{ changing dxdx to dydx }, When ? and How ?

[AbuYusufEg]

for integration in two variables, say x and y

let the original formula have dxdy
when can i change it into dydx? and how to do that ?

And also: Why ?

the only reason for doing that i know till now is for example i have a function that is easy-to-integrate w.r.t. y but not w.r.t x, so i do make dy before dx to integrate the function w.r.t y then x, instead if x then y

But how i do that ? and When ? And Why ( Another reasons .. ) ?

 

[Mute]

In most cases you encounter in practice you'll find that you can interchange the order of integration. The problem arises when the integral does not converge absolutely (that is, integrating the absolute value of the integrand yields an infinite result). See the following page for more information and an example of a function whose values differ by a sign under swapping the order:

http://en.wikipedia.org/wiki/Fubini's_theorem

 

[AbuYusufEg]

In most cases you encounter in practice you'll find that you can interchange the order of integration. The problem arises when the integral does not converge absolutely (that is, integrating the absolute value of the integrand yields an infinite result). See the following page for more information and an example of a function whose values differ by a sign under swapping the order:

http://en.wikipedia.org/wiki/Fubini's_theorem

Can you write some example please ? as i can't get it from the wikipedia's page !

 

[AbuYusufEg]

i'm still needs help here!, I've exam after about 10 hours ..

 

[Mute]

What do you mean you "can't get it from wikipedia's page"? Do you mean the page isn't working or you don't understand the example? I'll assume the latter and explain wikipedia's example:

The example given is

\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx

which the page claims you cannot change the order of integration in order to perform the integral. You can see this directly by noting that if you relabel the variable x by y and the variable y by x, the integrand looks exactly the same except that the terms in the numerator are reversed and the dydx is now dxdy. Pulling a minus sign out of the numerator you see that the integrand is again the same as before except for changing the order of integration - hence by changing the order of integration the value you calculate will change by a minus sign (in THIS particular example - this does NOT mean that in general changing the order of integration for such functions will just change the result by a minus sign).

The reason you can't change the order without changing the value you calculate is that the function you are integrating is not absolutely convergent, meaning

\int_0^1\int_0^1<br /> \left|\frac{x^2-y^2}{(x^2+y^2)^2}\right|\,dy\,dx=\infty

Observe the evaluation done on the page:

\int_0^1\int_0^1<br /> \left|\frac{x^2-y^2}{(x^2+y^2)^2}\right|\,dx\,dy=\int_0^1\left[\int_0^y<br /> \frac{y^2-x^2}{(x^2+y^2)^2}\,dx+\int_y^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\right]\,dy
=\int_0^1\left(\frac{1}{2y}+\frac{1}{2y}-\frac{1}{y^2+1}\right)\,dy=\int_0^1 \frac{1}{y}\,dy-\int_0^1\frac{1}{1+y^2}\,dy.

To quote wikipedia,

When

\int_a^b\int_c^d \left|f(x,y)\right|\,dy\,dx=\infty

then the two iterated integrals

\int_a^b\int_c^d f(x,y)\,dy\,dx\ \mbox{and}\ \int_c^d\int_a^b f(x,y)\,dx\,dy

may have different finite values.

The example given above is one of those cases.

 

[NoMoreExams]

The way you would switch is by drawing a picture and seeing how that would change your bounds. Unless you are integrating to and from actual values, your inner integral goes from a function to a function, which is what you would need to adjust for.