§ 3. Some Applications of Cauchy’s Method
129
Let r and R be the radii of the two circles and ±δ be the ordinates of the
two segments of the horizontals that μ consists of. The segment with ordinate
+δ contributes
R
r
(x + iδ)
s dx/
1 + (x + iδ)
2
(15.6)
to the integral, where the limits r
< r and R
< R tend to r and R as δ
tends to 0. Since the argument of x + iδ approaches 0 as δ tends to 0, for all
real x > 0, the function
F (x, δ) = (x + iδ)
s /
1 + (x + iδ)
2
tends to F (x, 0+) = x
s (1 + x
2 )
−1 , where x
s = e
s log x takes its usual value
for real x > 0 (Chap. IV). The limit of integral (6) can therefore be assumed
to be the extended integral over all of the interval [r, R] of F (x, 0+), but this
requires justification, which will be provided by the theorem of dominated
convergence.
First, it is clear that r
> r/2 for δ sufficiently small. Integral (6) is, therefore, the one over the fixed interval (r/2, R) of the function equal to F (x, δ)
between r
and R
and 0 elsewhere. This new function tends to F (x, 0+) in
]r, R[ and to 0 elsewhere in the interval (r/2, R). On the other hand, the formula defining F (x, δ) continues to be well-defined for x > 0 and δ = 0 since
the result is obviously a continuous function on the product set R
∗
+ × R + . It
follows that f is bounded on the compact set {x ∈ [r/2, R] & 0 ≤ δ ≤ 1}. As
δ tends to 0, the modified function F (x, δ) integrated over [r/2, R] tends to
the function equal to F (x, 0+) on ]r, R[ and zero elsewhere, while remaining
dominated by a fixed constant. As integration is over a compact set, passing
to the limit is justified and finally
lim
R
r
F (x, δ)dx =
R
r
F (x, 0+)dx =
R
r
x
s dx/(1 + x
2 )
(15.7)
as expected.
The integral along the segment with ordinate −δ can be dealt with in
a similar way; it is necessary to change the direction followed and to take
into account that the argument of x − iδ tends to 2π, which introduces a
factor e
2πis = e(s) in the calculation. The limit value is, therefore, integral
(7) multiplied by −e(s).
Hence, for given r and R and δ tending to 0, the total contribution from
the segments of horizontals is equal to
(1 − e(s))
R
r
x
s dx/
1 + x
2
.
(15.8)
By (2), this expression tends to (1 − e(s))ϕ(s) as r and R tend to 0 and +∞.
129
Let r and R be the radii of the two circles and ±δ be the ordinates of the
two segments of the horizontals that μ consists of. The segment with ordinate
+δ contributes
R
r
(x + iδ)
s dx/
1 + (x + iδ)
2
(15.6)
to the integral, where the limits r
< r and R
< R tend to r and R as δ
tends to 0. Since the argument of x + iδ approaches 0 as δ tends to 0, for all
real x > 0, the function
F (x, δ) = (x + iδ)
s /
1 + (x + iδ)
2
tends to F (x, 0+) = x
s (1 + x
2 )
−1 , where x
s = e
s log x takes its usual value
for real x > 0 (Chap. IV). The limit of integral (6) can therefore be assumed
to be the extended integral over all of the interval [r, R] of F (x, 0+), but this
requires justification, which will be provided by the theorem of dominated
convergence.
First, it is clear that r
> r/2 for δ sufficiently small. Integral (6) is, therefore, the one over the fixed interval (r/2, R) of the function equal to F (x, δ)
between r
and R
and 0 elsewhere. This new function tends to F (x, 0+) in
]r, R[ and to 0 elsewhere in the interval (r/2, R). On the other hand, the formula defining F (x, δ) continues to be well-defined for x > 0 and δ = 0 since
the result is obviously a continuous function on the product set R
∗
+ × R + . It
follows that f is bounded on the compact set {x ∈ [r/2, R] & 0 ≤ δ ≤ 1}. As
δ tends to 0, the modified function F (x, δ) integrated over [r/2, R] tends to
the function equal to F (x, 0+) on ]r, R[ and zero elsewhere, while remaining
dominated by a fixed constant. As integration is over a compact set, passing
to the limit is justified and finally
lim
R
r
F (x, δ)dx =
R
r
F (x, 0+)dx =
R
r
x
s dx/(1 + x
2 )
(15.7)
as expected.
The integral along the segment with ordinate −δ can be dealt with in
a similar way; it is necessary to change the direction followed and to take
into account that the argument of x − iδ tends to 2π, which introduces a
factor e
2πis = e(s) in the calculation. The limit value is, therefore, integral
(7) multiplied by −e(s).
Hence, for given r and R and δ tending to 0, the total contribution from
the segments of horizontals is equal to
(1 − e(s))
R
r
x
s dx/
1 + x
2
.
(15.8)
By (2), this expression tends to (1 − e(s))ϕ(s) as r and R tend to 0 and +∞.