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

[cwbullivant]

Homework Statement

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

Homework Equations

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...

 

[SammyS]

Homework Statement

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

Homework Equations

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...

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

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

 

[cwbullivant]

You're not going to get far integrating e^-x2 . That result involves the "error function".

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

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

∫(sqrt(y)-1) e^x2/2x³ 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

 

[SammyS]

I gather that your original integral was

\displaystyle <br /> \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?

 

[cwbullivant]

I gather that your original integral was

\displaystyle <br /> \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?

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.

 

[SammyS]

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.

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 .

 

[cwbullivant]

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 .

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)?

 

[Mark44]

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)?

e^x2 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?

 

[SammyS]

In an integral such as

\displaystyle <br /> \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\ ?

 

[cwbullivant]

In an integral such as

\displaystyle <br /> \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\ ?

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)?

e^x2 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?

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.

 

[SammyS]

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)?

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.

 

[haruspex]

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.

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.

 

[cwbullivant]

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.

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

e^x2 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?

Rereading that, I think I realize the problem. I was trying to put the e^x2 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?

 

[haruspex]

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

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?

 

[SammyS]

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

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

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 .