Hi Hollywood and ibdutt,
Sorry, I tried.
But I still don't know to integrate even I know I should solve the inner dy integral first.
Can you show me how to do it?
.
When I tried to solve it, I let xy=tan(theatre),
then the question becomes integral limit from zero to infinity [Arctan(7x)-arctan(5x)]/x dx.
Which when I tried to continue to solve it,
It becomes the starting question (the question with 1/(1+(xy)^2) )
Oh............
It becomes a recurring problem -.-
Once you change the order of integration, you can integrate the inner integral first, treating x as a constant:
$\displaystyle \int_0^\infty \frac{1}{1+x^2y^2}\,dy$
substitute $\displaystyle y=\frac{1}{x}\tan{\theta}$, $\displaystyle dy=\frac{1}{x}\sec^2{\theta}\,d\theta$, $\displaystyle \frac{1}{1+x^2y^2}=\cos^2{\theta}$, and converting the limits of integration: $\displaystyle y=0$ to $\displaystyle \infty$ means $\displaystyle \theta=0$ to $\displaystyle \frac{\pi}{2}$.
$\displaystyle \int_0^\frac{\pi}{2} \cos^2{\theta}\frac{1}{x}\sec^2{\theta}\,d\theta =$
$\displaystyle \frac{1}{x}\int_0^\frac{\pi}{2} \,d\theta =$
$\displaystyle \frac{\pi}{2x}$.
Now we can do the outer integral:
$\displaystyle \int_5^7 \int_0^\infty \frac{1}{1+x^2y^2}\,dy\,dx =$
$\displaystyle \int_5^7 \frac{\pi}{2x}\,dx =$
$\displaystyle \frac{\pi}{2}\int_5^7 \frac{dx}{x} =$
$\displaystyle \frac{\pi}{2}(\ln{7}-\ln{5})$.
- Hollywood