§ 3. Some Applications of Cauchy’s Method
125
|s| =
σ
2 + t
2
1/2 = |t|
1 + O
t
−2
,
log |s| = log |t| + log
1 + O
t
−2
= log |t| + O
t
−2
.
So finally,
Re [(s − 1/2) Log s] = −π|t|/2 + (σ − 1/2) log |t| + σ + O
t
−2
,
and so
s
s−1/2
∼ e
−π|t|/2
|t|
σ−1/2 e
σ
at infinity in B
since the factor exp
O(|t|
−2 )
tends to exp(0) = 1. Returning to (29), we
finally get the evaluation sought (Remmert tells us it is due to the Italian
Salvatore Pincherle, 1889), namely
|Γ (σ + it)| ∼ (2π)
1/2
|t|
σ−1/2 e
−π|t|/2 .
(14.27)
It is reassuring to see that this result is compatible with formula (10.5.8)
|Γ (1/2 + it)|
2 = π/ cosh πt .
15 – The Fourier Transform of 1/ cosh πx
Like Γ (1/2 + it), the function 1/ cosh πx is in the space S(R). This can be
directly verified since its n-th derivative is obtained by dividing by cosh
2n πx
a polynomial in sinh πx and cosh πx all of whose monomials are of total
degree < 2
n ; now, cosh πx ∼
1
2 exp(π|x|) at infinity.
We show that it is identical to its Fourier transform. This result will later
give rise to the strange identity (17) that can be found at the end of this n
◦ .
The Fourier transform of 1/ cosh πx is the integral
2
exp(−2πiyt)
e πt + e −πt dt =
2
π
+∞
0
x
2iy
1 + x 2 dx
(15.1)
as shown by the change of variable e
−πt = x. More generally, the integral
ϕ(s) =
+∞
0
x
s
1 + x 2 dx , | Re(s)| < 1 ,
(15.2)
therefore, remains to be computed. It is the Mellin transform of
f (x) = x/(1 + x
2 ) = f (1/x) .
If
ϕ(s) = π/2 cos(πs/2)
is shown to hold, the Fourier transform sought will then be equal to