78
VIII – Cauchy Theory
follows. We will use this formula in n
◦ 15.
Exercise. Deduce from (2) the formulas
π
sin az
sin πz
= 2
n≥1
(−1)
n n. sin na
z 2 − n 2 (−π < a < π) ,
π
cosh az
sinh πz
= 1/z + 2z
n≥1
(−1)
n cos na
z 2 + n 2 , (−π < a < π) .
Show that the convergence of the first amounts to that of the series
(−1)
n sin
Can these formulas be obtained from the theory of Fourier series?
10 – The Gamma Function, The Fourier Transform of e
−x
x
s−1
+
and the Hankel Integral
Formula (8.15) which can also be written
(y − a)
−n e(xy)dy = (2πi)
n x
n−1
+ e(ax)/(n − 1)! ;
was proved above. It assumes that n is an integer ≥ 2, x ∈ R and that
Im(a) > 0. We intend to show that more generally
(y − a)
−s e(xy)dy = (2πi)
s x
s−1
+ e(ax)/Γ (s)
(10.1)
for x ∈ R, Re(s) > 1 and Im(a) > 0, where
Γ (s) =
+∞
0
e
−t t
s d
∗ t, Re(s) > 0
is Euler’s function (Chapter V, § 7, n
◦ 22) and where d
∗ t = dt/|t|. Since, in
(1), the function (y − a)
−s and that of the right hand side are continuous and
integrable for Re(s) > 1, this amounts (inversion formula) to showing that
(y − a)
−s is the Fourier transform of the function
ϕ(x) = e(ax)x
s−1
+
= exp(2πiax)x
s−1
+ ,
(10.2)
where Im(a) = c > 0. As | exp(2πiax)| = exp(−2πcx), Re(s) > 0 is sufficient
for ϕ ∈ L
1 (R) to hold. Then, setting w = 2πi(y − a),
ˆ
ϕ(y) =
x
s−1
+ e(ax − xy)dx =
+∞
0
exp(−wx)x
s d
∗ x
(10.3)
and so Re(w) = 2π Im(a) > 0. If w were real > 0, the following formula could
be obtained by a change of variable x → x/w:
(na)/n.