推 Jasy:nice 218.187.11.143 05/08
※ 引述《Jasy (面對現實)》之銘言：
: 1.Suppose that f is a conti function which satisfies the intergral equation
: x f(t)
: f(x)=2 + S ------------ dt, x>=0 then f(1)=?
: 0 (t+2)(t+3)
: -x^2(t^2+1)
: x -t^2 2 1 e
: 2.If f(x)={S e dt} , g(x)=S ------------- dt, then
: 0 0 2
: t +1
-x^2(t^2+1)
1 e
g(x) = S -------------------- dt
0 2
t + 1
1 x -y^2(t^2+1)
= S S -2y e dy dt + pi/4
0 0
x 1 -y^2(t^2+1)
= S S -2y e dt dy + pi/4
0 0
x 1 -(yt)^2 -y^2
= S S -2y e ( e ) dt dy + pi/4
0 0
x y -k^2 -y^2
= S S -2 e ( e ) dk dy + pi/4
0 0
x y -t^2 -y^2
= S S -2 e ( e ) dt dy + pi/4
0 0
x -t^2 -x^2
f'(x) = 2 S e dt ( e )
0
x -t^2 -x^2 x -t^2 -x^2
g'(x) = S -2 e ( e ) dt = -2 S e dt ( e )
0 0
= -f'(x)
-> g'(x) + f'(x) = 0
-> g(x) + f(x) = c = g(0) + f(0) = pi/4
-> g(∞) + f(∞) = pi/4
since lim g(x) = 0
x->∞
-> f(∞) = pi/4
∞ -t^2
-> S e dt = √π/2
0
: (1) Show that g'(x)+f'(x)=0 for all of x
: (2) Use (1) to show that g(x)+f(x)=π/4
: ∞ -t^2 dt
: (3) Use (2) to prove that S e = √π/2
: 0
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