauchy's Thm I_2=0,
     and thus we have    

     I = [2i pi e^(-6) + 0] / 2.

     Note that for I_2 we have to use the lower semicircle because
     f(z) is unbounded in the UHP.
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