# Need help in solving a triple integral

1. Aug 15, 2012

### IBY

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

This is an integral I came across while reading a book. It is:
$$\int_0^\infty \int_x^\infty \int_x^\infty cos(t^2-u^2)dt du dx$$

I know the solution is:
$$\frac{1}{2}\sqrt{\frac{\pi}{2}}$$

I want to know how it was solved.

3. The attempt at a solution

I don't know where to start. According to the book, the mathematician who solved it ignored the order of integration and went to integrate with x first. The book doesn't fill in the steps. So I tried to do it. Then, it would get:
$$\int_x^\infty \int_x^\infty cos(t^2-u^2)*x|_0^\infty dt du$$

But then, it would all be infinity minus zero, and that's just nonsense. So I am trapped.

Last edited: Aug 15, 2012
2. Aug 15, 2012

### andrien

you should try cos(t2-u2)=real[exp{i(t2-u2)}] then it will be separable.

3. Aug 15, 2012

### IBY

Wait, wouldn't $$cos(t^2-u^2)=\frac{e^{i(t^2-u^2)}+e^{-i(t^2-u^2)}}{2}$$?

Last edited: Aug 15, 2012
4. Aug 15, 2012

### clamtrox

Both are correct.

If you do the x-integral first, you need to massage the integration limits so t and u integrals do not have x-dependence in their boundaries.

5. Aug 15, 2012

### jackmell

Sometimes not solving it is the way to solve it. You remember:

$$\int_a^b \int_{p(x)}^{q(x)}\int_{g(u,x)}^{h(u,x)} f(t,u,x)dtdudx$$

Suppose all you had to do was draw nicely, the region of integration in the u,x, and t region with t pointing up and identify those limits of integration? You can do that nicely in Mathematica. Ok, now turn the plot by a quarter turn so that now the x-axis is point up. Now identify the new limits of integration in terms of:

$$\int_c^d \int_{f_1(u)}^{f_2(u)}\int_{s(t,u)}^{v(t,u)} f(t,u,x)dxdtdu$$

Don't worry about solving the integral just yet. Just get down clearly, the nice plot and carefully and clearly identifying the regions and the functions representing those limits of integration. You do that and I think (but have not worked it yet), the integral will fall out naturally.

6. Aug 16, 2012

### IBY

I don't have Mathematica. But I did follow one of the suggestion. So at least I know that:

$$\int_x^\infty \int_x^\infty \int_0^\infty e^{it^2}e^{-iu^2}+e^{-it^2}e^{iu^2}dxdtdu=\sqrt{\frac{\pi}{2}}$$

Now, another advice was to massage the limits of integration. Not exactly sure how I go about doing that. Even if I do, how will this get rid of the pesky infinity problem?

$$\int_x^\infty \int_x^\infty (e^{it^2}e^{-iu^2}+e^{-it^2}e^{iu^2})*x|_0^\infty dtdu=\sqrt{\frac{\pi}{2}}$$

7. Aug 16, 2012

### jackmell

You puttin' the horse before the cart (don't worry about the integral yet, just get a good handle on the region first) and must have missed the first day of class: anybody taking a math class in the 21 century should have three things: Mathematica, Latex, and PF.

I think the region is everything above that black triangle back there except I have a minor problem. When I flip the integration order, I get:

$$\int _0^{\infty }\int _0^u\int _0^t \cos(t^2-u^2)dxdtdu=1/4 \sqrt{\pi/2}$$

but that's really ok. It's a start. Can you understand how I got those limits even if they're not correct? Now how can we fix this start to give the correct answer?

I don't know. Obviously I don't have a good handle on it yet but if I keep working on it I would. So you try to do that.

#### Attached Files:

• ###### triple integral region.jpg
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Last edited: Aug 16, 2012
8. Aug 16, 2012

### clamtrox

What are you trying to do here? That makes no sense: you still have x in the integration limits for t and u integrals. How would you rewrite this integral if you, say, instead integrated u from 0 to infinity?

9. Aug 16, 2012

### jackmell

No offense, but I dont' think he's gettin' this until he plays around with my diagram for a few hours and if that one is a problem for him, he's got to go back and do some of the sample problems in the book first: sometimes if you can't work a problem, best to put it on the back burner and work on some simpler, related problems.

