Jacobian: how to change limits of integration?

[fatpotato]

Homework Statement

Find the new region of integration under a given change of variables.

Relevant Equations

Jacobian matrix and determinant

Hello,

I have to compute a double integral of the form $\int_{0}^{\infty} \int_{0}^{\infty} f(u,v) du dv$, where $f(u,v)$ is not relevant. The following change of variable is advised as a hint: $u = zt$ and $v = z(1-t)$.

From there, I can reformulate with respect to $z$ and $t$: $z = u+v$, $t = \frac{u}{u+v}$, but I am absolutely unable to find the limit of integrations of the new region, or to find someone that can clearly explain to me how to find the new bounds.

Sure, I can guess that the new region for $z$ is $[0, \infty[$, but how would one find the new region for $t$? Presence of both $u$ and $v$ in the denominator makes it awkward to work with limits.

I suppose that I could try finding $\lim_{(u,v) \rightarrow (\infty, \infty)} u + v = \infty$ for the $z$ variable, but this method does not work for $t$, as the limit is not defined (or at least, I cannot find it).

Can someone please help me? From several threads I looked at, either here or on Stackexchange, it seems that there is no systematic way of finding the new region, but there has to be at least a few heuristics one can follow to, isn't there?

 

[Orodruin]

Look at the expression for $t$ in terms of $u$ and $v$ and keep in mind that the original variables are both positive.

 

[pasmith]

Observe that $$(u,v) = (0,z) + (z,-z)t.$$ For fixed $z$, this is a line in the $(u,v)$ plane from $(0,z)$ at $t = 0$ to $(z,0)$ at $t = 1$. Thus the new region of integration is $(z,t) \in [0, \infty) \times [0,1]$.

 

[fatpotato]

Look at the expression for $t$ in terms of $u$ and $v$ and keep in mind that the original variables are both positive.

Yes, so from this information, I am only considering the first quadrant in the $(u,v)$ plane. Next, I tried to draw level curves for $v = z-u$ but I am not going anywhere. Which step would you take next?

Observe that $$(u,v) = (0,z) + (z,-z)t.$$ For fixed $z$, this is a line in the $(u,v)$ plane from $(0,z)$ at $t = 0$ to $(z,0)$ at $t = 1$. Thus the new region of integration is $(z,t) \in [0, \infty) \times [0,1]$.

While I understand your explanation, I have no idea how you figured this, could you please guide me through your thought process so I can apply this knowledge next time? I guess that you are making some sort of a 2D parametrization with the parameter $t$ right? I want to understand why, for example, you did not bother with limits at all, which is correct since I tried and failed, but how did you know?

Thank you both for answering.

 

[Orodruin]

Yes, so from this information, I am only considering the first quadrant in the (u,v) plane.

You knew that from u and v being positive. The question was what you can conclude about the quantity $t = u/(u+v) = 1/(1+v/u)$ when u and v are both positive.

 

[fatpotato]

You knew that from u and v being positive. The question was what you can conclude about the quantity $t = u/(u+v) = 1/(1+v/u)$ when u and v are both positive.

Oh right, nice circular reasoning from my part...Thank you for pointing it out.

The problem is that with $t = \frac{1}{1+ \frac{v}{u}}$ I struggle to extract anything. With a fixed $v$, $t=1$ when $u$ goes to infinity, but with a fixed $u$, $t=0$ when $v$ goes to infinity...

 

[Orodruin]

Oh right, nice circular reasoning from my part...Thank you for pointing it out.

The problem is that with $t = \frac{1}{1+ \frac{v}{u}}$ I struggle to extract anything. With a fixed $v$, $t=1$ when $u$ goes to infinity, but with a fixed $u$, $t=0$ when $v$ goes to infinity...

You don’t need to worry about which of u and v you fix. Look at the latter form of the expression. What are the possible values of v/u if u and v both can take any positive values?

 

[Orodruin]

Or, put slightly differently, you also need to look at how t behaves in the limit when you fix one of u and v and the other goes to zero as these are the other u and v integration bounds.

 

[fatpotato]

You don’t need to worry about which of u and v you fix. Look at the latter form of the expression. What are the possible values of v/u if u and v both can take any positive values?

Right, so indeed by looking at the ratio $v/u$, this will only result in values in $[0, \infty[$, so $\frac{1}{1 + \frac{v}{u}}$ can only take values in $[0, \infty[$, got it!

Or, put slightly differently, you also need to look at how t behaves in the limit when you fix one of u and v and the other goes to zero as these are the other u and v integration bounds.

So now I would have $\lim_{u \rightarrow \infty} \frac{u}{u+v} = 1$ and $\lim_{v \rightarrow \infty} \frac{u}{u+v} = 0$, but there is still something I do not understand...

How do we know that the integration bounds are $\int_0^1 dt$ and not the other way around, $\int_1^0dt$? In single variable calculus, this is immediate while doing "u-substitution" since the relationship is straightforward, but how would we figure this out here?

