Ištrintas turinys Pridėtas turinys
46 eilutė:
46 eilutė:
:Vadinasi,
:Vadinasi,
:<math>G(0)=\int_0^\infty e^{-0\cdot t} \frac{\sin t}{t}\; \mathbf{d}t=\int_0^\infty e^0 \frac{\sin t}{t}\; \mathbf{d}t=\int_0^\infty 1\cdot \frac{\sin t}{t}\; \mathbf{d}t=\int_0^\infty \frac{\sin t}{t}\; \mathbf{d}t.</math>
:<math>G(0)=\int_0^\infty e^{-0\cdot t} \frac{\sin t}{t}\; \mathbf{d}t=\int_0^\infty e^0 \frac{\sin t}{t}\; \mathbf{d}t=\int_0^\infty 1\cdot \frac{\sin t}{t}\; \mathbf{d}t=\int_0^\infty \frac{\sin t}{t}\; \mathbf{d}t.</math>
:Bet taip pat:
:<math>G(\infty)=\int_0^\infty e^{-\infty\cdot t} \frac{\sin t}{t}\; \mathbf{d}t=0,</math>
:nes <math>\frac{\sin(t)}{t}</math> yra konverguojanti eilutė ir garantuotai:
:<math>\lim_{t\to infty} (e^{-t \cdot t} \frac{\sin t}{t} )=0.</math>
==Nuorodos==
==Nuorodos==
14:01, 31 gruodžio 2011 versija
S
i
(
x
)
=
∫
0
x
sin
t
t
d
t
.
{\displaystyle {\rm {Si}}(x)=\int _{0}^{x}{\frac {\sin t}{t}}\,dt.}
s
i
(
x
)
=
−
∫
x
∞
sin
t
t
d
t
.
{\displaystyle {\rm {si}}(x)=-\int _{x}^{\infty }{\frac {\sin t}{t}}\,dt.}
S
i
(
∞
)
=
∫
0
∞
sin
t
t
d
t
=
π
2
.
{\displaystyle {\rm {Si}}(\infty )=\int _{0}^{\infty }{\frac {\sin t}{t}}\,dt={\frac {\pi }{2}}.}
Sinuso Integralo užrašymas Teiloro eilute Kadangi sinuso Teiloro eilutetė yra
sin
t
=
t
−
t
3
3
!
+
t
5
5
!
−
t
7
7
!
+
t
9
9
!
−
t
11
11
!
+
⋯
,
{\displaystyle \sin t=t-{\frac {t^{3}}{3!}}+{\frac {t^{5}}{5!}}-{\frac {t^{7}}{7!}}+{\frac {t^{9}}{9!}}-{\frac {t^{11}}{11!}}+\cdots ,}
tai gauname, kad
sin
t
t
=
1
−
t
2
3
!
+
t
4
5
!
−
t
6
7
!
+
t
8
9
!
−
t
10
11
!
+
⋯
;
{\displaystyle {\frac {\sin t}{t}}=1-{\frac {t^{2}}{3!}}+{\frac {t^{4}}{5!}}-{\frac {t^{6}}{7!}}+{\frac {t^{8}}{9!}}-{\frac {t^{10}}{11!}}+\cdots ;}
toliau integruodami šią eilutę gauname
S
i
(
x
)
=
∫
0
x
sin
t
t
d
t
=
∫
0
x
(
1
−
t
2
3
!
+
t
4
5
!
−
t
6
7
!
+
t
8
9
!
−
t
10
11
!
+
⋯
)
d
t
=
{\displaystyle {\rm {Si}}(x)=\int _{0}^{x}{\frac {\sin t}{t}}\,dt=\int _{0}^{x}(1-{\frac {t^{2}}{3!}}+{\frac {t^{4}}{5!}}-{\frac {t^{6}}{7!}}+{\frac {t^{8}}{9!}}-{\frac {t^{10}}{11!}}+\cdots )\,dt=}
=
(
t
−
t
3
3
!
⋅
3
+
t
5
5
!
⋅
5
−
t
7
7
!
⋅
7
+
t
9
9
!
⋅
9
−
t
11
11
!
