# Ngô Quốc Anh

## September 29, 2007

### Lời giải 1 bài tích phân bội, 2

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Ngô Quốc Anh @ 3:42

Prove that $\displaystyle\int_0^1 \int_0^1 \dfrac{dxdy}{(\ln x+\ln y)\cdot(1+x^2 y^2)^2}=\boxed{-\frac{1}{8}(2+\pi)}$.

Solution 1. We shall use fact that that $\displaystyle\int \cos ^2(\theta ) d\theta = \frac {\theta }{2} + \frac {1}{4} \sin (2 \theta )$.

Assuming $x = e^{ - \phi }$ and $y = e^{\phi } \tan (\theta )$ with $0 < \theta < \frac {\pi }{4}$ and $0 < \phi < - \log (\tan (\theta ))$,

one has $\displaystyle\left|\left( \begin{array}{ll} \frac {\partial x}{\partial \theta } & \frac {\partial x}{\partial \phi } \\\frac {\partial y}{\partial \theta } & \frac {\partial y}{\partial \phi } \end{array} \right)\right| = \sec ^2(\theta )$.

The integral becomes $\displaystyle P = \int _0^{\frac {\pi }{4}}\int _0^{ - \log (\tan (\theta ))}\frac {\cos ^2(\theta )}{\log (\tan (\theta ))}d\phi d\theta = - \int_0^{\frac {\pi }{4}} \cos ^2(\theta ) \, d\theta = - \frac {1}{8} (2 + \pi )$.

Solution 2. We use the following substitution $\displaystyle u=xy, \quad v=\frac{x}{y}$.

Then we have Jacobian $J=-\frac{1}{2v}$ and $\displaystyle\begin{gathered} I = - \int_0^1 d u\int_u^{\frac{1}{u}} {\frac{{dv}}{{2v}}} \frac{1}{{\ln \,u\cdot{{(1 + {u^2})}^2}}} \hfill \\\quad= - \int_0^1 {\frac{{du}}{{\ln u\cdot{{(1 + {u^2})}^2}}}} \int_u^{\frac{1}{u}} {\frac{{dv}}{{2v}}} \hfill \\\quad= - \int_0^1 {\frac{{du}}{{\ln u\cdot{{(1 + {u^2})}^2}}}} \ln u \hfill \\\quad = - \int_0^1 {\frac{{du}}{{{{(1 + {u^2})}^2}}}}. \hfill \\ \end{gathered}$

Then we use substitution $u=\tan t$ and $\displaystyle\int_0^1 \frac{du}{(1+u^2)^2}=\int_0^\frac{\pi}{4} \frac{\dfrac{dt}{\cos^2 t}}{\left(\dfrac{1}{\cos^2 t}\right)^2}=\int_0^\frac{\pi}{4} \cos^2 t dt=\frac{2+\pi}{8}$.

### 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

### BT tính tích phân của hàm 1 biến

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 3:32

Tính $\int^{\pi}_{-\pi} \frac{sin3x}{(1+2^{x})sinx} \,dx$ .

Lời giải: You can use $\boxed {\ \int_{ - a}^af(x)g(x)\ \mathrm {dx} = \int_0^a\left[f(x) + f( - x)\right]g(x)\ \mathrm {dx}\ }$,

where $g$ is even. Therefore, for the functions $f(x) = \frac {1}{1 + 2^x}$ and $g(x) = \frac {\sin 3x}{\sin x}$ obtain :  and $\int^{\pi}_{ - \pi} \frac {1}{1+2^x}\cdot\frac {\sin 3x}{\sin x}\ \mathrm {dx} = \int^{\pi}_0 \frac {\sin 3x}{\sin x}\ \mathrm {dx}$ a.s.o.  Here is another problem from Japanese university entrance exam. Keio University entrance exam/Medical 1974. Evaluate $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sin ^ 2 x}{1+e^{-x}}\ dx$.