Functions are from R to N or
R
×
R
R \times R
to N, where R denotes the real numbers and N denotes a normed complete ring. If S, T and G are functions from
R
×
R
R \times R
to N, each of
S
(
p
−
,
p
)
,
S
(
p
−
,
p
−
)
,
T
(
p
−
,
p
)
S({p^ - },p),S({p^ - },{p^ - }),T({p^ - },p)
and
T
(
p
−
,
p
−
)
T({p^ - },{p^ - })
exists for
a
>
p
⩽
b
a > p \leqslant b
, each of
S
(
p
,
p
+
)
,
S
(
p
+
,
p
+
)
,
T
(
p
,
p
+
)
S(p,{p^ + }),S({p^ + },{p^ + }),T(p,{p^ + })
and
T
(
p
+
,
p
+
)
T({p^ + },{p^ + })
exists for
a
⩽
p
>
b
a \leqslant p > b
, G has bounded variation on [a, b] and
∫
a
b
G
\smallint _a^bG
exists, then each of
\[
∫
a
b
S
[
G
−
∫
G
]
T
and
∫
a
b
S
[
1
+
G
−
∏
(
1
+
G
)
]
T
\int _a^b S \left [ {G - \int G } \right ]T\quad {\text {and}}\quad \int _a^b {S\left [ {1 + G - \prod {(1 + G)} } \right ]} \;T
\]
exists and is zero. These results can be used to solve integral equations without the existence of integrals of the form
\[
∫
a
b
|
G
−
∫
G
|
=
0
and
∫
a
b
|
1
+
G
−
∏
(
1
+
G
)
|
=
0.
\int _a^b {\left | {G - \int G } \right | = 0} \quad {\text {and}}\quad \int _a^b {\left | {1 + G - \prod {(1 + G)} } \right |} = 0.
\]
This is demonstrated by solving the linear integral equation
\[
f
(
x
)
=
h
(
x
)
+
(
L
R
)
∫
a
x
(
f
G
+
f
H
)
f(x) = h(x) + (LR)\int _a^x {(fG + fH)}
\]
and the Riccati integral equations
\[
f
(
x
)
=
w
(
x
)
+
(
L
R
L
R
)
∫
a
x
(
f
H
+
G
f
+
f
K
f
)
f(x) = w(x) + (LRLR)\int _a^x {(fH + Gf + fKf)}
\]
without the existence of the previously mentioned integrals.