Question

Let $f:R_+^* ×R_+^*$ $-> R$ be a function defined by

$f(x,y)=∫_0^\frac{\pi}{2} ln⁡(x^2 sin^2⁡t+y^2 cos^2⁡t)dt$ 

(a). Show that for all $x,y >0 : \bigtriangledown f(x,y)= (\frac{\pi}{x+y},\frac{\pi}{x+y})$

(b). Deduce that for all $x,y>0 : f(x,y)=\pi ln(\frac{x+y}{2})$

Answer 1

Solution

(a). Show that for all $x,y >0 : \bigtriangledown f(x,y)= (\frac{\pi}{x+y},\frac{\pi}{x+y})$

 $\frac{\partial f}{\partial x} =\int_0^\frac{\pi}{2} \frac{2xsin^2t}{x^2sin^2t+y^2cos^2t} dt$ ; $\frac{\partial f}{\partial y} =\int_0^\frac{\pi}{2} \frac{2ycos^2t}{x^2sin^2t+y^2cos^2t} dt$

$x \frac{\partial f}{\partial x}+y\frac{\partial f}{\partial x}= \int_0^\frac{\pi}{2} 2dt=\pi$  (1) and $y \frac{\partial f}{\partial x}+x\frac{\partial f}{\partial x}=2xy \int_0^\frac{\pi}{2} \frac{1}{x^2sin^2t+y^2cos^2t}dt =2xy \int_0^\frac{\pi}{2} \frac{1/cos^2t}{y^2+x^2tan^2t}dt$

$u=tant => du=\frac{1}{cos^2t}dt$ , then

$t=0=>u->0 , t=\frac{\pi}{2} =>u-> \infty$

$y \frac{\partial f}{\partial x}+x\frac{\partial f}{\partial x}=2xy \int_0^\infty \frac{1}{y^2+x^2u^2}du =\frac{2y}{x}\int_0^\infty\frac{1}{(\frac{y}{x})^2 +u^2}du$

                     $=\frac{2y}{x} \frac{1}{\frac{y}{x}}arctan\left.(\frac{u}{\frac{y}{x}})\right |_0^ \infty$

                     $=\pi$   (2)

from(1)&(2) : \( \begin{cases} xf_x +yf_y =\pi \times x & \quad \\ yf_x +xf_y =\pi \times(-y) & \quad \ \end{cases} \)$=> (x^2-y^2)f_x=\pi(x-y)$

so, $=>f_x= \frac{\pi(x-y)}{x^2-y^2}=\frac{\pi}{x+y}$ 

similarly, $f_y =\frac{\pi}{x+y}$

Therefore, $\bigtriangledown f(x,y)= (\frac{\pi}{x+y},\frac{\pi}{x+y})$

(b). Deduce that for all $x,y>0 : f(x,y)=\pi ln(\frac{x+y}{2})$

we have \( \begin{cases} \frac{\partial f}{\partial x}(x,y) = \frac{\pi}{x+y} ,(1)& \quad \\ \frac{\partial f}{\partial y}(x,y) = \frac{\pi}{x+y} ,(2) & \quad \ \end{cases} \)

(1) : $\int\frac{\partial f}{\partial x}(x,y)dx = \int\frac{\pi}{x+y} dx$

$=>f(x,y)=\pi ln(x+y)+c(y)$

$=>\frac{\partial f}{\partial y}(x,y)=\frac{\pi}{x+y}+c'(y)$ , (3) 

from(2&(3) : $\frac{\pi}{x+y}=\frac{\pi}{x+y}+c'(y) =>c(y)=k\in R$

Thus, $f(x,y)=\pi ln(x+y)+k$

we take $(x=1,y=1)$ $=>f(1,1)=\pi ln2+k$

we have $f(1,1)=0, Then , \pi ln2+k=0=>k=-\pi ln2$

Therefore, $f(x,y)=\pi ln(x+y)-\pi ln2=\pi ln(\frac{x+y}{2})$

Solution

(a). Show that for all $x,y >0 : \bigtriangledown f(x,y)= (\frac{\pi}{x+y},\frac{\pi}{x+y})$

Therefore, $\bigtriangledown f(x,y)= (\frac{\pi}{x+y},\frac{\pi}{x+y})$

(b). Deduce that for all $x,y>0 : f(x,y)=\pi ln(\frac{x+y}{2})$

Therefore, $f(x,y)=\pi ln(x+y)-\pi ln2=\pi ln(\frac{x+y}{2})$