Դիցուք
f
(
x
)
{\displaystyle f(x)}
[
−
π
;
π
]
{\displaystyle [-\pi \;;\pi \;]}
f
(
π
)
=
f
(
−
π
)
{\displaystyle f(\pi )=f(-\pi )}
f
(
x
)
{\displaystyle f(x)}
lim
n
→
∞
1
π
∫
0
δ
sin
(
2
n
+
1
)
u
2
⋅
f
(
x
0
+
u
)
+
f
(
x
0
−
u
)
sin
u
2
d
u
{\displaystyle \lim \limits _{n\to \infty }{1 \over \pi }\int \limits _{0}^{\delta }{\sin(2n+1){u \over 2}}\cdot {{f(x_{0}+u)+f(x_{0}-u)} \over {\sin {u \over 2}}}du}
Եթե այն գոյություն ունի, ապա Ֆուրյեի շարքի գումարը հենց դա է, ընդ որում`
x
∈
(
x
0
−
δ
,
x
0
+
δ
)
;
∀
δ
>
0
{\displaystyle x\in \left({x_{0}-\delta ,x_{0}+\delta }\right);\forall \delta >0}
Մնացած կետերում արժեքները դեր չեն խաղում:
Ապացույց.
f
(
x
)
{\displaystyle f(x)}
u
=
t
−
x
,
{\displaystyle u=t-x,}
S
n
(
x
0
,
f
)
=
1
2
π
∫
−
π
π
f
(
t
)
D
n
(
t
−
x
0
)
d
t
=
1
2
π
∫
−
π
−
x
0
π
−
x
0
f
(
x
0
+
u
)
D
n
(
u
)
d
u
=
1
2
π
∫
−
π
π
f
(
x
0
+
u
)
D
n
(
u
)
d
u
:
{\displaystyle S_{n}(x_{0},f)={1 \over {2\pi }}\int \limits _{-\pi }^{\pi }{f(t)D_{n}(t-x_{0})dt=}{1 \over {2\pi }}\int \limits _{-\pi -x_{0}}^{\pi -x_{0}}{f(x_{0}+u)D_{n}(u)du=}{1 \over {2\pi }}\int \limits _{-\pi }^{\pi }{f(x_{0}+u)D_{n}(u)du:}}
Եթե տրված է
T
{\displaystyle T}
F
(
t
)
{\displaystyle F(t)}
∫
a
a
+
T
F
(
t
)
d
t
=
∫
b
b
+
T
F
(
t
)
d
t
{\displaystyle \int \limits _{a}^{a+T}{F(t)dt=}\int \limits _{b}^{b+T}{F(t)dt}}
Ապացույց.
նշանակենք
u
=
t
−
T
{\displaystyle u=t-T}
∫
a
a
+
T
F
(
t
)
d
t
=
∫
a
b
F
(
t
)
d
t
+
∫
b
b
+
T
F
(
t
)
d
t
+
∫
b
+
T
a
+
T
F
(
t
)
d
t
=
∫
a
b
F
(
t
)
d
t
+
∫
b
b
+
T
F
(
t
)
d
t
+
∫
b
a
F
(
u
+
T
)
d
t
=
∫
b
b
+
T
F
(
t
)
d
t
{\displaystyle \int \limits _{a}^{a+T}{F(t)dt=}\int \limits _{a}^{b}{F(t)dt+}\int \limits _{b}^{b+T}{F(t)dt+}\int \limits _{b+T}^{a+T}{F(t)dt=}\int \limits _{a}^{b}{F(t)dt+}\int \limits _{b}^{b+T}{F(t)dt+}\int \limits _{b}^{a}{F(u+T)dt=}\int \limits _{b}^{b+T}{F(t)dt}}
Հավասարության աջ մասը ներկայացնելով երկու ինտեգրալների գումարի տեսքով, առաջին ինտեգրալի մեջ կատարելով
y
=
−
u
{\displaystyle y=-u}
D
n
(
−
u
)
=
D
n
(
−
u
)
{\displaystyle D_{n}(-u)=D_{n}(-u)}
S
n
(
x
0
,
f
)
=
1
2
π
∫
−
π
0
f
(
x
0
+
u
)
D
n
(
u
)
d
u
+
1
2
π
∫
0
π
f
(
x
0
+
u
)
D
n
(
u
)
d
u
=
{\displaystyle S_{n}(x_{0},f)={1 \over {2\pi }}\int \limits _{-\pi }^{0}{f(x_{0}+u)D_{n}(u)du+}{1 \over {2\pi }}\int \limits _{0}^{\pi }{f(x_{0}+u)D_{n}(u)du}=}
=
1
2
π
∫
0
π
f
(
x
0
−
y
)
D
n
(
y
)
d
y
+
1
2
π
∫
0
π
f
(
x
0
+
u
)
D
n
(
u
)
d
u
=
1
π
∫
0
π
D
n
(
u
)
f
(
x
0
+
u
)
+
f
(
x
0
−
u
)
2
d
u
{\displaystyle ={1 \over {2\pi }}\int \limits _{0}^{\pi }{f(x_{0}-y)D_{n}(y)dy+}{1 \over {2\pi }}\int \limits _{0}^{\pi }{f(x_{0}+u)D_{n}(u)du}={1 \over \pi }\int \limits _{0}^{\pi }{D_{n}(u){{f(x_{0}+u)+f(x_{0}-u)} \over 2}du}}
Ստացված արտահայտության մեջ տեղադրելով Դիրիխլեի կորիզի արժեքը, կստանանք`
S
n
(
x
0
,
f
)
=
1
π
∫
0
π
sin
(
2
n
+
1
)
u
2
⋅
[
f
(
x
0
+
u
)
+
f
(
x
0
−
u
)
]
2
sin
u
2
d
u
