Proving the Double Integral Problem: Reversing the Order of Integration

[cholyoake]

The question is:

Show that:

\int_0^1\int_x^1e^\frac{x}{y}dydx=\frac{1}{2}(e-1)

I've tried reversing the order of integration then solving from there:

\int_0^1\int_y^1 e^{\frac{x}{y}}dxdy

=\int_0^1[ye^\frac{x}{y}]_y^1dy

=\int_0^1ye^\frac{1}{y}-ye^1dy

But I can't integrate ye^\frac{1}{y}

So either I've done something wrong when changing the order of integration or something else but I can't see how to go on from here.

Thanks,
Chris.

 

[Curious3141]

The question is:

Show that:

\int_0^1\int_x^1e^\frac{x}{y}dydx=\frac{1}{2}(e-1)

I've tried reversing the order of integration then solving from there:

\int_0^1\int_y^1 e^{\frac{x}{y}}dxdy

=\int_0^1[ye^\frac{x}{y}]_y^1dy

=\int_0^1ye^\frac{1}{y}-ye^1dy

But I can't integrate ye^\frac{1}{y}

So either I've done something wrong when changing the order of integration or something else but I can't see how to go on from here.

Thanks,
Chris.

Note that the integral can also equally well be represented by \int_0^1\int_0^xe^\frac{x}{y}dydx = \int_0^1\int_0^ye^\frac{x}{y}dxdy.

 

[cholyoake]

Oh of course, thank you very much for your help.

 

[DryRun]

The question is:
\int_0^1\int_x^1e^\frac{x}{y}dydx=\frac{1}{2}(e-1)

I have no idea why but my browser shows e^{z/y} instead of e^{x/y} as the itex code is written. I am using Firefox 14.0.1
I tried clearing cache, etc but the problem still persists. Is this some bug with the forum?

 

[Curious3141]

I have no idea why but my browser shows e^{z/y} instead of e^{x/y} as the itex code is written. I am using Firefox 14.0.1
I tried clearing cache, etc but the problem still persists. Is this some bug with the forum?

Try blowing up the font (Ctrl-'plus', using the Numeric Pad plus). It's an 'x', but looks like a 'z' at small font sizes.

If you still see a 'z', then there's a real problem.

 

[DryRun]

Try blowing up the font (Ctrl-'plus', using the Numeric Pad plus). It's an 'x', but looks like a 'z' at small font sizes.

If you still see a 'z', then there's a real problem.

That fixed it. But it is a really annoying problem. I don't want to leave my display font-size bigger than it should be in my browser in order to correctly read a problem/solution. This font-size bug will certainly cause a lot of doubts and confusion all around the forum.

 

[Curious3141]

That fixed it. But it is a really annoying problem. I don't want to leave my display font-size bigger than it should be in my browser in order to correctly read a problem/solution. This font-size bug will certainly cause a lot of doubts and confusion.

It's not a bug. It's just insufficient visual acuity to distinguish the cursive 'z' from the 'x' at that scale.

 

[DryRun]

It's not a bug. It's just insufficient visual acuity to distinguish the cursive 'z' from the 'x' at that scale.

I guess there's no 'fixing' it then. Anyway, good to know it's not just me.

 

[HallsofIvy]

The question is:

Show that:

\int_0^1\int_x^1e^\frac{x}{y}dydx=\frac{1}{2}(e-1)

I've tried reversing the order of integration then solving from there:

\int_0^1\int_y^1 e^{\frac{x}{y}}dxdy

This is not a correct "reversal". In the original integral, x ranges from 0 to 1 and, for each x, y ranges from x up to 1. That is, it is the triangle above the line y= x. Reversing the order of integration, y goes from 0 to 1 and then, for each y, x goes from 0 to y. The integral is
\int_0^1\int_0^y e^{\frac{x}{y}}dxdy

=\int_0^1[ye^\frac{x}{y}]_y^1dy

=\int_0^1ye^\frac{1}{y}-ye^1dy

But I can't integrate ye^\frac{1}{y}

So either I've done something wrong when changing the order of integration or something else but I can't see how to go on from here.

Thanks,
Chris.

 

[Curious3141]

Oh of course, thank you very much for your help.

Please note that I just realized there is an error in my post. The first integral (\int_0^1\int_0^xe^\frac{x}{y}dydx) should not be there as it defines the region below the line y = x (i.e. y ≤ x). So the only correct reversal is \int_0^1\int_0^ye^\frac{x}{y}dxdy, as HallsofIvy pointed out.