108
VIII – Cauchy Theory
Γ f (s) = π/2 cos(πs/2) ,
which is much more precise, but we rarely have the chance of being able to
calculate everything explicitly.
If f decreases rapidly at infinity, i.e. if f (x) = O(x
−N ) for all N , there is
no need to think: integral (5”) converges on all of C and is an entire function
of s.
As in the previous example, these two types of results apply if f admits
unbounded asymptotic expansions in the neighbourhood of 0 or for large x,
obviously provided that the integral defining Γ f is to start with convergent
on a strip of non-zero width. Otherwise “ gluing ” the meromorphic functions
defined by (5’) and (5”) would be impossible: the function 1 has the nicest
asymptotic behaviour in the world, but its Mellin transform is not defined;
in this case, Γ
−
f (s) = 1/s and Γ
+
f (s) = −1/s, and so Γ f (s) = 0 if Γ f were
well-defined : absurd assumptions lead to absurd conclusions.
The fact that the Mellin transform does not have simple poles is due to
the nature of the asymptotic expansions that we have admitted. In more
complicated cases, there may be multiple poles. Suppose for example that,
in the neighbourhood of 0, f is the sum of a function of type (6’) and of a
finite number of terms x
p log
q x, with p > 0. The contribution made by these
terms to (5’) converges for Re(s) > −p since log x = O(x
−r ) for all r > 0; it
is equal to c q (s + p), where
c q (s) =
1
0
log
q x.x
s−1 dx , Re(s) > 0 .
(13.9)
Integrating by parts, we get
sc q (s) = [1 − qc q−1 (s)] .
As c 0 (s) = 1/s, c 1 (s) = 1/s − 1/s
2 , it follows that c 2 (s) = 1/s − 2/s
2 + 2s
3
and more generally
c q (s) = 1/s − q/s
2 + q (q − 1) /s
3
− . . . + (−1)
q q!/s
q+1 .
(13.10)
Replacing s by s + p, we see that the presence of a term in x
p log
q x in the
asymptotic expansion introduces a pole of order q + 1 at the point −p.
(iii) Example : the Riemann zeta function. The function exp(−πu
2 ) being
equal to its Fourier transform, that of u → exp(−πxu
2 ),where x > 0, is
v → x
−1/2 exp(−πv
2 /x); at infinity, these functions approach 0 sufficiently
rapidly for Poisson’s formula to be written
exp(−πn
2 x) = x
−1/2
exp(−πn
2 /x) ,
where summation is over Z (Chap. VII, n
◦ 28). As will be seen below, the
results of section (ii) apply to the Jacobi function
VIII – Cauchy Theory
Γ f (s) = π/2 cos(πs/2) ,
which is much more precise, but we rarely have the chance of being able to
calculate everything explicitly.
If f decreases rapidly at infinity, i.e. if f (x) = O(x
−N ) for all N , there is
no need to think: integral (5”) converges on all of C and is an entire function
of s.
As in the previous example, these two types of results apply if f admits
unbounded asymptotic expansions in the neighbourhood of 0 or for large x,
obviously provided that the integral defining Γ f is to start with convergent
on a strip of non-zero width. Otherwise “ gluing ” the meromorphic functions
defined by (5’) and (5”) would be impossible: the function 1 has the nicest
asymptotic behaviour in the world, but its Mellin transform is not defined;
in this case, Γ
−
f (s) = 1/s and Γ
+
f (s) = −1/s, and so Γ f (s) = 0 if Γ f were
well-defined : absurd assumptions lead to absurd conclusions.
The fact that the Mellin transform does not have simple poles is due to
the nature of the asymptotic expansions that we have admitted. In more
complicated cases, there may be multiple poles. Suppose for example that,
in the neighbourhood of 0, f is the sum of a function of type (6’) and of a
finite number of terms x
p log
q x, with p > 0. The contribution made by these
terms to (5’) converges for Re(s) > −p since log x = O(x
−r ) for all r > 0; it
is equal to c q (s + p), where
c q (s) =
1
0
log
q x.x
s−1 dx , Re(s) > 0 .
(13.9)
Integrating by parts, we get
sc q (s) = [1 − qc q−1 (s)] .
As c 0 (s) = 1/s, c 1 (s) = 1/s − 1/s
2 , it follows that c 2 (s) = 1/s − 2/s
2 + 2s
3
and more generally
c q (s) = 1/s − q/s
2 + q (q − 1) /s
3
− . . . + (−1)
q q!/s
q+1 .
(13.10)
Replacing s by s + p, we see that the presence of a term in x
p log
q x in the
asymptotic expansion introduces a pole of order q + 1 at the point −p.
(iii) Example : the Riemann zeta function. The function exp(−πu
2 ) being
equal to its Fourier transform, that of u → exp(−πxu
2 ),where x > 0, is
v → x
−1/2 exp(−πv
2 /x); at infinity, these functions approach 0 sufficiently
rapidly for Poisson’s formula to be written
exp(−πn
2 x) = x
−1/2
exp(−πn
2 /x) ,
where summation is over Z (Chap. VII, n
◦ 28). As will be seen below, the
results of section (ii) apply to the Jacobi function