⋅
11
+
⋯
)
|
0
x
=
{\displaystyle =(t-{\frac {t^{3}}{3!\cdot 3}}+{\frac {t^{5}}{5!\cdot 5}}-{\frac {t^{7}}{7!\cdot 7}}+{\frac {t^{9}}{9!\cdot 9}}-{\frac {t^{11}}{11!\cdot 11}}+\cdots )|_{0}^{x}=}
=
x
−
x
3
3
!
⋅
3
+
x
5
5
!
⋅
5
−
x
7
7
!
⋅
7
+
x
9
9
!
⋅
9
−
x
11
11
!
⋅
11
+
⋯
.
{\displaystyle =x-{\frac {x^{3}}{3!\cdot 3}}+{\frac {x^{5}}{5!\cdot 5}}-{\frac {x^{7}}{7!\cdot 7}}+{\frac {x^{9}}{9!\cdot 9}}-{\frac {x^{11}}{11!\cdot 11}}+\cdots .}
Belieka pasakyti, kad Sinuso Integralo išvestinės lygios nuliui nuo nulio (antra ir trečia, pavyzdžiui, o pirmos riba lygi 1), todėl negalima gauti šios eilutės įprastai naudojantis Teiloro eilutės formule.
Įrodymas, kad
S
i
(
∞
)
=
π
/
2
{\displaystyle {\rm {Si}}(\infty )=\pi /2}
Užrašykime
S
i
(
∞
)
=
∫
0
∞
sin
t
t
d
t
,
{\displaystyle {\rm {Si}}(\infty )=\int _{0}^{\infty }{\frac {\sin t}{t}}\;\mathbf {d} t,}
tuomet turime funkciją
G
(
x
)
=
∫
0
∞
e
−
x
t
sin
t
t
d
t
,
x
>
0.
{\displaystyle G(x)=\int _{0}^{\infty }e^{-xt}{\frac {\sin t}{t}}\;\mathbf {d} t,\quad x>0.}
Diferencijuodami per x (
x
>
0
{\displaystyle x>0}
G(x) , gauname
G
′
(
x
)
=
−
∫
0
∞
e
−
x
t
sin
t
d
t
,
x
>
0.
{\displaystyle G'(x)=-\int _{0}^{\infty }e^{-xt}\sin t\;\mathbf {d} t,\quad x>0.}
Toliau integruodami nuo t , gauname
G
′
(
x
)
=
−
∫
0
∞
e
−
x
t
sin
t
d
t
=
−
e
−
x
t
(
x
sin
(
t
)
+
cos
(
t
)
)
x
2
+
1
|
0
∞
=
{\displaystyle G'(x)=-\int _{0}^{\infty }e^{-xt}\sin t\;\mathbf {d} t=-{\frac {e^{-xt}(x\sin(t)+\cos(t))}{x^{2}+1}}|_{0}^{\infty }=}
=
−
e
−
x
⋅
∞
(
x
sin
(
∞
)
+
cos
(
∞
)
)
x
2
+
1
−
(
−
e
−
x
⋅
0
(
x
sin
(
0
)
+
cos
(
0
)
)
x
2
+
1
)
=
{\displaystyle =-{\frac {e^{-x\cdot \infty }(x\sin(\infty )+\cos(\infty ))}{x^{2}+1}}-\left(-{\frac {e^{-x\cdot 0}(x\sin(0)+\cos(0))}{x^{2}+1}}\right)=}
=
−
e
−
x
⋅
∞
(
x
sin
(
∞
)
+
cos
(
∞
)
)
x
2
+
1
+
e
0
(
x
⋅
0
+
1
)
x
2
+
1
=
{\displaystyle =-{\frac {e^{-x\cdot \infty }(x\sin(\infty )+\cos(\infty ))}{x^{2}+1}}+{\frac {e^{0}(x\cdot 0+1)}{x^{2}+1}}=}
=
−
e
−
x
⋅
∞
(
x
sin
(
∞
)
+
cos
(
∞
)
)
x
2
+
1
+
1
⋅
(
0
+
1
)
x
2
+
1
=
{\displaystyle =-{\frac {e^{-x\cdot \infty }(x\sin(\infty )+\cos(\infty ))}{x^{2}+1}}+{\frac {1\cdot (0+1)}{x^{2}+1}}=}
=
−
x
sin
(
∞
)
+
cos
(
∞
)
e
x
⋅
∞
(
x
2
+
1
)
+
1
x
2
+
1
.
