Transcribed Text
3. (i) Prove that for
8 eAr II
n
I(A) = 1+12
dt = on(A) + Rn(A),
k0
where
n(A) = 2xn + 1 ,
72 = 1,2,
Write down an expression for the remainder term Rn(A) in the form of an integral.
[9 marks]
(ii) Show that On (A) is an asymptotic sequence as 1
00.
[2 marks]
(iii) Prove that Rn (A) = O(On+1 (A)) for A > 0.
[4 marks]
(iv) Prove that Rn(A) = o(on(A)) as &
80. Hence deduce an asymptotic expansion for I(A) as
X
00.
[3 marks]
do
(v) For what values of 1 does the series
E
Pt converge?
k1
4. (i) Show that the Mellin transform M[h(x); 8] defined by
M/h (x): 8]= = ... x° 
of the function
= + 1 I
with X > 0, is given by
= ,
0 < Re(s) < 1.
1
7T
[You may use without proof the result that the Mellin transform of
is
for 0 < R(s) < 1.
1 + I sin TS
]
(ii) Write down the Mellin transform of eI. Hence using the result that
[2 marks]
Fioo

where C lies in the common strip of analyticity of M[h; 8] and M[f; 1  s] , show that
0 A+r e= dx = 1
ds
(2)
where L is a vertical line and
G(s) = 
sin TTS
Explain clearly where the line L needs to be located.
[5 marks]
(iii)
Use the result (2), stating clearly any assumptions that you make, to show that
eI
dx~Apln(A) + A1 + O(A, Aln À) as X
0+
0 X + I
where the constant terms Ao.A, need to be calculated explicitly. A formal proof of the O(A, AIn A)
term is not required.
[
You may also use without proof the results
 12 + as Z
0,
where 72 = 0,1,2, and On, bn are nonzero. ]
5. Consider
I = 1 / (z) dz,
2i CN zu 
1
where f(z) = cosec( 2)   and I is not a pole of f (z) and you may assume that I does not lie on
CN. The contour CN is formed by the square (taken counter clockwise) with vertices at the locations
ze  I(N + 1 i (N + 1/2 ) 7T in the complex plane and N is a positive integer.
(i) Show that
N
1
I = f(x) + E (1)*
k:  I
k=N
kyo
[4 marks]
(ii) Show also that
N
(1)  k
I
I 
+
f ( z) dz.
k:
k=N
2mi Cx =( 2  x)
kyo
[4 marks]
(iii) Prove that f(x) is bounded on CN.
[5 marks]
(iv) Using the result from (iii) show that
/
(z)
da = O
Cx
20

I)
(
N. 1
as N
80.
[5 marks]
(v) Deduce that
80
cosec(x)  1/1  + (1)*
1
1
+
T  k:
)
in
k:
k=do
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