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Math. 4103, homework 8, solutions
- Compute the integral (^) ∫ (^) ∞
−∞
x^2 + 2x + 2 dx
We compute using the residue theorem: ∫ (^) ∞ −∞
x^2 + 2x + 2 dx^ =
−∞
(x + 1 + i)(x + 1 − i) dx =2πi Res−1+i
(z + 1 + i)(z + 1 − i)
= 2πi (^) (−1 + i^1 + 1 + i) = π.
- Compute the integral (^) ∫ (^) ∞
0
x^3 + 1 dx.
By the residue theorem,
(2.1)
0
x^3 + 1 dx^ −
0
(e^2 πi/^3 x)^3 + 1 e
2 πi/ (^3) dx = 2πi Reseπi/ 3
z^3 + 1
Notice that ∫ (^) ∞ 0
(e^2 πi/^3 x)^3 + 1 e
2 πi/ (^3) dx =
0
x^3 + 1 e
2 πi/ (^3) dx.
Hence, (2.1) may be rewritten as
(2.2) (1 − e^2 πi/^3 )
0
x^3 + 1 dx^ = 2πi Reseπi/^3
z^3 + 1
Since, z^3 + 1 = (z + 1)(z − eπi/^3 )(z − e−πi/^3 ),
we see that (2.3) Reseπi/ 3
z^3 + 1
= (^) (eπi/ (^3) + 1)(e^1 πi/ (^3) − e−πi/ (^3) ) = (^) (e 2 πi/ (^3) + eπi/ (^31) )(1 − e− 2 πi/ (^3) ). 1
By combining (2.2) and (2.3) we compute ∫ (^) ∞ 0
x^3 + 1 dx^ =^
2 π 3 √ 3.
- For a > 0 compute the integral ∫ (^) ∞ −∞
x sin(ax) x^4 + 1 dx.
Notice that
(3.1)
−∞
x sin(ax) x^4 + 1 dx^ =^ Im
−∞
x eiax x^4 + 1 dx.
By Jordan’s Lemma,
(3.2)
−∞
x eiax x^4 + 1 dx^ = 2πi
Reseπi/ 4 z e
iaz z^4 + 1 +^ Resieπi/^4
z eiaz z^4 + 1
Furthermore,
z^4 + 1 = (z − eπi/^4 )(z − ieπi/^4 )(z − e−πi/^4 )(z + ie−πi/^4 ).
Hence, Reseπi/ 4 z e
iaz z^4 + 1 =
4 i e
− √a 2 + √ia 2
and
Resieπi/ 4 z e
iaz z^4 + 1 =^ −^
4 i e
− √a 2 − √ia (^2).
Therefore (3.2) is equal to
πi e−^ √a^2 sin( √a 2 )
and finally, (3.1) is equal to π e−^ √a^2 sin( √a 2 )
Hence, Rouche’s Theorem implies that the number of zeros of (6.1) is the same as the number zeros of f (z), which is 2.
Next we compute the number of zeros in the disc |z| < 2. Let f (z) = z^5 and let g(z) = − 6 z^2 + z + 1. Then for any z with |z| = 2,
|g(z)| ≤ 26 < 32 = |f (z)|.
Hence, Rouche’s Theorem implies that the number of zeros of (6.1) is the same as the number of zeros of f (z), which is 5.
The number of zeros with in the the annulus {z ∈ C; 1 < |z| < 2 }, counted with multiplicities, is equal to the difference
5 − 2 = 3.
