# Technique to solve integral

1. Jan 26, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
What technique would you use to do the integral:

$$\int_0^1 \int_0^1 dy dz \frac{1}{(1+y^2+z^2)^{5/2}}$$

?
2. Relevant equations

3. The attempt at a solution

2. Jan 26, 2008

### Rainbow Child

Change to polar coordinates. Then it's trivial.

3. Jan 26, 2008

### arildno

Not utterly trivial, Rainbow Child:
Although the "radial" integration will go easy enough, the limits for the unit square, as represented in polar coordinates are somewhat nasty.

4. Jan 26, 2008

### Rainbow Child

Yes. But after the integration, he can change back to cartesian coordinates.

5. Jan 26, 2008

### arildno

Eeh, no.

We get:
$$\int_{0}^{2\pi}\int_{ugly}^{UGLY}\frac{rdrd\d\theta}{(1+r^{2})^{\frac{5}{2}}}=\int_{0}^{2\pi}(|_{ugly}^{UGLY}-\frac{1}{3}(1+r^{2})^{-\frac{3}{2}})d\theta$$
You can't shift this "back" into Cartesian coordinates in any simple, valid manner.

6. Jan 26, 2008

### ehrenfest

I get

$$(\theta_2 - \theta_1)(-1/3 (1+r^2)^{-3/2})|^{r_2}_{r_1}$$

How do I change back to rectangular coordinates?

EDIT: arildno beat me

7. Jan 26, 2008

### malawi_glenn

I would treat 1 + z^2 as A

The solving $$\int dy\frac{1}{(A+y^2)^{5/2}} = \frac{2y^3+3Ay}{3A^2(y^2+A)^{3/2}}$$ (used mathematica..) I think you can use integral by parts.

8. Jan 26, 2008

### arildno

Okay, I'll do a bit more:

"ugly" is 0, so it wasn't too ugly after all.

Here's UGLY:
$$0\leq\theta\leq\frac{\pi}{4}, r=\frac{1}{\cos\theta}, \frac{\pi}{4}\leq\theta\frac{3\pi}{4}, r=\frac{1}{\sin\theta}$$
and so on around the unit square.

This is not at all simple to go further with.

9. Jan 26, 2008

### ehrenfest

Maybe there is no reasonable way to do it. Mathematica gave a nice simple answer so I thought there would be. But I was probably wrong.

10. Jan 26, 2008

### arildno

No, it is probably something far more cleverer than trivial coordinate change that is needed here.

11. Jan 26, 2008

### malawi_glenn

Where did you find this integral?

I got by doing the whole in mathematica:

$$\left[ 2z + \sqrt{z^2+1} \left( \frac{2z}{3(z^2+1)} + \frac{z}{3(z^2+1)^2} \right) \right] _{z=0}^1$$

12. Jan 26, 2008

### arildno

That must be wrong, malawi, since the given integral is clearly less than 1 in value, whereas yours is greater than 2.

13. Jan 26, 2008

### coomast

I see no substitution that makes this an easy one. However the direct calculation is not that extreme. However I have to admit I used a little help from the integrator. I changed the variables a bit, z became x, sorry for that. Call $$1+y^2=A^2$$, the inner integral becomes:

$$\int_0^1 \frac{dx}{\left[A^2+x^2\right]^{\frac{5}{2}}}$$

This can be solved using:

$$x=A \cdot sinh(t)$$

giving then:

$$\int_0^{arcsinh \left(\frac{1}{A}\right)} \frac{dt}{cosh^4(t)}$$

I did this integral with the integrator. After filling in the limits and replacing A again with y,
I got:

$$\frac{1}{3 \cdot (1+y^2)^2 \sqrt{2+y^2}}\cdot \left(\frac{5+3y^2}{2+y^2}\right)$$

We are left with the following:

$$I=\frac{1}{3}\int_0^1 \frac{5+3y^2} {(1+y^2)^2 \cdot \left(2+y^2\right)^{\frac{3}{2}}}dy$$

This integral can also be found using the integrator. After filling in the limits I got:

$$I=\frac{1}{18}\left(2\sqrt{3}+\pi\right)$$

Sorry for the integrator use, but it's late and I don't have the time right now to solve them by hand. Can this result be confirmed by anyone?

14. Jan 26, 2008

### Troels

Yes. I cheated a bit too (used Maple 11) and obtained

$$\frac{\pi}{18}+\frac{\sqrt{3}}{9}$$

15. Jan 27, 2008

### Rainbow Child

You are right, this

was stupid statement, since I didn't paid much attention to the limits of the integration. Even though the resulting can be evaluted without by hand (but not trivially !!!).

The region of integration in polar coordinates becames

$$\mathcal{D}_1=\left\{(r,\theta):0<\theta<\frac{\pi}{4},0<r<\frac{1}{\cos\theta}\right\},\, \mathcal{D}_2=\left\{(r,\theta):\frac{\pi}{4}<\theta<\frac{\pi}{2},0<r<\frac{1}{\sin\theta}\right\}$$

thus

$$I=\int_0^1 \int_0^1 dx dy \frac{1}{(1+x^2+y^2)^{5/2}}=\int\int_{\mathcal{D}_1}d\theta\,dr\frac{r}{(1+r^2)^{5/2}}+\int\int_{\mathcal{D}_2}d\theta\,dr\frac{r}{(1+r^2)^{5/2}}=I_1+I_2$$

The first integral $$I_1$$ reads

$$I_1=\int_0^{\frac{\pi}{4}}d\theta\int_0^{1/\cos\theta}dr\frac{r}{(1+r^2)^{5/2}}=-\frac{1}{3}\int_0^{\frac{\pi}{4}}d\theta\left(\frac{1}{(1+1/\cos^2\theta)^{3/2}}-1\right)=-\frac{1}{3}\int_0^{\frac{\pi}{4}}d\theta\frac{\cos^3\theta}{(1+\cos^2\theta)^{3/2}}+\frac{\pi}{12}=-\frac{1}{3}\,J_1+\frac{\pi}{12}$$

The integral $$J_1$$ reads

$$J_1=\int_0^{\frac{\sqrt{2}}{2}}d(\sin\theta)\frac{\cos^2\theta}{(1+\cos^2\theta)^{3/2}}=\int_0^{\frac{\sqrt{2}}{2}}d(\sin\theta)\frac{1-\sin^2\theta}{(2-\sin^2\theta)^{3/2}}}}=\int_0^{\frac{\sqrt{2}}{2}}dt\frac{2-t^2}{(2-t^2)^{3/2}}-\int_0^{\frac{\sqrt{2}}{2}}dt\frac{1}{(2-t^2)^{3/2}}=\arcsin\frac{t}{\sqrt{2}}\Big|_0^{\frac{\sqrt{2}}{2}}-\frac{t}{2\,\sqrt{2-t^2}}\Big|_0^{\frac{\sqrt{2}}{2}}\Rightarrow$$
$$J_1=\frac{\pi}{6}-\frac{\sqrt{3}}{6}\Rightarrow I_1=\frac{\sqrt{3}}{18}+\frac{\pi}{36}$$

Similary for the 2nd integral $$I_2$$ we have

$$I_2=\frac{\sqrt{3}}{18}+\frac{\pi}{36}$$

yielding to

$$I=\frac{\pi}{18}+\frac{\sqrt{3}}{9}$$

as coomast and Troels posted.

Obviously not trivial!

16. Jan 27, 2008

### arildno

Supercool, RainbowChild, tenacity's reward to you!