Double Integral: Tricky Exponential Integration

[soofjan]

Homework Statement

Calculate the following:
integrate [ (e^(x/y)) dy y=x^(1/3)..x ] dx x=1..8

Homework Equations

The Attempt at a Solution

In this current state the dy part is not integrable, so there must be some trick with changing the variables or something else. Couldn't find a proper transformation.
I also tried changing it from dy dx to dx dy, to no avail. I'm not sure that way is correct since it is not integrable.

Any help would be appreciated.

 

[micromass]

My suggestion would be to change the dx and the dy. (but don't forget to change your ranges as well!). That way, you could at least solve one of the two integrals...

 

[Dick]

I don't think you have to change the variables, just changing the order of integration seems to do it. You can integrate e^(x/y)*dx just fine, so you could handle that factor if the dx were first.

 

[soofjan]

I tried that but I got something that is just wrong.. am I supposed to get a sum of 2 integrals when changing the order?

 

[micromass]

No you should get something of the form

\int_a^b\int_{c(y)}^{d(y)}e^{x/y}dxdy

where the ranges are yet to be determined...

 

[soofjan]

Ok, so how do I determine the new ranges? Couldn't find anything about that in our study material.

 

[micromass]

I feel that the best way to explain this, is with an example. Assume that you have the following integral

\int_0^1\int_{x^2}^{x}f(x,y)dydx

So the x-values range from 0 to 1. An for every x-value, the y values range from x² to x.
So we have the following area A=\{(x,y)~\vert~x\in [0,1], y\in [x^2,x]\}. It would be useful to sketch this area.

So, the first question is, what are the y-ranges. To answer this, we should see what y-values are possible in A. I claim that the possible y-values are [0,1]. Indeed, take a y in [0,1]. Then (y,y) is in A. Thus every y-value in [0,1] is attained. Thus the y-ranges are [0,1].

Now for the x-ranges. This will crucially depend on the y-ranges. So fix a y in [0,1]. What are the possible corresponding x-ranges. I claim that it is [y,\sqrt{y}]. Indeed, take an x in this interval, then (x,y) is in A.

So the new integral becomes

\int_0^1\int_y^{\sqrt{y}}f(x,y)dxdy

I hope you can make sense out of this...
Also see http://rutherglen.science.mq.edu.au/wchen/lnmvafolder/mva05.pdf page 13...

 

[soofjan]

First of all, thanks for your informative post.

Now, I drew the functions and it looks like y should be from 1 to 8. As for dx, how exactly to you know the range? I tried this:
x^(1/3) = y ==> x = y^3
x=y ==> x=y
so the new range for dx is y..y^3

That's far from what wolfram gives, so I have no clue what I'm doing wrong here.

Thanks.

 

[Dick]

First of all, thanks for your informative post.

Now, I drew the functions and it looks like y should be from 1 to 8. As for dx, how exactly to you know the range? I tried this:
x^(1/3) = y ==> x = y^3
x=y ==> x=y
so the new range for dx is y..y^3

That's far from what wolfram gives, so I have no clue what I'm doing wrong here.

Thanks.

Your x limits are good. The y limits aren't. You are getting there. Just keep going.

 

[micromass]

It would be best if you draw the area involved.

You are right that the y-values range is [1,8].
Now take y arbitrary, what x values are involved. You could say (and it is my first guess to) that the x-values range is [x,x^3]. But take y=3, then our range would be [3,9], but our x can only range from 1 to 8. So we'll have to cut out the interval [8,9]. (you can easily see that if you draw the figure)
In general, take y arbitrary, then the x-values range is [x,\min{x^3,8}]. We can simplify this by considering different values for y:
1) if y is in [1,2], then the x-values range from [x,x^3].
2) if y is in [2,8], then the x-values range from [x,8].

 

[soofjan]

edit: didnt see post above.

Thank you very much. It works =)

 

[soofjan]

Well, I guess that's not the end of it.