{\displaystyle S_{n}(x_{0},f)={1 \over \pi }\int \limits _{0}^{\pi }{\sin(2n+1){u \over 2}\cdot {{\left[{f(x_{0}+u)+f(x_{0}-u)}\right]} \over {2\sin {u \over 2}}}du}}
Այժմ ֆիքսենք որևէ
δ
,
0
<
δ
<
π
{\displaystyle \delta ,0<\delta <\pi }
S
n
(
x
0
,
f
)
=
1
π
∫
0
δ
sin
(
2
n
+
1
)
u
2
⋅
[
f
(
x
0
+
u
)
+
f
(
x
0
−
u
)
]
2
sin
u
2
d
u
+
1
π
∫
δ
π
sin
(
2
n
+
1
)
u
2
⋅
[
f
(
x
0
+
u
)
+
f
(
x
0
−
u
)
]
2
sin
u
2
d
u
{\displaystyle S_{n}(x_{0},f)={1 \over \pi }\int \limits _{0}^{\delta }{\sin(2n+1){u \over 2}\cdot {{\left[{f(x_{0}+u)+f(x_{0}-u)}\right]} \over {2\sin {u \over 2}}}du}+{1 \over \pi }\int \limits _{\delta }^{\pi }{\sin(2n+1){u \over 2}\cdot {{\left[{f(x_{0}+u)+f(x_{0}-u)}\right]} \over {2\sin {u \over 2}}}du}}
Քանի որ
1
2
sin
u
2
{\displaystyle {1 \over {2\sin {u \over 2}}}}
[
δ
,
π
]
{\displaystyle \left[{\delta ,\pi }\right]}
իսկ
f
(
x
0
+
u
)
+
f
(
x
0
−
u
)
{\displaystyle f(x_{0}+u)+f(x_{0}-u)}
x
∈
[
−
π
;
π
]
{\displaystyle x\in \left[{-\pi ;\pi }\right]}
2
π
{\displaystyle 2\pi }
u
{\displaystyle u}
x
∈
[
−
π
;
π
]
{\displaystyle x\in \left[{-\pi ;\pi }\right]}
f
(
x
0
+
u
)
+
f
(
x
0
−
u
)
2
sin
u
2
{\displaystyle {{f(x_{0}+u)+f(x_{0}-u)} \over {2\sin {u \over 2}}}}
[
δ
,
π
]
{\displaystyle \left[{\delta ,\pi }\right]}
lim
n
→
∞
∫
δ
π
sin
(
2
n
+
1
)
u
2
⋅
[
f
(
x
0
+
u
)
+
f
(
x
0
−
u
)
]
2
sin
u
2
d
u
=
0
{\displaystyle \lim \limits _{n\to \infty }\int \limits _{\delta }^{\pi }{\sin(2n+1){u \over 2}\cdot {{\left[{f(x_{0}+u)+f(x_{0}-u)}\right]} \over {2\sin {u \over 2}}}du}=0}
Թեորեմը ապացուցված է՛: -357-
![{\displaystyle [-\pi \;;\pi \;]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc409c7da7171825e301c33cbf345ea4c61ca42b)
![{\displaystyle S_{n}(x_{0},f)={1 \over \pi }\int \limits _{0}^{\pi }{\sin(2n+1){u \over 2}\cdot {{\left[{f(x_{0}+u)+f(x_{0}-u)}\right]} \over {2\sin {u \over 2}}}du}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b0174bd004d748cea6fd0bd781366bcaf415873)
![{\displaystyle S_{n}(x_{0},f)={1 \over \pi }\int \limits _{0}^{\delta }{\sin(2n+1){u \over 2}\cdot {{\left[{f(x_{0}+u)+f(x_{0}-u)}\right]} \over {2\sin {u \over 2}}}du}+{1 \over \pi }\int \limits _{\delta }^{\pi }{\sin(2n+1){u \over 2}\cdot {{\left[{f(x_{0}+u)+f(x_{0}-u)}\right]} \over {2\sin {u \over 2}}}du}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48079a1d5daa92e5590537d4b9b48726c474fd0e)
![{\displaystyle \left[{\delta ,\pi }\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f5d87b4c91639d829d98229c69ef2c231b08286)
![{\displaystyle x\in \left[{-\pi ;\pi }\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32d26fb14f71309a583f8e22474fb5d20ee36e48)
![{\displaystyle \lim \limits _{n\to \infty }\int \limits _{\delta }^{\pi }{\sin(2n+1){u \over 2}\cdot {{\left[{f(x_{0}+u)+f(x_{0}-u)}\right]} \over {2\sin {u \over 2}}}du}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d328990bea52d9c5e5852c57913b708e521c5e6a)