 

[pasmith]

While I understand your explanation, I have no idea how you figured this, could you please guide me through your thought process so I can apply this knowledge next time? I guess that you are making some sort of a 2D parametrization with the parameter $t$ right? I want to understand why, for example, you did not bother with limits at all, which is correct since I tried and failed, but how did you know?

This is the thought process which led me to that change of variable in the case $$\int_0^\infty \int_0^\infty u^{p-1}v^{q-1} e^{-(u + v)}\,du\,dv.$$ Clearly I want something to be $u + v$, but the only part of the line $u + v = z$ that falls in the domain of integration is the segment between $(0,z)$ and $(z,0)$; setting $(u,v) = (0,z)(1-t) + (z,0)t = (zt,z(1-t))$ with $0 \leq t \leq 1$ parametrizes this segment.

In general when presented with a proposed change of variable I take a geometric approach: What does a curve of constant $t$ look like? Here it's a line through the origin. Fortunately the boundary of our domain of integration consists of two such lines, so what values does $t$ take on them?

 

[Orodruin]

Right, so indeed by looking at the ratio $v/u$, this will only result in values in $[0, \infty[$,

So far so good.

so $\frac{1}{1 + \frac{v}{u}}$ can only take values in $[0, \infty[$, got it!

Either you wrote something different from what you intended or you need to think again here.

 

[WWGD]

Haven't been following 100% , but your left formula below is defined for all $u \neq 0$ ( as long as $v \neq 0$ ), while the one on the RH side is not.

Oh right, nice circular reasoning from my part...Thank you for pointing it out.

The problem is that with $t = \frac{1}{1+ \frac{v}{u}}$ I struggle to extract anything. With a fixed $v$, $t=1$ when $u$ goes to infinity, but with a fixed $u$, $t=0$ when $v$ goes to infinity...

 

[SammyS]

I have to compute a double integral of the form $\int_{0}^{\infty} \int_{0}^{\infty} f(u,v) du dv$, where $f(u,v)$ is not relevant. The following change of variable is advised as a hint: $u = zt$ and $v = z(1-t)$.

From there, I can reformulate with respect to $z$ and $t$: $z = u+v$, $t = \frac{u}{u+v}$, but I am absolutely unable to find the limit of integrations of the new region, or to find someone that can clearly explain to me how to find the new bounds.

It looks like you're still struggling with the resulting bounds, especially in regards to the variable $t$ .

It appears that you realize that $u,\,v, \text{ and} ,\, z$ are all non-negative. Hopefully, then you also see that $t$ is also non-negative.

A couple of easy ways to get un upper bound for $t$.

$v = z(1-t)$, so $1-t\ge 0$. Right what does this say about $t$ ?

Also, $(u+v)\ge u$, so what does this say regarding $t$ in the form of $\dfrac{u}{u+v}$ ?

Maybe next, we look at level curves.

 

[Steve4Physics]

How do we know that the integration bounds are $\int_0^1 dt$ and not the other way around, $\int_1^0dt$? In single variable calculus, this is immediate while doing "u-substitution" since the relationship is straightforward, but how would we figure this out here?

EDIT. What I wrote in the post below is likely wrong (as pointed out in Post #16). Anyone reading may want to skip it.
____________
May I join in? Perhaps there is a problem with the suggested transformation $z = u+v$ and $t = \frac u{u+v}$.

$(u, v) → (z,t)$
For small changes: $(u+δu, v+δv) → (z+δz,t+δt)$

Consider the transformations of (for example) the point $(u=100, v=100)$ and some nearby points.

$(100,100)→ (100+100, \frac {100}{100+100}) = (200, 0.5)$
$(101,100)→ (101+100, \frac {101}{101+100}) ≈(201,0.5025)$
$(100,101)→(100+101, \frac {100}{100+101}) ≈(201, 0.4975)$

Note that an increase in $u$ alone, moving from $(100, 100)$ to $(101,100)$, gives correpoding increases in both $z$ and $t$.

An increase in $v$ alone, moving from $(100, 100)$ to $(100,101)$, gives a corresponding increase in $z$ but a decrease in $t$.

So $δt$ can be positive or negative for positive values of $δu$ and $δv$. This introduces ambiguity. The direction of integration, $\int_0^1 dt$ or $\int_1^0 dt$, can't be decided.

As a non-mathematician I may have misunderstood something and, if so, welcome correction.

 

[pasmith]

How do we know that the integration bounds are $\int_0^1 dt$ and not the other way around, $\int_1^0dt$? In single variable calculus, this is immediate while doing "u-substitution" since the relationship is straightforward, but how would we figure this out here?

We don't; we're looking for an expression for the unsigned area element $du\,dv = |J|\,dz\,dt$.

 

[Orodruin]

So $δt$ can be positive or negative for positive values of $δu$ and $δv$. This introduces ambiguity. The direction of integration, $\int_0^1 dt$ or $\int_1^0 dt$, can't be decided.

As a non-mathematician I may have misunderstood something and, if so, welcome correction.

By this argument we cannot do integrals in polar coordinates.

 

[Steve4Physics]

By this argument we cannot do integrals in polar coordinates.

Aha. Interesting. I'll need to go away and think about that.