# Integral with nested roots

1. Jan 28, 2005

### msmith12

I'm working with a professor on a project and we are going through a bunch of integrals in the "table of integrals series and products" and there is one that we cannot seem to get--the answer in the book is wrong after doing some numerical calculations, but we can't find a "solution" to the integral...

So, here is the crazy integral...

$$\int_{0}^\infty \frac{dx}{(1+x^2)^\frac{3}{2}*\sqrt(1+\frac{4x^2}{3(1+x^2)^2}+\sqrt(1+\frac{4x^2}{3(1+x^2)^2}))}$$

the answer that the book gives is
$$\frac{\pi}{2\sqrt6}$$

but, like I said, this is not the value of the above integral. We are trying to determine two things--first, what the given integral equals, and second, what integral corresponds to the answer given...

a few interesting things to note that I have come across that might help someone out...

$$\int_{0}^\infty \frac{dx}{(1+x^2)^\frac{3}{2}*\sqrt(1+\frac{4x^2}{3(1+x^2)^2}+\sqrt(1+\frac{4x^2}{3(1+x^2)^2}))} = \int_{0}^\infty \frac{x}{(1+x^2)^\frac{3}{2}*\sqrt(1+\frac{4x^2}{3(1+x^2)^2}+\sqrt(1+\frac{4x^2}{3(1+x^2)^2}))}$$

also...

$$\frac{\pi}{2\sqrt6} = \frac{\sqrt(\sum_{n=1}^\infty \frac{1}{n^2})}{2}$$

And, Im not sure if it will help any, but the one other thing that i've gotten is that
$$\int_{0}^{1}\int_{0}^{1} \frac{3dxdy}{4(1-x^2y^2)} = \frac{\pi^2}{6}$$

any help or comments would be greatly appreciated...

cheers
~matt

2. Jan 28, 2005

### arildno

If I am to make a suggestion (I haven't tried it out), you should try to utilize the residue theorem from complex analysis.
Given a length "L" along the real axis, you can make a closed contour by say, considering a half-circle with radius L/2, with the origin at the real axis at L/2.

Then let "L" go to infinity, and use the residue theorem.
Hopefully, the integral along the other part of the contour is easy to evaluate.

Welcome to PF!

Last edited: Jan 28, 2005
3. Jan 28, 2005

### dextercioby

Is this the integral;
$$\int_{0}^{+\infty} \frac{dx}{(1+x^{2})^{\frac{3}{2}}\sqrt{1+\frac{4x^{2}}{(1+x^{2})^{2}}+\sqrt{1+\frac{4x^{2}}{(1+x^{2})^{2}}}}}$$

Daniel.

P.S.One thing is certain...You cannot apply the FTC... :tongue2:

4. Jan 28, 2005

### msmith12

yeah... that is the integral...

~matt

5. Jan 28, 2005

### dextercioby

I'll go to the library & look it up in Gradsteyn & Rytzhik.

Awful object... :yuck:

Daniel.

6. Jan 28, 2005

### msmith12

we're using the gradsteyn & rytzhik -- the answer that they list is incorrect (see my first post)

~matt

7. Jan 28, 2005

### dextercioby

How do you know that?
What nomber does it have??Page...??

Daniel.

8. Jan 28, 2005

### NateTG

Something about
$$\frac{2x}{1+x^2}}$$
Makes me think trig substitution. Not that it's necessarily a good idea...
$$x=\tanh(\theta)$$

9. Jan 28, 2005

### dextercioby

I made the obvious one
$$x=\sinh u$$

And and got to a dead end... :yuck:

Daniel.

10. Jan 28, 2005

### msmith12

I really don't remember the exact page/number... my professor said that he would email that info to me...

the answer that the book gave, however, is
$$\frac {\pi}{2\sqrt6}$$
which, if you play around a little on your preferred math computer program, you can find out that this answer is not correct...

