2
$\begingroup$

I have two questions.

First:

Is my proof "strong" enough?

I am being asked to prove that $$\int_{0}^\infty\int_{0}^x e^{-sx}f(x-y,y) dydx = \int_{0}^\infty\int_{0}^\infty e^{-s(u+v)}f(u,v) dudv$$

where $u =x-y$ and $v = y$.

We will rewrite the left side:$$\int_{0}^\infty\int_{0}^x e^{-sx}f(x-y,y) dydx = \lim \limits_{n \to \infty}\int_{0}^n\int_{0}^xe^{-sx}f(x-y,y) dydx$$

Since $u + v = (x-y) + y = x$ and $v = y$, we can define the following transformation: $$(x,y) = T(u,v) = (u+v,v)$$

The Jacobian of T is : $$\frac{\partial(x,y)}{\partial(u,v)} =\begin{vmatrix} 1 & 1 \\ 0 & 1\\ \end{vmatrix} = 1.$$

If we consider the $XY$ plane, we will notice that we are integrating over a triangle whose legs, which are parallel to the coordinate axis, are n units long. Thus, $y\in[0,n] \Leftrightarrow v\in[0,n]$. We can also observe that the segment, $x=n$ and $y\in[0,n]$, is generated by evaluating T when by the line $u = -v$. This last statement reveals to us that if $v\in[0,n]$ then $u\in[0,n-v]$.

By the Change of Variables Theorem: \begin{align}\lim \limits_{n \to \infty}\int_{0}^n\int_{0}^xe^{-sx}f(x-y,y) dydx & = \lim \limits_{n \to \infty}\int_{0}^n\int_{0}^{n-v}e^{-s(u+v)}f((u+v)-v,v) * \frac{\partial(x,y)}{\partial(u,v)} dudv\\ & = \lim \limits_{n \to \infty}\int_{0}^n\int_{0}^{n-v}e^{-s(u+v)}f(u,v) dudv\\ & = \int_{0}^\infty\int_{0}^{\infty-v}e^{-s(u+v)}f(u,v) dudv\\ & = \int_{0}^\infty\int_{0}^{\infty}e^{-s(u+v)}f(u,v) dudv\\ \end{align}


Second:

I am asked to determine if the improper integral $$\int\int \frac{1}{x^4 + y^2} dA$$ over the region R = $\text{{(x,y) : x $\ge$ 1, y $\ge x^2$}}$ is convergent, and find calculate its value if it is convergent.

I know that the given improper integral is convergent. However, I have been having trouble proposing a region $B_{n}$ such that $B_{n} \subset B_{n+1}$ and $Area(R) = \lim \limits_{n \to \infty}Area(B_{n})$. The $B_{n}$ I have proposed thus far have been $$B_{n} = \text{{(x,y) : $x\in[1,n]$, $y\in[x^2,n^2]$}}$$ and $$B_{n} = \text{{(x,y) : $y\in[1,n]$, $y\in[1,\sqrt{x}]$}}$$

Basically, the first $B_{n}$ represents an approximation of R as a type I region while the second one approximates R as a type II region. The second one presents a few problems when we try to integrate with respect to $y$, which is why I have tried to use the first one.

Using the first $B_{n}$, we have: \begin{align}\int_{1}^n\int_{x^2}^{n^2} \frac{1}{x^4 + y^2} dydx & = \int_{1}^n \frac{1}{x^2} arctan(\frac{n^2}{x^2}) - \frac{1}{x^2} arctan(\frac{x^2}{x^2}) dx\\ & = \int_{1}^n \frac{1}{x^2} arctan(\frac{n^2}{x^2}) - \frac{pi}{4} \frac{1}{x^2} dx \end{align}. This is where I get stuck. Any recommendations?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.