pembuktian gamma(1/2)=akar.pi
4
FORMULA ESSENTIAL FUNCTIONS OF THE GAMMA At previous meetings, we have learned the definition of the gamma function is (n)= ∫ 0 x 1−n e −x dx. We also have obtained the values of the gamma function when n positive integers and in fact form the factorial function. In this discussion, we tried to find the value of the gamma function is not unanimously positive, in this case is the gamma function when n = 1 / 2. Of course if we put on the definition of gamma function, then we will write it as follows. ( 1 2 ) = ∫ 0 x 1− 1 2 e −x dx = ∫ 0 x −1 2 e −x dx To solve the above functions are not as easy as completing the gamma function is the value of n is a positive integer. Integral on the above functions can be completed by setting up two equations in u and v, namely ( 1 2 ) = ∫ 0 u −1 2 e −u du … (1) ( 1 2 ) = ∫ 0 v −1 2 e −v dv … (2) If equation (1) and (2) multiplied by the obtained (( 1 2 )) 2 = ∫ 0 u −1 2 e −u du ∫ 0 v −1 2 e −v dv = ∫ 0 ∫ 0 u −1 2 e −u v −1 2 e −v dudv … (3) Next, use the following form
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Transcript of pembuktian gamma(1/2)=akar.pi
FORMULA ESSENTIAL FUNCTIONS OF THE GAMMA
At previous meetings, we have learned the definition of the gamma function is (n)=
∫0
x1−n e−x dx. We also have obtained the values of the gamma function when n positive integers
and in fact form the factorial function.
In this discussion, we tried to find the value of the gamma function is not unanimously positive,
in this case is the gamma function when n = 1 / 2. Of course if we put on the definition of gamma
function, then we will write it as follows.
(12
) = ∫0
x1−
12 e−x dx = ∫
0
x−1
2 e−x dx
To solve the above functions are not as easy as completing the gamma function is the value of n
is a positive integer. Integral on the above functions can be completed by setting up two
equations in u and v, namely
(12
) = ∫0
u−12 e−udu … (1)
(12
) = ∫0
v−12 e−vdv … (2)
If equation (1) and (2) multiplied by the obtained
((12
))2 = ∫0
u−12 e−udu ∫
0
v−12 e−vdv = ∫
0∫0
u−12 e−u v
−12 e−v dudv … (3)
Next, use the following form
(Γ ( 12 ))
2
=∫0
∞
∫0
∞
u−1/2e−u v−1 /2 e−v dudv
u=x2→ du=2x dx
v= y2→ dv=2 y dy
Substitution results in the form of the equation (3) and if the way you are correct then you will
get the following equation
(Γ ( 12 ))
2
=∫0
∞
∫0
∞
(x2 )−12 e−x
2
¿
(Γ ( 12 ))
2
=∫0
∞
∫0
∞
x−1 e−x2
y−1 e− y2
2 xdx2 ydy
(Γ (12 ))
2
=4∫0
∞
∫0
∞
e− x2
e− y2
dx dy… (4)
To solve the above integral, we need the help of polar integral in the first quadrant. So the next
step is you have to change the integral model
Cartesian above the polar integral. As such, then you must change the x, y, and dy dx into polar
form, ie x=r cosθ
y=rsinθ
dx dy=r dr dθ
Substitution of the above values in equation (4) and use the first quadrant is the limit for θ = 0 to
θ = π / 2. If you do it right, then obtained the following form
(Γ (12 ))
2
=4∫0
∞
∫0
∞
e−(x¿¿2+ y2)dx dy¿
(Γ (12 ))
2
=4 ∫θ=0
π /2
∫r=0
∞
e−[(r cosθ )¿¿2+(r sin θ)2 ]r dr dθ ¿
(Γ (12 ))
2
=4 ∫θ=0
π /2
∫r=0
∞
e−(r 2 cos2θ+r2 sin2θ )r dr dθ
(Γ (12 ))
2
=4 ∫θ=0
π /2
∫r=0
∞
e−r2
r dr dθ…(5)
for ,∫0
∞
4 r e−r2
dr dθ=4∫0
∞
r e−r2
dr dθ
4∫0
∞
r e−r2
dr dθ=4 .−12∫0
∞
d (e−r2
)
¿−2∫0
∞
d (e−r2
)
¿2[e−r2
]0∞
¿−2[ 1
e∞2 −
1
e02 ] ¿−2[0−1]
¿2
Now, come back to equation (5) to solve the integral and get its solution
(Γ (12 ))
2
=4 ∫θ=0
π /2
∫r=0
∞
e−r2
r dr dθ
¿∫0
π2
2dθ
¿¿
¿2( π2 )−2 (0 )
¿ π
(Γ (12 ))
2
=π
Γ ( 12 )=√π
You can easily solve the integral form above. Have you guys done to get the value of (1 / 2)
Yes, we have. The value of (1 / 2) is √π
9 th GROUP
NOOR AZIZAH WALIANA MAPPA GISTRA ESTI MELISA