After integrating:
e^(x/y) dx dy, x=y..8, y=2..8
I get:
( y*e^(8/y) - y*e ) dy y=2..8

What do I do about the bolded part? It's not integrable.

 

[Dick]

Well, I guess that's not the end of it.

After integrating:
e^(x/y) dx dy, x=y..8, y=2..8
I get:
( y*e^(8/y) - y*e ) dy y=2..8

What do I do about the bolded part? It's not integrable.

I see your point. I was just looking at the parts where the boundaries of the dx integral were y^3 and y and things were working so well I didn't bother to draw a careful diagram. The parts where the x boundaries are constant do look like a problem. I wonder if the person that created this problem was as sloppy as I was?

 

[stevenb]

Well, I guess that's not the end of it.

After integrating:
e^(x/y) dx dy, x=y..8, y=2..8
I get:
( y*e^(8/y) - y*e ) dy y=2..8

What do I do about the bolded part? It's not integrable.

I think Dick has a good hypothesis that the answer is messier than intended by its author. However, I think Dick's method is the one you want and you can make progress on the simplification process and work down to a series solution. I quickly sketched out that a substitution of u=1/y will allow the bolded integral to get simplified through several steps with the final one leading to an infinite series. I think an infinite series summation is acceptable as part of a final solution. Our transcendental equations (sin, cos, exp, ln ... etc. ) are really just series solutions. We generally feel comfortable when the series is a familiar one, but still ... a series is a series. Perhaps this series has an accepted name, but I didn't recognize it myself.

 

[soofjan]

This homework was written by the Vice Dean of Mathematics, so I'm almost sure it is supposed to be messy like that... however, I'm not sure I even remember what to do in case of a series.. but I'll give it some more thought.

Thanks everyone =)

 

[stevenb]

This homework was written by the Vice Dean of Mathematics, so I'm almost sure it is supposed to be messy like that... however, I'm not sure I even remember what to do in case of a series.. but I'll give it some more thought.

Thanks everyone =)

The case of a series is just a matter of expanding your function as a series and then integrating it term by term. In the case of an infinite series, you just write out the general form of the terms with a summation symbol and integrate the general form, by moving the summation symbol outside the integral. I worked through various steps (u=1/y substitution, and then a couple of integration by parts) before doing the series integration.

I did some digging and (assuming I didn't make a mistake) the answer can be related to the "exponential integral" function Ei(x). I don't know if you are required to get an answer in terms of "official" named functions, but if so, this appears to be one way to do it.

As an engineer, I usually don't worry about digging up official names, and just put the series in a computer and crank out the numerical answer when needed. Joining in here has taught me the error of my ways. I was not at all familiar with the "exponential integral" function, and now I see that it is common enough to be included in Matlab as the "expint" function. So, a little digging can save time (both personal-time and program-run-time) in the long run. Perhaps your Vice Dean of Mathematics does indeed have a method to his madness.

http://en.wikipedia.org/wiki/Exponential_integral

 

[soofjan]

The problem with Exponential Intergal is that we don't learn it in Calculus 2, thus we can't use something we never learned in class. I will ask a teaching assistant about this, since I'm obviously missing something here.

 

[stevenb]

The problem with Exponential Intergal is that we don't learn it in Calculus 2, thus we can't use something we never learned in class. I will ask a teaching assistant about this, since I'm obviously missing something here.

I would suggest following the problem through, as it will be a good exercise for you. It's only about 1/2 hour of work if you go slowly to make sure you don't make any mistakes. I just sketched it out quickly, so I could have made a mistake. If you can verify that I'm correct, then I think you will either have to express the answer in terms of a function that you haven't covered (Ei, or another similar function), or using an infinite series.

It's quite possible that Dick is correct that there is a mistake in the original problem, or perhaps you are expected to just express the answer with a term that includes an infinite series. Note that the infinite series is in terms of functions you know, and at the Calc II level, infinite series should be something you covered.