Re: [積分] 一題積分
看板trans_math作者josephw (Shower the People)時間16年前 (2007/11/19 14:18)推噓5(5推 0噓 6→)留言11則, 5人參與討論串32/92 (看更多)
※ 引述《SCJhang (SCJhang)》之銘言：
: 請問如何證明下式：
: p-1
: ∞ x π
: ∫ ────dx = ────
: 0 1+x sinpπ
: 煩請板友指導，謝謝。
Suppose 0 < p < 1
x = t/(1-t) = 1/(1-t) - 1 => dx = 1/(1-t)^2
1 (t/(1-t))^(p-1) 1
=> ∫ ────────── . ──── dt
0 (1/(1-t)) (1-t)^2
1 (t)^(p-1)
= ∫ ────────── dt
0 (1-t)^p
1
= ∫ t^(p-1) (1-t)^(-p) dt
0
= B(p,-p) (by beta function)
= gamma(p) gamma(1-p) ( gamma(.) : gamma function)
= ....= π/sin(pπ)
最後一項卡關了....以前有證出來 抱歉 請高手再寫好了
複變的方法不太行 也請高手寫一下
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