Download Complex Variables & Integral Transforms: MATH-6640, Spring 2005, Assignment 3 Solutions and more Assignments Mathematics in PDF only on Docsity!
MATH-6640 Spring 2005 COMPLEX VARIABLES AND INTEGRAL TRANSFORMS WITH APPLICATIONS
Assignment 3 Solutions
- Show that
I =
0
x 6
(a^4 + x^4 )^2
dx =
2 π
16 a
where a is real and positive.
Consider
J =
C
z^6
(a^4 + z^4 )
dz,
where the contour C is traversed in the positive direction and consists of the straight segment from z = −R to z = R and the upper semicircle of radius R centered at z = 0. We take R > a > 0.
-R R
C
Re z
Im z
O
o o
The integrand has double poles at z^4 + a^4 = 0, i.e., at z = ±aeiπ/^4 and z = ±ae^3 iπ/^4. Only those corresponding to the upper signs are within C. To find the residues it is best to expand the integrand in a Laurent series. If z 0 denotes one of the poles, then on setting z = z 0 + t and expanding about t = 0 one obtains, after some algebra, the Laurent expansion
f (z) =
z^6
(a^4 + z^4 )
(z 0 + t)^6
(a^4 + (z) + t)^4
16 t^2
16 z 0 t
Thus the residue, being the coefficient of 1/t, is 3/(16z 0 ). By the residue theorem, then,
J = 2πi
[
Res
f (z), z = ae iπ/ 4
f (z), z = ae i 3 π/ 4
)]
= 2πi
16 a
[
e −iπ/ 4
]
2 π
8 a
Let JR be the contribution to J from the straight portion [−R, R] of C and JΓ from the curved portion Γ. Since zf (z) → 0 as z → ∞ on Γ, JΓ → 0 as R → ∞. At the same time, JR → 2 I. Therefore, J = JR + JΓ → 2 I and we obtain
i =
π
16 a
- For a real and positive, show that
I 1 =
0
(1 + x^2 ) cos ax
1 + x^2 + x^4
dx =
π √ 3
exp(−
3 a/2) cos(a/2),
and
I 2 =
0
x sin ax
1 + x^2 + x^4
dx =
π √ 3
exp(−
3 a/2) sin(a/2).
Hint: Do both parts together. Note that
I 1 =
0
[g(x) + g(−x)] cos ax dx =
−∞
g(x) cos ax dx
and similarly,
I 2 =
−∞
g(x) sin ax dx,
where
g(x) =
1 − x + x^2
Consider, therefore,
J =
C
g(z)e iaz dz
on an appropriate contour C.
We follow the hint and consider
J =
C
f (z) dz
where
f (z) = g(z)e iaz =
eiaz
1 − z + z^2
and C is the same contour as in the preceding example. The only singularity of f (z) within C is the simple pole at z 0 = (1 +
3 i)/ 2 , where the residue is
Res (f (z), z = z 0 ) =
e iaz 0
2 z 0 − 1
3 i
exp
ia
2
3 a
2
Therefore, according to the residue theorem,
J =
2 π √ 3
exp
ia
2
3 a
2
Since g(z) → 0 uniformly on the semicircular part of the contour, Jordan’s lemma applies and JΓ → 0 as R → ∞. One is then left with
I 1 + I 2 =
−∞
g(x)e
iax dx
J
π √ 3
e −
√ 3 a/ 2
cos
a
2
a
2
The values of I 1 and I 2 follow.
- Show that for 0 < a < 2 ,
I 1 =
0
x a− 1
1 + x + x^2
dx =
2 π √ 3
cos
π+2πa 6
sin πa
and
I 2 =
0
xa−^1
1 − x + x^2
dx =
2 π √ 3
sin
π+2πa 3
sin πa
Hint: Consider
J =
C
z a− 1
1 + z + z^2
dz
Comparing real and imaginary parts,
I 2 =
2 π √ 3
sin
(2a+1)π 3 sin aπ
I 1 = I 2 cos aπ −
2 π √ 3
cos
(2a + 1)π
3
2 π √ 3
cos
(2a+1)π 6 sin aπ
- Show that (^) ∫ ∞
0
dx
xα(x + 1)
π
sin πα
Hint: Integrate f (z) = 1 zα(z+1) around the contour shown below. The contour suggests how to select the branch cut.
-1 O^ R Re z
Im z
C L
L
Γ
ε +
-
Figure for problem 4.
We note that for the integral
I =
0
dx
xα(x + 1)
to converge, α must satisfy 0 < α < 1. Consider the contour integral
J =
C
z −α
1 + z
dz,
where C is the contour shown above. There are branch points at z = 0 and z = ∞. We have selected the cut to lie on the positive real axis, and we let z = re iθ , with 0 ≤ θ < 2 π. We shall use the branch z−α^ = r−αe−iαθ^.
The only singularity within the contour is the pole at z = −1 where the residue is
Res
z−α
1 + z
, z = − 1
= e −iπα .
Then, by the residue theorem, J = 2iπe −iπα .
Since z z−α 1+z →^0 both as^ z^ →^0 and^ z^ → ∞,^ the large circle^ Γ^ and the small circle^ C^ do not contribute as R → ∞ and → 0. We are only left with contributions from the line segments L+ and L−. On L+, z = re i 0 , so that in the limit,
JL+ =
0
r−α
(r + 1)
dr = I.
On L−, z = re^2 iπ^ , so that in the limit,
JL− = −
0
e − 2 iπα r −α
(r + 1)
dr = −e − 2 iπα I.
With J = J+ + J−, we get, following simple algebra,
I =
π
sin πα
- Prove that
I =
0
sin ax
e^2 πx^ − 1
dx =
coth
a
2
2 a
, a > 0.
Hint: Integrate eiaz^ /(e^2 πz^ − 1) around the rectangle of sides x = 0, R; y = 0, 1 , suitably indented at 0 and i.
Re z
Im z
O
i
R
R+i
γ
γ
1
2
Consider
J =
C
f (z) dz
where
f (z) =
e iaz
e^2 πz^ − 1
and C is the rectangular indented contour shown above. The circular segments γ 1 and γ 2 are each of vanishingly small radius .
The singularities of f (z) are simple poles at ni; n = 0, ± 1 , ± 2 , · · ·. Since none of the singularities lie within C, J = 0. Therefore,
∫ (^) R
e iax
e^2 πx^ − 1
dx +
R
e −a+iax
e^2 πx^ − 1
dx +
0
ie ia(R+iy)
e^2 π(R+iy)^ − 1
dy +
1 −
ie −ay
ei^2 πy^ − 1
dy +
γ 1
f dz +
γ 2
f dz = 0.
Note that ∫
γ 1
f dz = −
2 πi
4
Res (f (z), z = 0) = −
i
4
γ 2
f dz = −
2 πi
4
Res (f (z), z = i) = −
ie−a
4
