Evaluating Double Integral Involving Imaginary Error Function

[lockedup]

Homework Statement

Evaluate
\int\int x^{2}e^{x^{2}y} dx dy
over the area bounded by y=x^{-1}, y=x^{-2}, x=ln 4

Homework Equations

The Attempt at a Solution

\int^{1}_{(ln 4)^{-2}}\int^{y^{-1}}_{y^{\frac{-1}{2}}}x^{2}e^{x^{2}y}dx dy

I got this far before I realized that this wasn't a straightforward integral. There is nothing like it in the tables. I put \int x^{2}e^{x^{2}y} dx into Mathematica's online integrator and I got something involving the imaginary error function... Help please?

 

[Mark44]

Homework Statement

Evaluate
\int\int xe^{x^{2}y} dx dy
over the area bounded by y=x^{-1}, y=x^{-2}, x=ln 4

Homework Equations

The Attempt at a Solution

\int^{1}_{(ln 4)^{-2}}\int^{y^{-1}}_{y^{\frac{-1}{2}}}x^{2}e^{x^{2}y}dx dy

There are several mistakes here.
1. Your integrand here is different from the one you wrote at the beginning of the post. You switched from xe^x2y to x²e^x2y. The first one will probably be easier to integrate.
2. Your limits of integration don't agree with the order of integration. The limits of integration on the inner integral need to be x values, because you are going to integrate with respect to x first, then y later.

The integral should look something like this.
\int_{y = ?}^1\int_{x = ?}^? xe^{x^{2}y}dx dy

If you haven't done so already, draw a careful sketch of the region over which integration takes place. You need to find where the two curves intersect, since that point defines one of your limits of integration. From your sketch you should be able to fill in the ? placeholders I have in my revision of your integral.

I got this far before I realized that this wasn't a straightforward integral. There is nothing like it in the tables. I put \int xe^{x^{2}y} dx into Mathematica's online integrator and I got something involving the imaginary error function... Help please?

 

[LCKurtz]

Adding to Mark44's comments, you haven't told us whether x = ln(4) is the right or left boundary nor what the other x boundary is, although I don't agree with Mark that you inner x limits are wrong. And the two curves don't intersect.

 

[lockedup]

It is supposed to be x^2. Sorry about that.

The limits of the second integrand are x=1/y and x = y^-1/2. I just didn't put the "x =" part in there.

I did draw a sketch. x = 1/y is on top. They intersect at 1,1. So the values of the first integrand should be 1 on top and y = (ln 4)^-2 on bottom.

Right?

 

[Mark44]

y = 1/x and y = 1/x^2 intersect at (1, 1).

I take back what I said about the x limits of integration. I mistook them for y values.

The upshot of this is that the limits of integration look fine.

 

[LCKurtz]

y = 1/x and y = 1/x^2 intersect at (1, 1).

I take back what I said about the x limits of integration. I mistook them for y values.

Heh heh. And I take back that the curves don't intersect.

 

[LCKurtz]

To the OP. I don't think you want to integrate in the x direction first, or at least, if you do, you must break the problem into two parts as the right boundary is not just y = 1/x². The upper boundary is also part of the right boundary. But I have to run...the wife calls for dinner