SOLUTION TO PROBLEM SET 5, FYS3140
Problem 5.1 (Residue theory)
a) 14.7.17
Let us substitute x → z and sin x → eiz and start by evaluating the integral
in the upper half plane,
Z ∞
I
x sin xdx
zeiz dz
→
2
z 2 + 4z + 5
−∞ x + 4x + 5
I
I
zeiz dz
zeiz dz
=
z 2 + 4z + 5
(z + (i + 2))(z − (i − 2))
z = i − 2 is the only pole in the upper half-plane, and its residue becomes
R(z = i − 2) =
lim (z − (i − 2))
z→(i−2)
zeiz
(z + (i + 2))(z − (i − 2))
(i − 2)ei(i−2)
2i
(i − 2)ei(i−2)
2i
(−2 cos 2 + sin 2) + i(cos 2 + 2 sin 2)
2ie
=
=
=
and
I
zeiz dz
(z + (i + 2))(z − (i − 2))
(−2 cos 2 + sin 2) + i(cos 2 + 2 sin 2)
= 2πi ·
2ie
(−2 cos 2 + sin 2) + i(cos 2 + 2 sin 2)
= π
e
Now the complex integral splits into integral along x-axit and along the
semicircle. Along x-axis, z → x and on the semicircle z → ρeiθ
I
zeiz dz
=
z 2 + 4z + 5
Z
ρ
−ρ
xeix dx
+
x2 + 4x + 5
Z
0
π
ρeiz ρieiθ dθ
ρ2 ei2θ + 4ρeiθ + 5
taking the limit ρ → ∞,
Z
∞
=
−∞
xeix dx
+
x2 + 4x + 5
Z
1
0
π
eiz ρ2 ieiθ dθ
ρ2 ei2θ + 4ρeiθ + 5
The second part of RHS is evaluated as follows,
Z π
Z π
eiz ρ2 ieiθ dθ
ei(ρ cos θ+iρ sin θ) ie−iθ dθ
= lim
lim
ρ→∞ 0
ρ→∞ 0 ρ2 ei2θ + 4ρeiθ + 5
Z π
= lim i
ei(−θ+ρ cos θ) e−ρ sin θ dθ
ρ→∞
0
= 0
where we know that |eiy | = 1, and e−ρ sin θ vanishes as ρ → ∞ (sin θ is
positive in upper half plane). You can also apply Jordan’s lemma. Therefore,
Z
∞
−∞
xeix dx
(−2 cos 2 + sin 2) + i(cos 2 + 2 sin 2)
=π
x2 + 4x + 5
e
Equating the real and imaginary parts of the two sides, we get
Z
∞
−∞
xeix dx
x2 + 4x + 5
Z ∞
x cos xdx
x sin xdx
=
+i
2 + 4x + 5
2 + 4x + 5
x
x
−∞
−∞
(−2 cos 2 + sin 2) + i(cos 2 + 2 sin 2)
= π
e
Z ∞
x sin xdx
π
⇒
= (cos 2 + 2 sin 2)
2
e
−∞ x + 4x + 5
Z
∞
b) 14.7.24
We use same procedure here as the previous problem. Making substitutions
x → z and sin x → eiz , we start by evaluating the following complex integral
in the upper half-plane. i.e
Z ∞
I
x sin πx
zeiπz dz
dx
→
2
1 − z2
−∞ 1 − x
here the poles z = 1 and z = −1 are at the boundary. Residue method is not
generally correct when multiple poles are present at the boundary. Thus, I
will evaluate the integral directly by excluding the poles from our region by
creating small circles of radius r about each pole.
I
zeiπz dz
=0
(∗)
1 − z2
Now we let the small circles to vanish by taking the limit r → 0.
For semicircle around z = 1 the following substitutions will be made: z =
1+reiθ ; dz = rieiθ dθ = i(z −1)dθ. When r → 0; then z → 1; and eiπz → −1.
the integral around the small semicircle about z = 1 is thus
2
Z
lim
r→0 C 0
zeiπz dz
(1 − z)(1 + z)
Z
i
dθ
C0 2
Z 0
i
=
dθ
π 2
−iπ
=
2
=
similarly, for semicircle around z = −1, we make substitutions: z = −1 +
reiθ ; dz = rieiθ dθ = i(z + 1)dθ. When r → 0; then z → −1; and eiπz → −1,
hence
Z
zeiπz dz
−iπ
lim
=
0
r→0 C (1 − z)(1 + z)
2
Now we resolve the integral in eqn (*) into six components, i.e. along xaxis; along the two small circles of radius r; and along the larger semicircle
of radius ρ. i.e.