{\displaystyle =-{\frac {x\sin(\infty )+\cos(\infty )}{e^{x\cdot \infty }(x^{2}+1)}}+{\frac {1}{x^{2}+1}}.}
Kadangi
sin
(
∞
)
{\displaystyle \sin(\infty )}
cos
(
∞
)
,
{\displaystyle \cos(\infty ),}
−
x
sin
(
∞
)
+
cos
(
∞
)
e
x
⋅
∞
(
x
2
+
1
)
=
0.
{\displaystyle -{\frac {x\sin(\infty )+\cos(\infty )}{e^{x\cdot \infty }(x^{2}+1)}}=0.}
Arba kitaip užrašius
lim
t
→
∞
(
−
x
sin
(
t
)
+
cos
(
t
)
e
x
t
(
x
2
+
1
)
)
=
0.
{\displaystyle \lim _{t\to \infty }\left(-{\frac {x\sin(t)+\cos(t)}{e^{xt}(x^{2}+1)}}\right)=0.}
Todėl turime
G
′
(
x
)
=
−
∫
0
∞
e
−
x
t
sin
t
d
t
=
1
x
2
+
1
.
{\displaystyle G'(x)=-\int _{0}^{\infty }e^{-xt}\sin t\;\mathbf {d} t={\frac {1}{x^{2}+1}}.}
Integruodami nuo x turime:
G
(
x
)
=
∫
G
′
(
x
)
d
x
=
∫
1
x
2
+
1
d
x
=
arctan
(
x
)
+
C
.
{\displaystyle G(x)=\int G'(x)\mathbf {d} x=\int {\frac {1}{x^{2}+1}}\;\mathbf {d} x=\arctan(x)+C.}
Gauname kam lygi funkcija G(x) , kai
x
=
∞
:
{\displaystyle x=\infty :}
G
(
∞
)
=
arctan
(
∞
)
+
C
=
π
2
+
C
.
{\displaystyle G(\infty )=\arctan(\infty )+C={\frac {\pi }{2}}+C.}
Taipogi gauname kam lygi funkcija G(x) , kai
x
=
0
:
{\displaystyle x=0:}
G
(
0
)
=
arctan
(
0
)
+
C
=
C
.
{\displaystyle G(0)=\arctan(0)+C=C.}
Vadinasi,
G
(
0
)
=
∫
0
∞
e
−
0
⋅
t
sin
t
t
d
t
=
∫
0
∞
e
0
sin
t
t
d
t
=
∫
0
∞
1
⋅
sin
t
t
d
t
=
∫
0
∞
sin
t
t
d
t
.
{\displaystyle G(0)=\int _{0}^{\infty }e^{-0\cdot t}{\frac {\sin t}{t}}\;\mathbf {d} t=\int _{0}^{\infty }e^{0}{\frac {\sin t}{t}}\;\mathbf {d} t=\int _{0}^{\infty }1\cdot {\frac {\sin t}{t}}\;\mathbf {d} t=\int _{0}^{\infty }{\frac {\sin t}{t}}\;\mathbf {d} t.}
Bet taip pat:
G
(
∞
)
=
∫
0
∞
e
−
∞
⋅
t
sin
t
t
d
t
=
0
,
{\displaystyle G(\infty )=\int _{0}^{\infty }e^{-\infty \cdot t}{\frac {\sin t}{t}}\;\mathbf {d} t=0,}
nes
sin
(
t
)
t
{\displaystyle {\frac {\sin(t)}{t}}}
lim
t
→
i
n
f
t
y
(
e
−
t
⋅
t
sin
t
t
)
=
0.
{\displaystyle \lim _{t\to infty}(e^{-t\cdot t}{\frac {\sin t}{t}})=0.}
Nuorodos 
