Ngô Quốc Anh

September 29, 2007

Lời giải 1 bài tích phân bội

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Ngô Quốc Anh @ 3:38

Chứng minh \int_{0}^{1}\;\int_{0}^{1}\;\;\left(\,\frac{\ln^{2}\,(xy)}{1+xy}\,\right)^{2}\;\;\textbf dx\;\textbf dy\;\;=\;\;\boxed{\frac{225}{2}\,\zeta{(5)}}.

Lời giải.

P =\int_{0}^{1}\int_{0}^{1}\left(\frac{\log^{2}(x y)}{x y+1}\right)^{2}dxdy

a substituition x = e^{-\phi } , y = e^{\phi }\theta. Using mister Jacob

\left(\frac{\log^{2}(x y)}{x y+1}\right)^{2}\left|\left(\begin{array}{ll}\frac{\partial x}{\partial\theta }&\frac{\partial x}{\partial\phi }\\ \frac{\partial y}{\partial\theta }&\frac{\partial y}{\partial\phi }\end{array}\right)\right| d\theta d\phi =\frac{\log^{4}(\theta )}{(\theta+1)^{2}}(d\theta d\phi ).

Solving the inequalities we get 0 <\theta < 1\land 0 <\phi <-\log (\theta )

P =\int_{0}^{1}\int_{0}^{-\log (\theta )}\frac{\log^{4}(\theta )}{(\theta+1)^{2}}d\phi d\theta =-\int_{0}^{1}\frac{\log^{5}(\theta )}{(\theta+1)^{2}}\, d\theta

\int_{0}^{1}\frac{\log^{5}(\theta )}{(\theta+1)^{2}}\, d\theta =\int_{0}^{1}\left(\sum_{i = 0}^{\infty }\log^{5}(\theta )\theta^{i}(i+1) (-1)^{i}\right)\, d\theta =\sum_{i = 0}^{\infty }(i+1) (-1)^{i}\int_{0}^{1}\log^{5}(\theta )\theta^{i}\, d\theta

it’s very trivial to demonstrate that 

\int_{0}^{1}\log^{c_{1}}(\theta )\theta^{c_{2}}\, d\theta =-\left(-\frac{1}{c_{2}+1}\right)^{c_{1}+1}\Gamma\left(c_{1}+1\right)

\int_{0}^{1}\log^{5}(\theta )\theta^{i}\, d\theta =-\frac{120}{(i+1)^{6}}

P =\sum_{i = 0}^{\infty }\frac{120 (-1)^{i}}{(i+1)^{5}}= 120\sum_{i = 1}^{\infty }\frac{(-1)^{i-1}}{i^{5}}

\zeta (s)=\frac{\sum_{i=1}^{\infty }\frac{(-1)^{i-1}}{i^{s}}}{1-2^{1-s}}

P =\frac{225\zeta (5)}{2}

and we are done

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