# Can anyone help me with these integral?

### 1 Answer

- az_lenderLv 74 weeks agoFavorite Answer
(2) Think of the integral as the volume over the unit square of an object whose lower boundary is at z = 0 and whose upper boundary is at

z = (x - y)*sin(xy)/(x^2 + y^2). You might have to investigate whether the volume is finite, but it clearly does not matter whether the volume is evaluated by taking slices parallel to the xz-plane or slices parallel to yz-plane. Slicing parallel to the xz-plane means doing the "dx" integration first; slicing parallel to the yz-plane means doing the "dy" integration first.

(3) I'll do the "dx" integration first. The indefinite integral of

xe^(-kx^2) dx

would best be attacked by letting u = kx^2, du = 2kx dx, so you'd have a new integrand of

(1/(2k))e^(-u) du

and the indefinite integral would be

-(1/(2k))e^(-u) = -(1/(2k))e^(-kx^2).

At x = infinity you get zero; at x = 0 you get -1/(2k), but it is to be subtracted, so it's 1/(2k).

In your problem the "k" would be (1+y^2), so now you have

the integral from y = 0 to infinity of

[1/(2+2y^2)] dy.

Note that the exponential disappeared entirely!

Looks like you're going towards some answer with arctangent in it. Shouldn't be too hard?

(1) I don't know.