Evaluating a double integral with an e^(x^2) term

1. Mar 27, 2013

cwbullivant

1. The problem statement, all variables and given/known data

Evaluate ∫(0-1)∫(sqrt(y)-1) (ye^(x^2))/x^3 dx dy

2. Relevant equations

3. The attempt at a solution

First, factor out the y for the inner integral, making

∫(0-1) y∫(sqrt(y)-1) (e^(x^2))/x^3 dx dy

And evaluating the inner integral first:

y∫(sqrt(y)-1) (e^(x^2))/x^3 dx

And I'm not sure where to go with this one... I initially suspected integration by parts, which breaks down to:

y[-(e^(x^2))/2x^2 - ∫-(e^(x^2))/x dx]

But I have no idea how to integrate that second integral. U-substitution doesn't work as the x is in the wrong place, parts leads to a ∫(-2x)(ln|x|)(e^(x^2))dx term, and none of the other techniques I can think of seem applicable. I would try switching the order of the integrals, but given the sqrt(y) bound on the dx integral, I'm not sure that would lead to a coherent answer.

Apologies if it's difficult to read, but I don't know how to use latex yet...

2. Mar 27, 2013

SammyS

Staff Emeritus
You're not going to get far integrating e-x2 .

Sketch the area of integration, and change the order of integration.

Last edited: Mar 28, 2013
3. Mar 28, 2013

cwbullivant

I reversed the order of integration, and the single integral after integrating the dy integral is

∫(sqrt(y)-1) ex2/2x3 dx

Which seems to be just as problematic as the first integral, as the only difference after using parts is that the next integral will have a -1/2x rather than -1/x

4. Mar 28, 2013

SammyS

Staff Emeritus
I gather that your original integral was
$\displaystyle \int_0^1\int_{\sqrt{y}}^{1}\,y\frac{e^{x^2}}{x^3}dx\,dy$​

What do you get for the limits of integration.for y and then for x after swichng the order of integration?

5. Mar 28, 2013

cwbullivant

Though I'm not 100% sure I understand the question, the limits for y seem like they should still be 0-1, and I'm not sure how to change them for x after switching the order (I didn't know they would change at all, as I originally thought the y term in the bounds of the original integral made it impossible to change them).

I have sketched the area, and do see the area to be integrated, though I'm not sure where to go from there.

EDIT: Do the x bounds become 0 to 1? Given that sqrt(y) and x = 1 meet at y = 1 (and the coordinates would be (1,1)), and sqrt(y) exists only in quadrant I, it appears to make sense.

Last edited: Mar 28, 2013
6. Mar 28, 2013

SammyS

Staff Emeritus
Have you sketched the region of integration for the original integral ? It's essential that the region of integration is the same, no matter which order of integration is used.

Solve $\ x = \sqrt{y\,}\$ for y .

Yes, x will go from 0 to 1 .

7. Mar 28, 2013

cwbullivant

It sounds like the y bounds would need to change as well... Would I want to leave them at 0 to 1, or perhaps change it to 0 to e(x2)?

8. Mar 28, 2013

Staff: Mentor

ex2 is part of the integrand - it has nothing to do with the region over which integration is taking place.

Can you describe, in words, what this region looks like?

9. Mar 28, 2013

SammyS

Staff Emeritus
In an integral such as
$\displaystyle \int_0^1f(y)\left(\int_{\sqrt{y}}^{1}\,f(x)\,dx\,\right)dy$
what does it mean for the limits of integration for x to be $\displaystyle \ \sqrt{\,y\phantom{I}\,}\$ and $\displaystyle \ 1\ ?$

10. Mar 28, 2013

cwbullivant

That the x component goes from left to right, from 0 to $\displaystyle \ \sqrt{\,y\phantom{I}\,}\$, stopping at 1 (Presumably simultaneous with the infinitesimal increments that bring y from 0 to 1)?

I don't think I have it sketched correctly, but I can try. What I have looks like a very small incline in the first quadrant between x = 0 and x = 1.

11. Mar 28, 2013

SammyS

Staff Emeritus
No.

It means that x goes from $\displaystyle \ \sqrt{\,y\,}\$ to 1 .

I still haven't seen your result for solving $\displaystyle \ x=\sqrt{\,y\,}\$ for y.

12. Mar 28, 2013

haruspex

Try expressing the bounds as algebraic inequalities relating 0, 1, x and y. That should give you a form which is independent of the order, so is a stepping stone to reversing the order.

13. Mar 29, 2013

cwbullivant

For x = sqrt(y), y = x2 (though obviously only the part of x2 that exists in quadrant one for the purposes of this integral)?

Rereading that, I think I realize the problem. I was trying to put the ex2 as part of the region. I'm almost positive I've got the right picture now; it looks marginally similar to the bottom right corner of a box with a circle inscribed inside (with the region itself being the area below the curve of the circle, though the curve I have is given by x = sqrt(y), and is clearly not of uniform curvature like a circle would be).

If we change the order, does the y order become 0 to x2?

Last edited: Mar 29, 2013
14. Mar 29, 2013

haruspex

Yes. The algebra is straightforward:
Original bounds: y from 0 to 1, x from sqrt(y) to 1.
In algebra: 0 <= y <= 1, sqrt(y) <= x <= 1;
0 <= y <= 1, y <= x2 <= 1;
Since sqrt() is defined to be nonnegative, x >= 0. Conversely, for any x in (0, 1) there exists y s.t. 0 < y < x2. Thus the range for x is 0 to 1, and for y it is 0 to x2.
Can you do the integral wrt y now?

15. Mar 29, 2013

SammyS

Staff Emeritus
Yes, y goes from 0 to x2.

Hopefully, now you can complete the integration by changing the order of integration.

Below are some details addressing what we were trying to get at in helping you change the order of integration.

A graph of $\displaystyle \ x=\sqrt{\,y\phantom{I}\,}\$ along with x = 1 :

So for the original integral, for each value of y between 0 and 1, x goes from $\displaystyle \ \sqrt{\,y\phantom{I}\,}\$ to $\displaystyle \ 1\ .$

If the order of integration is reversed, then for each value of x between 0 and 1, y goes from o to x2.

haruspex was trying to get you to write the region of integration as a set of inequalities.
$\displaystyle \sqrt{\,y\phantom{I}\,}\le x\$

x ≤ 1

0 ≤ y​

Notice that if $\displaystyle \ \sqrt{\,y\phantom{I}\,}\le x\,,\$ then y ≤ x2 .

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