Re: [分析] 初微(42)
※ 引述《TaiwanBank (澳仔金控台灣分行)》之銘言：
: +∞
: 若 ∫ |f(x)| dx < ∞, 請證明:
: -∞
: +∞ +∞
: ∫ f(x) dx = ∫ f(x-1/x) dx.
: -∞ -∞
令 t = x - 1/x => dt = (1 + 1/x^2)dx
x^2 - tx - 1 = 0 => x = [t±√(4+t^2)]/2
1/x^2 = 2/[t^2 + 2 ±t√(4+t^2)]
1 + 1/x^2 = [t^2 + 4 ±t√(4+t^2)] / [t^2 + 2 ±t√(4+t^2)]
取 A = [t^2 + 4 + t√(4+t^2)] / [t^2 + 2 + t√(4+t^2)]
B = [t^2 + 4 - t√(4+t^2)] / [t^2 + 2 - t√(4+t^2)]
直接的計算 => 1/A + 1/B = 1
+oo 1 +oo
∫ f(x-1/x) dx = ∫ f(x-1/x) dx + ∫ f(x-1/x) dx
-oo 0 1
0 -1
+ ∫ f(x-1/x) dx + ∫ f(x-1/x) dx
-1 -oo
0 +oo +oo 0
= ∫f(t)/A dt + ∫f(t)/A dt + ∫f(t)/B dt + ∫f(t)/B dt
-oo 0 0 -oo
+oo
由假設條件可推得上面四個暇積分都收斂, 故 ∫ f(x-1/x) dx 收斂
-oo
0 +oo
= ∫f(t)(1/A + 1/B) dt + ∫f(t)(1/A + 1/B) dt
-oo 0
0 +oo +oo
= ∫f(t) dt + ∫f(t) dt = ∫f(t) dt ... Q.E.D.
-oo 0 -oo
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