1

I was asked to determine whether the integrals $\int_{0}^{\infty}dy\int_{0}^{\infty}e^{-xy}\sin(x)dx$ and $\int_{0}^{\infty}dx\int_{0}^{\infty}e^{-xy}\sin(x)dy$ converge, and if they do, calculate them.

I'm not sure what this notation means. Is $\int_{0}^{\infty}dy\int_{0}^{\infty}e^{-xy}\sin(x)dx$ supposed to mean $\int_{0}^{\infty}\int_{0}^{\infty}e^{-xy}\sin(x)dxdy$? Or does this literally mean the product of the integrals: $\int_{0}^{\infty} dy\cdot \int_{0}^{\infty}e^{-xy}\sin(x)dx$? where the $y$ in the second integral is just some unknown parameter now, not related to the first integral.

Anyone seen this notation before?

Oria Gruber
  • 12,739

1 Answers1

1

$\int_{0}^{\infty}dy\int_{0}^{\infty}e^{-xy}\sin(x)dx$ means $\int_{0}^{\infty}\left(\int_{0}^{\infty}e^{-xy}\sin(x)dx\right)dy$. It is very common to put the $dx$ (or $dy$) at the end of an integral (thus signifying the end of the integrand). While still common, it is less common to do so in repeated integrals.

Probably, the purpose of this is to make it clearer which integral sign (with bounds) belongs to which variable. In this case both $x$ and $y$ go from $0$ to $\infty$, but that will not always be so, and then it is important to knwo which variable has which bounds. Sticking the $d$ right next to the $\int$ it belongs to makes that easier.

Arthur
  • 199,419
  • and $\int_{0}^{\infty}(\int_{0}^{\infty}e^{-xy}\sin(x)dx)dy = \lim_{a \to \infty} \int_{0}^{a}(\lim_{b \to \infty}\int_{0}^{b}e^{-xy}\sin(x)dx)dy$ right? It's the iterated limits, not the multivariable one $\lim_{(a,b) \to (\infty,\infty)}$ – Oria Gruber Jan 02 '19 at 09:26
  • @OriaGruber Exactly. It's one limit inside another, not a single multivariable limit. The single multivariable limit would correspond to $\iint_X f(x, y)dA$, where $X$ is the relevant region of the plane (in this case the first quadrant). – Arthur Jan 02 '19 at 09:31
  • there's a small problem though: $\int_{0}^{\infty}e^{-xy}\sin(x)dy =\lim_{a \to \infty}\frac{\sin(x)}{x}-\frac{\sin(x)}{x}e^{-ax}$ according to my calculation. Without knowing $x$ what can we do? – Oria Gruber Jan 02 '19 at 09:36
  • 1
    @OriaGruber There shouldn't be any $x$ left over from $\int_{0}^{\infty}e^{-xy}\sin(x)dx$. They ought to disappear, and give you a final integral of the form $\int_0^\infty g(y)dy$ without any $x$ whatsoever. – Arthur Jan 02 '19 at 09:37
  • Im doing the $dy$ first because I think it's easier to integrate. Perhaps I'm wrong. – Oria Gruber Jan 02 '19 at 09:38
  • @OriaGruber You can't really do that. It's akin to swapping the order of the two limits (the ones we discussed above), and swapping the order of limits shouldn't be done without careful consideration. But for argument's sake let's say you want to calculate $\int_{0}^{\infty}\left(\int_{0}^{\infty}e^{-xy}\sin(x)dy\right)dx$ instead. Then the inner integral is the limit you've writtten. Evaluating that to something of the form $g(x)$ shouldn't be too hard. So now you've found that $\int_{0}^{\infty}\left(\int_{0}^{\infty}e^{-xy}\sin(x)dy\right)dx = \int_0^\infty g(x)dx$. Where's the problem? – Arthur Jan 02 '19 at 09:45
  • The problem is we can't evaluate that limit because we don't know whether $x$ is positive negative or zero. – Oria Gruber Jan 02 '19 at 09:47
  • @OriaGruber We know that for almost all of the relevant region, $x$ is positive (in fact, it's improper in both ends, so $x$ is actually positive evrywhere, and then we take limits to $0$ and to $\infty$). – Arthur Jan 02 '19 at 09:48
  • Yes but do we know that? Can we use that? after all, the inner integral ($dy$) is unrelated to the outer integral ($dx$), so when we compute the limit can we use our knowledge of $x$? – Oria Gruber Jan 02 '19 at 09:50
  • @OriaGruber In some sense they are independent, sure. The inner integral will have a value (or not) for any given value of $x$. For $x = 0$, the limit is easily calculated. And for $x>0$, the limit is easily calculated. That gives you a function which is defined for any $x\geq 0$. Since the outer integral doesn't care about $x<0$, there is no problem. – Arthur Jan 02 '19 at 09:54
  • @OriaGruber Sorry, I misspoke. For $x = 0$ the limit doesn't exist, but the integral itself is easily calculated directly. For $x> 0$ the limit is easily calculated. And from there you get your integrand for the outer integral. – Arthur Jan 02 '19 at 10:00
  • Let us continue this discussion in chat. – Oria Gruber Jan 02 '19 at 10:03
  • @OriaGruber I prefer not to go in the chat. I know the site wants it, but I don't. – Arthur Jan 02 '19 at 10:04
  • Suppose then that we can do as you say and turn this into $\int_{0}^{\infty}\frac{\sin(x)}{x}dx$ which is not an easy integral. After a bit of digging around, I found an answer on this site to this integral here https://math.stackexchange.com/questions/1681341/integration-question-sinx-x but it uses change of order of integration, which we don't know we can do yet. I'm struggling here a bit. – Oria Gruber Jan 02 '19 at 10:07
  • @OriaGruber If you read that answer, you will see that what they actually calculate in the end is the integral you were originally given (with a slightly different $y$ bound). Why not use that post for what it's worth? Solve your original problem without any theoretical issues. It even gives you a hint on how to compute the inner $dx$ integral, and which result you should get from it. – Arthur Jan 02 '19 at 10:10
  • I see that, great. I totally agree that $\int_{0}^{\infty}\int_{0}^{\infty}e^{-xy}\sin(x)dxdy$ converges. I don't think that means that $\int_{0}^{\infty}\int_{0}^{\infty}e^{-xy}\sin(x)dydx$ converges. To show it converges, we need to calculate the integral of $\frac{\sin(x)}{x}dx$ and to do that we need to switch the order of integration - which we don't know we can do. – Oria Gruber Jan 02 '19 at 10:13