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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]

[tex]\int_{c-i\infty}^{c+i\infty}\int_0^{av}g(u)\frac{u+v}{v}dudv[/tex]

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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]

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[tex]\frac{1}{v}[/tex]

and the function becomes

[tex]g(u) v[/tex]

so the integral should be

[tex]\int_{c-i\infty}^{c+i\infty} \int_{0}^{av} g(u) du dv [/tex]

Me personally, I'd scrap the complex part, check it with just a real double integral, make the substitutions you guys are talking about using a real integrand, calculate the integral, and then do it another way and verify the two answers agree. I'd request this be done here but don't want you guys jumping on me.

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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] ?

I tell you what eljoise, you guys can discuss all these substitutions as you wish, but until I see a real example with a real integrand worked and comparred with the answer worked using no substitution, either here, or I'll probably do one myself, I'm just flat-out not convinced these substitutions work.

Alright, I'm convinced. Here, I'll put two in LaTex for punish work for doubting you guys:

[tex]\int_0^1\int_0^1\frac{(xy)^2}{2}ydxdy=\int_0^1\int_0^v\frac{u^2}{2}dudv[/tex]

[tex]\int_0^1\int_0^1 e^{-(xy)}ydxdy=\int_0^1\int_0^v e^{-u}dudv[/tex]

I'll spend time reviewing the transformations. Thanks.

[tex]\int_0^1\int_0^1\frac{(xy)^2}{2}ydxdy=\int_0^1\int_0^v\frac{u^2}{2}dudv[/tex]

[tex]\int_0^1\int_0^1 e^{-(xy)}ydxdy=\int_0^1\int_0^v e^{-u}dudv[/tex]

I'll spend time reviewing the transformations. Thanks.

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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] ?

[tex]\int_1^{\infty}\int_0^a g(xy)h(y)dydx=\int_0^1\int_v^{\infty}\frac{g(u)h(v)}{v}dudv[/tex]

For example:

[tex]\int_1^{\infty}\int_0^1 e^{-(xy)}y^2dydx=\int_0^1\int_v^{\infty}e^{-u}vdudv[/tex]

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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]

[tex]\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx=\int_1^{\infty}\int_v^{\infty} e^{-u}dudv[/tex]

The complex one, can't help. Wish I could take a class because I consider them elegant and would like to know more and also, Hurkyl tells me Complex Analysis holds the keys to the secrets of the Universe.

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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.

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.

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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]

[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]

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]

[tex]y\rightarrow(0,\infty)\quad\text{so}\quad v\rightarrow(0,\infty)[/tex]

[tex]\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx=\int_0^{\infty}\int_v^{\infty} e^{-u}dudv[/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

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