Example

⌠ ⌡ 2π 0  dθ 2 + cos θ = ⌠ (⎜) ⌡ |z|=1  dz iz 2 + 1 2 (z + 1/z) = ⌠ (⎜) ⌡ |z|=1  2 dz i(z2 + 4 z + 1) = 2 i ⌠ (⎜) ⌡ |z|=1  1 z2 + 4 z + 1 dz = 2 i 2πi ∑ Res ⎛ ⎝ 1 z2 + 4 z + 1 ;ai ⎞ ⎠ . (1)

The roots of the equation z² + 4 z + 1 = 0 are −2 ±√3 of which only −2+√3 is inside the unit circle so

= 2 i 2πi Res ⎛ ⎝ 1 z2 + 4 z + 1 ;−2 + √3 ⎞ ⎠ = 4π 1 2 z + 4 |z = −2 + √3 = 2π √3 . (2)

Note that we used the short-cut to find a residue at a first order pole.

Example

⌠ ⌡ 2π 0  cos6 θdθ.

(Solution)

⌠ ⌡ 2π 0  cos6 θdθ = ⌠ (⎜) ⌡ |z|=1  ⎛ ⎝ 1 2 (z + 1 z ) ⎞ ⎠ 6   d z iz = 1 64i ⌠ (⎜) ⌡ |z|=1  (1 + z2)6 z7 dz = 1 64i ⌠ (⎜) ⌡ |z|=1  z12 + 6 z10 + 15 z8 + 20 z6 + 15 z4 + 6 z2 + 1 z7 dz = 1 64i ⎛ ⎝ …+ ⌠ (⎜) ⌡ |z|=1  20 dz z + … ⎞ ⎠ = 1 64i 20 ⌠ (⎜) ⌡ |z|=1  dz z = 1 64i ×20 ×2πi = 5 8 π. (3)

Example 1:

⌠ ⌡ ∞ −∞  dx 1 + x2 .

The integrand satisfies the above two conditions (the difference in the polynomial order is 2 and z = ±i are not on the real axis. Note that only z = i is in the upper half plane.) so

⌠ ⌡ ∞ −∞  dx 1 + x2 = 2πi Res( 1 1 + z2 ;i) = 2πi 1 2 z |z=i = π. (4)

Of course this result can be also obtained by directly integrating 1/(1 + x²) (=arctanx).

Example 2:

⌠ ⌡ ∞ −∞  dx 1 + x4 dx.

(5)

The difference in degree of P(x)/Q(x) is 4 and all the zeros to 1 + z⁴ = 0 are off the real axis. The zeros to z⁴ + 1 = 0 are

z = eπi/4 + n πi/2        n=0,1,2,3,

of which z's for n = 0 and n = 1 are in the upper half plane.

⌠ ⌡ ∞ −∞  dx 1 + x4 = 2πi ∑ upper half plane  Res( 1 z4+1 ;ai) = 2πi ⎛ ⎝ Res( 1 z4 + 1 ; eπi/4) +Res( 1 z4 + 1 ; e3πi/4) ⎞ ⎠ = 2πi ⎛ ⎝ 1 4z3 |z=eπi/4 + 1 4z3 |z=e3πi/4 ⎞ ⎠ = 2πi 4 ⎛ ⎝ 1 e3πi/4 + 1 eπi/4 ⎞ ⎠ = πi 2 (e−3πi/4 + e−π/4 i) = πi 2 ( −√2i ) = π √2 . (6)

Example 3:

⌠ ⌡ ∞ −∞  x2 dx 1 + x4 = 2πi ∑ upper half plane  Res( z2 z4 + 1 ;ai) = 2πi ⎛ ⎝ z2 4 z3 |z=exp(πi/4) + z2 4 z3 |z=exp(3πi/4) ⎞ ⎠ = πi 2 ⎛ ⎝ 1 z |z=exp(πi/4) + 1 z |z=exp(3πi/4) ⎞ ⎠ = πi 2 (e−πi/4 + e−3πi/4) = πi 2 ( −√2 i) = π √2 . (7)

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File translated from T_EX by T_TH, version 4.03.
On 18 Jan 2021, 22:02.