Is this function integrable?

**HenryB** · March 7th 2009, 05:45 AM

Is the function $f(x,y) = \exp(-xy)$ integrable over the region

$\{(x,y) : 0 < x < y < x+x^2\}$ ?

I can show that f is not integrable over the whole positive quadrant, but this seems a lot tricker...I'm not even sure what the answer "should" be (I'm assuming it is integrable though). I guess we could use comparison though or some other technique or trick.

**Mentia** · March 7th 2009, 08:01 AM

So integrate over y first:

$\int_{ x}^{x+x^2 } e^{-xy} =$

$\frac{e^{-x^2 (x+1)} \left(-1+e^{x^3}\right)}{x}$

Now show that this goes to zero faster than something you know is integrable from zero to infinity, like E^-(x+1)

Limit x->Infinity of $\frac{ \left ( \frac{e^{-x^2 (x+1)} \left(-1+e^{x^3}\right)}{x} \right ) }{e^{-(x+1)} }$

Do L'Hoptial on the numerator once, then bring down the negative powers and you get an E^(x^3) on top and an E^(x^4) on the bottom. Then this will converge to zero.

Then the integral is integrable.

I suppose there are also problems at zero, but take the limit of the function at zero and you get zero so all is well.

**Mentia** · March 7th 2009, 08:04 AM

I made a mistake, you dont get E^(x^4), thats obvious. It will still converge though due to the other terms. There might be something better to compare it to though besides E^-(x+1).

**Mentia** · March 7th 2009, 08:22 AM

It is even more clear if you just write it as:

$\frac{e^{-x^2 (x+1)} \left(-1+e^{x^3}\right)}{x} = - \frac{1 }{ xe^{x^{2}(1+x)}} + \frac{ 1}{ xe^{x^{2}}}$

and then just compare it to 1/(x^2) or something similar.

**HenryB** · March 7th 2009, 09:00 AM

Comparison to $1/x^2$ will only show integrability over $[1,\infty)$ . I think the real difficulty is showing integrability over $(0,1)$ . Indeed I'm not even sure if it integrable over this interval.

In fact all someone needs to be able to do is to prove whether or not $1/(x\exp(x^2))$ is integrable over $(0,1)$ and then we'll be done.

**Amanda1990** · March 7th 2009, 09:21 AM

I think you can show that it isn't integrable by making the substitution $x^2 = t$ and then using the fact that $1/(t\exp(t))$ is not integrable.

**Laurent** · March 7th 2009, 09:32 AM

Originally Posted by Mentia

I suppose there are also problems at zero, but take the limit of the function at zero and you get zero so all is well.

Originally Posted by HenryB

Comparison to $1/x^2$ will only show integrability over $[1,\infty)$ . I think the real difficulty is showing integrability over $(0,1)$ . Indeed I'm not even sure if it integrable over this interval.

Mentia is right, there should be no problem at 0: $\int_0^1 \int_x^{x+x^2} e^{-xy} dy dx$ amounts to integrating a continuous function on a compact subset of $\mathbb{R}^2$ , so it is finite, without computation. It only remains to show that $\int_1^\infty \int_x^{x+x^2} e^{-xy} dy dx$ is finite, and a comparison with $\frac{1}{x^2}$ works fine. In order to simplify the computation, you can even say: for $x\geq 1$ ,

$0\leq \int_x^{x+x^2} e^{-xy} dy\leq \int_x^\infty e^{-xy} dy = \frac{e^{-x^2}}{x}\leq \frac{C}{x^2}$ ,

where $C$ is some constant. So that the integral on $x\in[1,+\infty)$ converges.

Just for the sake of not letting a question answered: the function $\frac{e^{-x^2}}{x}$ is clearly not integrable on $(0,1]$ : the numerator converges to 1 as $x\to0^+$ , hence the function is greater than $\frac{1/2}{x}$ for small enough $x$ (or we can say we have the asymptotic equivalence $\frac{e^{-x^2}}{x}\sim_{x\to 0^+} \frac{1}{x}$ ), and $\frac{1}{x}$ is not integrable on $(0,1]$ .
So how come I proved the function is integrable ? Just because there is another term than $\frac{e^{-x^2}}{x}$ , and their difference tends to 0 when $x\to 0$ : no divergence.

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