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Past exam of the mathematics course of the University of Cambridge
/
2018
/
ib
/
Paper 1
/
13A
/
a
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Mathematics course of the University of Cambridge
Past exam of the mathematics course of the University of Cambridge
2018
ib
Paper 1
13A
Created
2026-09-24
Updated
2026-09-24
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Solution
a
Solution
0
0
0
a
The integrand
F
(
z
)
=
e
z
/2
−
e
−
z
/2
e
i
z
2
/
(
4
π
)
(1)
has one
pole
inside the
rectangle
, at
z
=
0
, with
residue
1
. The
residue theorem
therefore gives
∫
C
F
(
z
)
d
z
=
2
πi
. The vertical-side
integrals
tend to zero
as
R
→
∞
.
On the horizontal sides, direct substitution
z
=
x
±
πi
, with the upper side oppositely oriented, shows that their
sum
is
i
e
−
iπ
/4
∫
−
R
R
e
i
x
2
/
(
4
π
)
(
1
+
e
x
1
+
1
+
e
x
e
x
)
d
x
.
(2)
Taking the limit and comparing with
2
πi
gives
R
→
∞
lim
∫
−
R
R
e
i
x
2
/
(
4
π
)
d
x
=
2
π
e
iπ
/4
.
(3)
Ancestors
(10)
13A
Paper 1
ib
2018
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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