Integral Solution Technique for (1+y^2+z^2)^(-5/2)

[ehrenfest]

Homework Statement

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}}

?

Homework Equations

The Attempt at a Solution

 

[Rainbow Child]

Change to polar coordinates. Then it's trivial.

 

[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.

 

[Rainbow Child]

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

 

[arildno]

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

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.

 

[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

 

[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.

 

[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.

 

[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.

 

[arildno]

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.

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

 

[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

 

[arildno]

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

 

[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?

 

[Troels]

Can this result be confirmed by anyone?

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

<br /> \frac{\pi}{18}+\frac{\sqrt{3}}{9}<br />

 

[Rainbow Child]

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.

You are right, this

Then it's trivial.

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&lt;\theta&lt;\frac{\pi}{4},0&lt;r&lt;\frac{1}{\cos\theta}\right\},\, \mathcal{D}_2=\left\{(r,\theta):\frac{\pi}{4}&lt;\theta&lt;\frac{\pi}{2},0&lt;r&lt;\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!

 

[arildno]

Supercool, RainbowChild, tenacity's reward to you!