# 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

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