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Since I’m away this week, the solutions would be brief. Please e-mail me if you have a question and would like a more detailed answer.

G. Arfken and H. Weber, Mathematical Methods , Chapter 7.

(a) Show that Z 0 π dθ ( a + cos θ ) 2 = ( a 2 πa − 1) 3 / 2 , a > 1 .

(1) Solution: this problem can be done with a direct application of residue theorem with a variable substitution z = exp( iθ ). Then dθ = ( − i/z ) dz and cos θ = ( z 2 + 1) / (2 z ). The contour integral is over the unit circle in the complex z − plane. Note that the integrand is even, so se can calculate the same integral from 0 to 2 π and divide it by two.

Then, the contour integral is I C = − 2 i I zdz C ( z 2 + 2 az + 1) 2 = − 2 i I C zdz ( z − z 1 ) 2 ( z − z 2 ) 2 = 2 πi X Res [ f ( z i )] , i where z 1 , 2 = − a ± p ( a + 1)( a − 1) .

and 1 for f ( z ) in the integrand function. Note that a > | z 2 | = − a − p ( a + 1)( a − 1) > 1, so it lies outside of the integration contour (unit circle). Thus, we only need to compute a residue of the order-two pole at z = z 1 . Calculating it gives the result that we were supposed to prove.

(b) Show that Z 0 2 π dθ 1 − 2 a cos θ + a 2 = 1 2 π − a 2 , | a | < 1 .

1 (2)

What happens if | a | > 1? What happens if | a | = 1?

Solution: do the substitutions 1 − to reduce the integral to the integral a 2 = α and 2 a = β in the denominator I = Z 2 π 0 dθ a + b cos θ done in class. The result follows. If For a = | a | − 1 there are singularities for > θ 1 the integral equals to 2 π/ ( α 2 − 1).

= 0 and 2 π . For a = 1 there is singularity at θ = π . In both cases the integral does not exist.

(a) Show that ( a > 0) Z ∞ −∞ cos xdx x 2 + a 2 = π a e − a .

How is the right side modified if cos x is replaced with cos kx ?

(b) Show that ( a > 0) Z ∞ −∞ x sin xdx = πe − a .

x 2 + a 2 How is the right side modified if sin x is replaced with sin kx (3) (4) Solution: problem 2 was done in class.

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