
2 Attachment(s)
improper integral
Hello.
Attachment 26906
I don't know how to solve the above improper integral.
I tried to get rid of dy, but turn out to be Attachment 26907
which is even more difficult to me.
I don't know how to get rid of x^2y^2 in the denominator.
Can any teach me how to solve this question?

Re: improper integral
If you could change the order of integration, you would have $\displaystyle \int_5^7 \int_0^\infty \frac{1}{1+x^2y^2}\,dy\,dx$. Can you integrate that?
 Hollywood

Re: improper integral
Consider the inner integral first that is integral of 1 /(1 + x^2y^2) w r t y, thereafter proceed to integrate w r t x.

Re: improper integral
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 .

Re: improper integral
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