Ngô Quốc Anh

September 17, 2007

Một số bài tập tích phân bội

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Ngô Quốc Anh @ 6:02

  • Chứng minh 

\int_{0}^{1}\int_{0}^{1}\;\;\frac{x\cdot\ln(xy)}{1-x^{2}y^{2}}\;\;\textbf dx\;\;\textbf dy\;\;=\;\;\boxed{-\frac{1}{2}\,\zeta{(2)}}

\int_{0}^{1}\;\int_{0}^{1}\;\;\frac{x}{(1+x^{2}y^{2})\cdot\ln\,(xy)}\;\;\textbf dx\;\textbf dy\;\;=\;\;\boxed{\ln\left(\frac{\sqrt{\;2\pi^{3}\;}}{\Gamma^{2}\left(\frac{1}{4}\right)}\right)}

\int_0^1\;\int_0^1\;\;\dfrac{\textbf dx\;\textbf dy}{\left(-\,\ln\,(x\,y)\right)^{\frac{3}{2}}}\;\;=\;\;\boxed{\sqrt\pi}

  •  Tính

\int_{0}^{1}\;\int_{0}^{1}\;\;\frac{\textbf dx\;\;\textbf dy}{1+x^{2}+y^{2}}

Bất đẳng thức cho dãy Lalescu

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 6:00

Prove that for the T. Lalescu‘s sequence L_{n}=\sqrt [n+1]{(n+1) !}-\sqrt [n]{n!} exists the “clenching\left(\frac{n}{n+1}\right)^{n+1}< L_{n}<\left(\frac{n}{n+1}\right)^{n}.

1\blacktriangleright \mathrm{L.H.S.\ : } Here is a nice and dificult problem (if is possible, without derivatives) !
2\blacktriangleright \mathrm{R.H.S.\ :\ G.M.\le A.M.\ for\ }
x_{k}: =\left\{\begin{array}{ccc}\frac{\sqrt [n]{n!}}{n}&\mathrm{if}& k\in\overline{1,n}\\ \\ \left(\frac{n}{n+1}\right)^{n}&\mathrm{if}& k = n+1\end{array}\right\| 
\frac{n\cdot\frac{\sqrt [n]{n!}}{n}+\left(\frac{n}{n+1}\right)^{n}}{n+1}> \sqrt [n+1]{\left(\frac{\sqrt [n]{n!}}{n}\right)^{n}\cdot\left(\frac{n}{n+1}\right)^{n}}=

\frac{\sqrt [n+1]{(n+1)!}}{n+1}\implies \sqrt [n]{n!}+\left(\frac{n}{n+1}\right)^{n}>\sqrt [n+1]{(n+1)!}\implies \sqrt [n+1]{(n+1)!}-\sqrt [n]{n!}<\left(\frac{n}{n+1}\right)^{n} \implies L_{n}<\left(\frac{n}{n+1}\right)^{n}.

Remark.
\lim_{n\to\infty }\left(\frac{n}{n+1}\right)^{n}=\frac{1}{e}\implies\boxed{\ \lim_{n\to\infty}\left(\sqrt [n+1]{(n+1)!}-\sqrt [n]{n!}\right)=\frac{1}{e}\ }.

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