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