10. Aug 16, 2012

### andrien

there is a problem with the inner two integrals.both have same limits.I think one of it should be u to ∞ or t to ∞ in order to get the desired change.I am really suspicious about the range of integration drawn.

11. Aug 16, 2012

### jackmell

. . . hate when I can't solve something I feel I'm capable of solving. Got an idea though: How about we just look at solving the volume equivalent of a 5x5x5 cube:

$$\int_0^5 \int_x^5 \int_x^5 dtdudx=\frac{125}{3}$$

and when I do what I did above I get:

$$\int_0^5\int_0^u\int_0^t dxdtdu=\frac{125}{6}$$

You see what I mean? Ain't no reason at all to go to the cosine one until I figure out what's wrong with this simple one. So we get this one right, we are poke-a-poke. Got other things to do though. Will look at it later.

12. Aug 16, 2012

### jackmell

yeah fine, got it, poke-a-poke. Hope everyone watching me struggle with this learns the lesson: sometimes it's best to work on something else (not work on the problem), in order to solve the problem.

Now, what's up with the guy who started this anyway?

13. Aug 16, 2012

### LCKurtz

It isn't so simple and you need a good picture to get it correct. I have changed $t$ to $z$ and $u$ to $y$ so the axes seem more familiar. So your integral is$$\int_0^5\int_0^y\int_0^z dxdzdy$$The difficulty arises when you integrate in the $x$ direction first. So the rectangle in the $yz$ plane is the "ground" and the slanted planes form the "roof". The problem is that when you move from the $yz$ plane in the $x$ direction, you don't hit just one of the slanted planes. Sometimes you hit the plane $x=z$ and sometimes you hit $x=y$. It depends on whether $z > y$ or $z<y$ in the $zy$ plane, so you have to break the integral into two parts.

The correct setup for that order of integration is$$\int_0^5\int_0^y\int_0^z dxdzdy + \int_0^5\int_y^5\int_0^y dxdzdy$$Notice the $dz$ limits in the two integrals.

 I see you apparently solved it while I was posting this. Did you do it this way?

14. Aug 16, 2012

### jackmell

Yes, it's a pyramid. That's the key. Ok, now IBY, post some really nice pictures showing how it's two integrals, label everything, color-code it, the works. Try and find a machine running Mathematica:
Code (Text):

mypolygon =
Graphics3D[
Polygon[{{{0, 0, 0}, {0, 0, 5}, {5, 5, 5}}, {{0, 0, 5}, {5, 5,
5}, {0, 5, 5}}, {{0, 0, 0}, {5, 5, 5}, {0, 5, 5}, {0, 5, 0}}}]];
mypoly2 =
Graphics3D[Polygon[{{0, 0, 0}, {5, 0, 5}, {5, 5, 5}, {0, 5, 0}}]];
Show[{mypolygon, mypoly2}, PlotRange -> {{0, 5}, {0, 5}, {0, 5}},
AxesLabel -> {Style[x, 25, Bold], Style[y, 25, Bold],
Style[z, 25, Bold]}, Axes -> True]

I came up with the same integrals as LC.

Note now it is a simple matter to extrapolate this to the cosine integral, it falls right out.

Last edited: Aug 16, 2012
15. Aug 16, 2012

### IBY

Sorry. This is not from a class, so it is not top priority. It is just something I found while reading. Which is why I don't have Mathematica. The book tells me the answer, and kind of gives a vague description on what the mathematician did. I just want to know how he got that way. Anyways, I will go fiddle with the problem for a while.

16. Aug 17, 2012

### andrien

$$\int _0^{\infty }\int _0^u\int _0^t \cos(t^2-u^2)dxdtdu=1/4 \sqrt{\pi/2}$$
the initial integral in op is well defined.but in this transformed one if you do the x integral get a t,and then do the t integral which gives something like [ exp{iu2}-1] when multiplied with exp(-iu2) and integrated from 0 to ∞ .look out the first term is giving ∞.to get
$$1/4 \sqrt{\pi/2}$$
two limits should be 0 to ∞ and one from 0 to t or 0 to u.