[分析] 積分中的微分
Apostol的Mathematical analysis - 2e , p.167 , Theorem 7.40
請板友們看那頁證明後就知道我在講啥了(如果沒有書的話我這個定理我打在備註)
我把α(x)寫成x了 比較簡單
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想請問一下為何他這樣證是對的
b
因為他是利用 G(y) = ∫f_y(x,y)dx €C[c,d] 來證的 (因為f_y(x,y)€C[a,b]Ｘ[c,d])
a
可是在倒數第三行的式子
F(y)-F(y_0) b
───── = ∫f_y(x,y')dx , where y' is between y and y_0
y - y_0 a
如果今天y'與x無關，那這個證明就很OK
可是問題就在於y'與x有關
b
意思是 ∫f_y(x,y'(y,x))dx --- ●
a
這個東西根本跟G(y)毫無關係了，因為連是否存在一個y*值使得G(y*) = ●都不知道了
b
甚至即便對於任予g(x)€[c,d]的函數，∫f_y(x,g(x))dx 也不一定存在
a
所以，是否能證出此y'與x無關
或是這證明是錯的??
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P.S.
Theorem：
Let Q = {(x,y)：a≦x≦b , c≦y≦d}. Assume that the integral
b
F(y) = ∫ f(x,y)dx exists
a
If the partial derivative f_y is continuous on Q
then derivative F'(y) exists in (c,d) and is given by
b
F'(y) = ∫ f_y(x,y)dx
a
<pf> If y_0€(c,d) and y =/= y_0, we have
F(y)-F(y_0) b f(x,y)-f(x,y_0) b
───── = ∫──────── = ∫ f_y(x,y')dx
y - y_0 a y-y_0 a
where y' is between y and y_0. Since f_y is continuous on Q,
we obtain the conclusion by arguing as in the proof of Theorem 7.38
(Theorem 7.38就是在講如果f_y is continuous on Q
b
則G(y) = ∫ f_y(x,y)dx is continous on [c,d] )
a
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