What Are the New Limits of Integration for the Rare Integral with Substitution?

[eljose]

let be the integral:

$$\int_{c-i\infty}^{c+i\infty}\int_0^ag(xy)ydxdy$$

then i make the substitution xy=u y=v then what would be the new limits of integration? (the y are the same than v but what would happen with the u?)

 

[Zurtex]

I've only been doing double integrals for half a year now, but without knowing g(xy) then surely there is no way to tell?

 

[HallsofIvy]

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:
$$\int_{c-i\infty}^{c+i\infty}\int_0^{av}g(u)\frac{u+v}{v}dudv$$

 

[Zurtex]

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:
$$\int_{c-i\infty}^{c+i\infty}\int_0^{av}g(u)\frac{u+v}{v}dudv$$

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 courses

 

[quetzalcoatl9]

i believe that the Jacobian det is

$$\frac{1}{v}$$

and the function becomes

$$g(u) v$$

so the integral should be

$$\int_{c-i\infty}^{c+i\infty} \int_{0}^{av} g(u) du dv$$

 

[Hurkyl]

It's much easier (and avoids any issues that may arise from the outer integral being improper and complex) to simply recognize that we're only doing a substitution on the inner integral, and the outer integral is just a relabelling of the dummy variable.

 

[saltydog]

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.

 

[eljose]

First of all thanks to all who have answered my question :)

and what would happen for the integral.

$$\int_1^{\infty}\int_0^a g(xy)h(y)dydx$$ making the change xy=u y=v

would it be $$\int_1^{\infty}\int_0^{av} g(u)h(v)dudv/v$$ ?

 

[saltydog]

First of all thanks to all who have answered my question :)

and what would happen for the integral.

$$\int_1^{\infty}\int_0^a g(xy)h(y)dydx$$ making the change xy=u y=v

would it be $$\int_1^{\infty}\int_0^{av} g(u)h(v)dudv/v$$ ?

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.

 

[saltydog]

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

$$\int_0^1\int_0^1\frac{(xy)^2}{2}ydxdy=\int_0^1\int_0^v\frac{u^2}{2}dudv$$

$$\int_0^1\int_0^1 e^{-(xy)}ydxdy=\int_0^1\int_0^v e^{-u}dudv$$

I'll spend time reviewing the transformations. Thanks.

 

[saltydog]

First of all thanks to all who have answered my question :)

and what would happen for the integral.

$$\int_1^{\infty}\int_0^a g(xy)h(y)dydx$$ making the change xy=u y=v

would it be $$\int_1^{\infty}\int_0^{av} g(u)h(v)dudv/v$$ ?

Hello Eljoise. I get:

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

For example:

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

 

[eljose]

what about the change xy=u y=v with:

The following code was used to generate this LaTeX image:



$$\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx$$

and $$\int_{c-i\infty}^{c+i\infty}\int_0^{\infty} e^{-(xy)}ydydx$$

 

[saltydog]

what about the change xy=u y=v with:

The following code was used to generate this LaTeX image:



$$\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx$$

and $$\int_{c-i\infty}^{c+i\infty}\int_0^{\infty} e^{-(xy)}ydydx$$

Hey Eljoise, I get:

$$\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx=\int_1^{\infty}\int_v^{\infty} e^{-u}dudv$$

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.

 

[eljose]

thansk saltydog,i have checked several integral and get almost the same result as yours but this:

$$\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx=\int_1^{\infty}\int_0^{\infty} e^{-u}dudv$$

 

[James Jackson]

There's nothing to 'not work' in the substitutions, other than mistakes in the algebra (which are, unfortunately, all too easy to make).

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 into - ever wondered easy routes to evaluate horrific quotiant polynomial / trigonometric expressions? Complex analysis (sometimes) has the answer, in very mathematically beautiful forms.

 

[saltydog]

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 into - 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.

 

[saltydog]

thansk saltydog,i have checked several integral and get almost the same result as yours but this:

$$\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx=\int_1^{\infty}\int_0^{\infty} e^{-u}dudv$$

I don't get that Eljoise but rather:

$$\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx=\int_1^{\infty}\int_v^{\infty} e^{-u}dudv$$

Of course the Jacobian determinant is 1/v which cancels the y. Here's the switch of variables:

$$y\rightarrow(0,\infty)\quad\text{so}\quad v\rightarrow(1,\infty)$$

Now:

$$u=xv$$

with v never zero. So:

when x=1, u=v and when

$$x=\infty\rightarrow u=\infty$$

 

[eljose]

and what about $$\int_1^{\infty}\int_{-\infty}^{\infty}e^{-xy}ydydx$$

with the change of variable xy=u y=v what would be then the new integral.. ...thanx

 

[saltydog]

I don't get that Eljoise but rather:

$$\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx=\int_1^{\infty}\int_v^{\infty} e^{-u}dudv$$

Of course the Jacobian determinant is 1/v which cancels the y. Here's the switch of variables:

$$y\rightarrow(0,\infty)\quad\text{so}\quad v\rightarrow(1,\infty)$$

Now:

$$u=xv$$

with v never zero. So:

when x=1, u=v and when

$$x=\infty\rightarrow u=\infty$$

Dang it. I hate when that happens. I have an error above. It should read:

$$y\rightarrow(0,\infty)\quad\text{so}\quad v\rightarrow(0,\infty)$$

$$\int_1^{\infty}\int_0^{\infty} e^{-(xy)}ydydx=\int_0^{\infty}\int_v^{\infty} e^{-u}dudv$$

 

[saltydog]

and what about $$\int_1^{\infty}\int_{-\infty}^{\infty}e^{-xy}ydydx$$

with the change of variable xy=u y=v what would be then the new integral.. ...thanx

The inner integral doesn't converge right? You know, minus-minus infinity.

 

[eljose]

let,s supose the integral: $$\int_1^{\infty}\int_{-\infty}^{\infty}F(x/y)ydydx$$ converge,then we make the change of variable x/y=u y=v what would be the new limits?..thanx..