I
zeiπz dz
1 − z2
zZ
−1−r
=
xeiπx dx
1 − x2
−ρ
I
+
0
C1
|
2
zeiπz dz
0
Cbig
|
1−r
I
zeiπz dz
zeiπz dz
+
2
0
1 − z2
C−1 1 − z
{z
}
I
+
1
}|
{
Z ρ
xeiπx dx
xeiπx dx
+
+
2
2
−1+r 1 − x
1+r 1 − x
Z
1 − z2
{z
}
3
= 0
R∞
when we take the limit as r → 0 and ρ → ∞, integral (1) gives −∞ ();
integral (2) is −iπ; and integral (3) vanishes by the Jordan’s lemma or the
technique we used in the previous problem. Thus,
I
Z ∞
zeiπz dz
xeiπx dx
=
− iπ = 0
2
1 − z2
−∞ 1 − x
Z
∞
⇒
−∞
xeiπx dx
= iπ
1 − x2
by equating real and imaginary parts of the two sides, we get
Z ∞
x sin πx
dx = π
2
−∞ 1 − x
3
c) The poles are z = −1 and z = 2, both poles are with in |z| = 3. Assuming
simple pole, we get
I
cos(z − 1)dz
= 2πi · {R(−1) + R(2)}
C (z + 1)(z − 2)
cos 2 cos 1
= 2πi · {−
+
}
3
3
2πi
=
{cos 1 − cos 2}
3
d) The poles are z = 1 and z = −1. Both poles are with in |z| = 2.
Assuming poles of second order, we get
I
I
dz
dz
=
= 2πi · {R(−1) + R(1)}
z (z 2 − 1)2
z (z − 1)2 (z + 1)2
e
e
C
C
1
= 2πi · 0 −
2e
iπ
= −
e
Problem 5.2 (First order differential equations)
a)
2
dy + (2xy − xe−x )dx = 0
⇒
Q(x) = xe−x
P (x) = 2x ;
eI = e
R
0
y + 2xy = xe−x
2xdx
=
y
=
=
=
2
2
ex Z
e−I
eI Q(x)dx + C
Z
−x2
x2
−x2
e
e xe dx + C
2
x
−x2
e
+C
2
b)
0
y + y cos x
=
P (x) = cos x;
R
eI = e
cos xdx
sin 2x
Q(x) = sin 2x
=
4
2
esin x
the solution for y is computed as
Z
−I
I
y = e
e Q(x)dx + C
Z
−sinx
sin x
= e
e
sin 2xdx + C
Z
−sinx
sin x
= e
e
2 sin x cos xdx + C
making substitution u = sin x and du = cos xdx,
Z
−sinx
u
=e
2 e udu + C
integration by part gives
1). Thus
R
R
eu udu = ueu − eu du = eu (u − 1) = esin x (sin x −
y = e− sin x 2esin x (sin x − 1) + C
= 2 sin x − 2 + Ce− sin x
= 2 sin x + Ce− sin x − 2
c)
0
y cos x + y = cos2 x
1
;
cos
x
R dx
eI = e cos x
⇒
P (x)
0
y +
1
y = cos x
cos x
Q(x) = cos x
=
R
e
sec xdx
here
e
R
sec xdx
= e
= e
R
sec x·1dx
R
sec x· (sec x+tan x) dx
(sec x+tan x)
= eln | sec x+tan x|
= sec x + tan x =
1 + sin x
cos x
R 0 (x)
based on the property that ff (x)
dx = ln |f (x)|. The solution to y is thus,
Z
−I
I
y = e
e Q(x)dx + C
Z
cos x
1 + sin x
=
cos xdx + C
1 − sin x
cos x
cos x
=
{x − cos x + C}
1 − sin x
5
Extra problems: (Second order homogeneous DE with constant
coefficients)
00
0
a) The auxiliary equation for 4y + 12y + 9y = 0 is 4r2 + 12r + 9 = 0,
and its roots would be r = − 32 . Because the two roots are identical the
general solution is given by
3
y = (A + Bxe− 2 x )
Where A and B are arbitrary constants.
00
0
b) The auxiliary equation for y − 4y + 13y = 0 is r2 − 4r + 13 = 0, and its
roots would be r = 2 ± i3. The general solution is thus,
y = Ae(2+i3)x + Be(2−i3)x = e2x Aei3x + Be−i3x
This can also be written as,
y = e2x {C1 cos 3x + C2 sin 3x}
where C1 and C2 are new constants but can be related to constants A and
B. We can also write it as
y = Ce2x {sin(3x + α)}
again C and α are new constants, and can be related to the C1 and C2 .
6