I'm a little confused, you don't seem to have applied the appropriate substitution rules when transforming an integral, like multiplying by the determinate of J. Did I miss something in my coursesHallsofIvy said:Since y= v, xy= xv= u so x= u/v. x= u/v= 0 becomes u= 0, x= u/v= a becomes u= av. The integral becomes:
[tex]\int_{c-i\infty}^{c+i\infty}\int_0^{av}g(u)\frac{u+v}{v}dudv[/tex]
eljose said:First of all thanks to all who have answered my question :)
and what would happen for the integral.
[tex] \int_1^{\infty}\int_0^a g(xy)h(y)dydx[/tex] making the change xy=u y=v
would it be [tex] \int_1^{\infty}\int_0^{av} g(u)h(v)dudv/v [/tex] ?
Hello Eljoise. I get:eljose said:First of all thanks to all who have answered my question :)
and what would happen for the integral.
[tex] \int_1^{\infty}\int_0^a g(xy)h(y)dydx[/tex] making the change xy=u y=v
would it be [tex] \int_1^{\infty}\int_0^{av} g(u)h(v)dudv/v [/tex] ?
Hey Eljoise, I get:eljose said:what about the change xy=u y=v with:
The following code was used to generate this LaTeX image:
[tex]\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx[/tex]
and [tex]\int_{c-i\infty}^{c+i\infty}\int_0^{\infty} e^{-(xy)}ydydx[/tex]
Ok James. You're right. Allow me some defense though: My suggestion of "scrapping" the complex part is only an initial effort, just to get something working that's easy, then go to the hard part. Yes, I agree. It is beautiful.James Jackson said:To those requesting scrapping the complex part and comparing with a double real integral, I'm afraid you can't just write off the complex part because you don't like working with complex numbers! Complex analysis is indeed very beautiful mathematics, and worth looking in to - ever wondered easy routes to evaluate horrific quotiant polynomial / trigonometric expressions? Complex analysis (sometimes) has the answer, in very mathematically beautiful forms.
I don't get that Eljoise but rather:eljose said:thansk saltydog,i have checked several integral and get almost the same result as yours but this:
[tex]\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx=\int_1^{\infty}\int_0^{\infty} e^{-u}dudv[/tex]
Dang it. I hate when that happens. I have an error above. It should read:saltydog said:I don't get that Eljoise but rather:
[tex]\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx=\int_1^{\infty}\int_v^{\infty} e^{-u}dudv[/tex]
Of course the Jacobian determinant is 1/v which cancels the y. Here's the switch of variables:
[tex]y\rightarrow(0,\infty)\quad\text{so}\quad v\rightarrow(1,\infty)[/tex]
Now:
[tex]u=xv[/tex]
with v never zero. So:
when x=1, u=v and when
[tex]x=\infty\rightarrow u=\infty[/tex]
The inner integral doesn't converge right? You know, minus-minus infinity.eljose said:and what about [tex]\int_1^{\infty}\int_{-\infty}^{\infty}e^{-xy}ydydx [/tex]
with the change of variable xy=u y=v what would be then the new integral.. ...thanx