§ 3. Some Applications of Cauchy’s Method
73
As in (ii), the more subtle case when n = 1 can be handled; as we will see,
the integral defining ˆ
f (t) is then semi-convergent for all t, which we already
know to be the case for t = 0.
Note first that
f (z) = c/z + O
1/z
2
at infinity
as already remarked above. So only the term en 1/z poses a problem in the
calculations and estimations which were successfully done for n ≥ 2.
On the other hand, if the function f is real on R, which may be assumed,
its derivative only has finitely many real roots and hence its sign remains
constant for large |x|; f (x), therefore, monotonously tends to 0 as x ∈ R
tends to +∞ or −∞, so if the Fourier integral is defined by the formula
ˆ
f (t) = lim
R
−R
f (x)e(−tx)dx ,
(8.17)
it remains well-defined for t = 0 (Chapter V, n
◦ 24, Theorem 23, integrals
of “ oscillating ” functions), hence also for complex f by separating real and
imaginary parts. In fact (17) is also well-defined for t = 0, but for different
reasons as shown above in (ii).
Having said this, suppose first that t > 0 and integrate along the path
already used for absolutely converging integrals. Setting g(z) = f (z)e(−tz),
the contribution from the lower half-circle can be obtained by integrating the
function g(Re
iu )i Re
iu from u = 0 to u = −π. As |f (z)| ≤ M/|z|, where M
is a constant,
g
Re
iu
i Re
iu
= R
f
Re
iu
.
e
−Rte
iu
≤ M. exp (2πRt sin u) .
Therefore, the contribution from the lower half-circle is, in modulus, bounded
above by
M
0
−π
exp(2πRt sin u)du = M
π
0
exp (−ρ sin u) du ,
where ρ = 2πRt tends to +∞ since t is supposed to be strictly positive. To
show that this integral tends to 0, first note that the integrated function is
everywhere ≤ 1 and that it tends to 0 except for u = 0 or π since −ρ sin u
tends to −∞; the (real) Lebesgue theorem of dominated convergence takes
care of the question since the measure on the set reduces to the points 0
and π. If the event of a refusal to use it, note first that, for given t > 0 and
δ > 0, the continuous function being integrated decreasingly converges to a
limit function continuous on [δ, π − δ], namely 0; convergence is, therefore,
uniform in such an interval (Chapter V, § 2, n
◦ 10, Dini’s theorem), so that
its contribution to the integral tends to 0, and so is ≤ δ for large ρ; as those
of the two forgotten intervals are ≤ δ for all ρ, the total is ≤ 3δ for large ρ,
73
As in (ii), the more subtle case when n = 1 can be handled; as we will see,
the integral defining ˆ
f (t) is then semi-convergent for all t, which we already
know to be the case for t = 0.
Note first that
f (z) = c/z + O
1/z
2
at infinity
as already remarked above. So only the term en 1/z poses a problem in the
calculations and estimations which were successfully done for n ≥ 2.
On the other hand, if the function f is real on R, which may be assumed,
its derivative only has finitely many real roots and hence its sign remains
constant for large |x|; f (x), therefore, monotonously tends to 0 as x ∈ R
tends to +∞ or −∞, so if the Fourier integral is defined by the formula
ˆ
f (t) = lim
R
−R
f (x)e(−tx)dx ,
(8.17)
it remains well-defined for t = 0 (Chapter V, n
◦ 24, Theorem 23, integrals
of “ oscillating ” functions), hence also for complex f by separating real and
imaginary parts. In fact (17) is also well-defined for t = 0, but for different
reasons as shown above in (ii).
Having said this, suppose first that t > 0 and integrate along the path
already used for absolutely converging integrals. Setting g(z) = f (z)e(−tz),
the contribution from the lower half-circle can be obtained by integrating the
function g(Re
iu )i Re
iu from u = 0 to u = −π. As |f (z)| ≤ M/|z|, where M
is a constant,
g
Re
iu
i Re
iu
= R
f
Re
iu
.
e
−Rte
iu
≤ M. exp (2πRt sin u) .
Therefore, the contribution from the lower half-circle is, in modulus, bounded
above by
M
0
−π
exp(2πRt sin u)du = M
π
0
exp (−ρ sin u) du ,
where ρ = 2πRt tends to +∞ since t is supposed to be strictly positive. To
show that this integral tends to 0, first note that the integrated function is
everywhere ≤ 1 and that it tends to 0 except for u = 0 or π since −ρ sin u
tends to −∞; the (real) Lebesgue theorem of dominated convergence takes
care of the question since the measure on the set reduces to the points 0
and π. If the event of a refusal to use it, note first that, for given t > 0 and
δ > 0, the continuous function being integrated decreasingly converges to a
limit function continuous on [δ, π − δ], namely 0; convergence is, therefore,
uniform in such an interval (Chapter V, § 2, n
◦ 10, Dini’s theorem), so that
its contribution to the integral tends to 0, and so is ≤ δ for large ρ; as those
of the two forgotten intervals are ≤ δ for all ρ, the total is ≤ 3δ for large ρ,