Matematika/Sinuso Integralas
Sinuso Integralo užrašymas Teiloro eilute
keisti
Įrodymas, kad
S
i
(
∞
)
=
π
/
2
{\displaystyle {\rm {Si}}(\infty )=\pi /2}
keisti
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,}
Pastaba, kad
lim
t
→
0
e
−
∞
⋅
t
sin
t
t
=
e
0
⋅
1
=
1.
{\displaystyle \lim _{t\to 0}e^{-\infty \cdot t}{\frac {\sin t}{t}}=e^{0}\cdot 1=1.}
G
(
∞
)
=
0
{\displaystyle G(\infty )=0}
G
(
∞
)
=
1
{\displaystyle G(\infty )=1}
G
(
∞
)
=
e
−
1
.
{\displaystyle G(\infty )=e^{-1}.}
S
=
lim
t
→
0
t
e
−
∞
⋅
t
=
lim
t
→
0
t
e
∞
⋅
t
=
0
1
=
0
,
{\displaystyle S=\lim _{t\to 0}te^{-\infty \cdot t}=\lim _{t\to 0}{\frac {t}{e^{\infty \cdot t}}}={\frac {0}{1}}=0,}
G
(
∞
)
=
0
,
{\displaystyle G(\infty )=0,}
xOy kampo plotas lygus nuliui, kai t artėja į nulį. Galutinai turime, kad taip pat
S
G
=
lim
t
→
0
t
e
∞
⋅
t
⋅
sin
t
t
=
0
⋅
1
=
0
{\displaystyle S_{G}=\lim _{t\to 0}{\frac {t}{e^{\infty \cdot t}}}\cdot {\frac {\sin t}{t}}=0\cdot 1=0}
Todėl turime, kad
G
(
∞
)
=
arctan
(
∞
)
+
C
=
π
2
+
C
{\displaystyle G(\infty )=\arctan(\infty )+C={\frac {\pi }{2}}+C}
ir
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;}
vadinasi,
C
=
−
π
2
.
{\displaystyle C=-{\frac {\pi }{2}}.}
Bet
G
(
0
)
=
arctan
(
0
)
+
C
=
C
;
{\displaystyle G(0)=\arctan(0)+C=C;}
vadinasi,
G
(
0
)
=
∫
0
∞
sin
t
t
d
t
=
−
π
2
.
{\displaystyle G(0)=\int _{0}^{\infty }{\frac {\sin t}{t}}\;\mathbf {d} t=-{\frac {\pi }{2}}.}
Bet atsakymas panašesnis į ne su "-", o su "+", todėl įrodėme, kad
S
i
(
∞
)
=
∫
0
∞
sin
t
t
d
t
=
π
2
.
{\displaystyle {\rm {Si}}(\infty )=\int _{0}^{\infty }{\frac {\sin t}{t}}\;\mathbf {d} t={\frac {\pi }{2}}.}