(I'll post the page/number whenever I find them)

~matt

11. Jan 28, 2005

### dextercioby

I don't have mathematical sofware installed on the computer.I had Matlab 5.3 and disinstalled coz i didn't know how to work with it... I still have the Mathcad 7.0 installation kit,but i won't install it coz i don't know how to work with it...

So i can't check tha answer out...

Daniel.

P.S.I want the formula's # and page's # and from which edition of G&R...

12. Jan 28, 2005

### msmith12

3.248.5 p.321 (powers of x and binomials and polynomials) 6th edition

hope thats the info that u needed

~matt

13. Jan 29, 2005

### Integral

Staff Emeritus
I have the corrected and enlarged edition of G&R, neither your page number or equation number match up.

I set up a quick Simpson's method integrator VB, as well as plotting the function in Excel. This looks like something designed to give numeric headaches. It is really a very well behaved function, max of $\frac 1 {\sqrt 2}$at zero and by 10 is hugging zero. But it approaches zero very slowly so there is significant area under the nearly flat tail (we are talking infinity here)

The trouble is round off errors start getting significant so it is very hard to capture the area under the infinte tail.

What did your numeric method yield? Are you sure it is a more reliable answer then the one in G&R. I get .664.. for 0-100 with 30000 steps (Simpson's) the G&R result is .64... So I am already to big. But frankly I would be more inclined to suspect the numerical results then G&R.

It really bothers me that I my result is to big. With the step size and considering the behavior of the function I should have a very good value for 0 to 100.

for 0 to 5, I get .65.. With 30000 steps in that range I should be good to several decimals and I am already bigger then G&R... Humm...

Last edited: Jan 29, 2005
14. Jan 29, 2005

### arildno

I assume the integral is:
$$I=\int_{0}^{\infty} \frac{dx}{(1+x^{2})^{\frac{3}{2}}\sqrt{1+\frac{4x^{2}}{3(1+x^{2})^{2}}+\sqrt{1+\frac{4x^{2}}{3(1+x^{2})^{2}}}}}$$
The first thing to note, is that the integrand is even, so that I more conveniently may be written as:

$$I=\frac{1}{2}\int_{-\infty}^{\infty} \frac{dx}{(1+x^{2})^{\frac{3}{2}}\sqrt{1+\frac{4x^{2}}{3(1+x^{2})^{2}}+\sqrt{1+\frac{4x^{2}}{3(1+x^{2})^{2}}}}}$$

Next, the singularities are located in the complex plane:
$$\pm{i},\pm\frac{1+\sqrt{5}}{\sqrt{3}}i,\pm\frac{\sqrt{5}-1}{\sqrt{3}}i,\pm\frac{1+\sqrt{7}}{\sqrt{6}}i,\pm\frac{\sqrt{7}-1}{\sqrt{6}}i$$
as is readily verified.

Now, consider the closed contour in the complex plane:
1. The real line segment from (-L,L)
2. The half-circle in the upper complex half-plane with radius L centered at the origin.

As L goes to infinity, the integral along the half-circle will vanish, roughly as $$\frac{1}{L^{2}}$$

By the residue theorem therefore, the integral along the the real line will be given by calculating the residues associated with the enclosed singularities on the positive imaginary axis.
(Since it is some time since I did Laurent expansions, I'll check up in Bak&Newman, and get back on this)

If G&R is wrong, it is probable that the residues from some of the singularities were not taken into account.
(There are 5 of those..:yuck:)
EDIT:
Possibly, I made some mistake in identifying the singularities; I must check that as well..sigh

Last edited: Jan 29, 2005
15. Jan 31, 2005

### Integral

Staff Emeritus
Where is the OP on this? I have scoured my G&R for this form and cannot find it. There are about 80 pages in the "Power and Algebraic Functions" section I have gone through the section page by page several times the given form is no where to be found. What edition are you using?

Perhaps that form was part of the "corrected" that appears on the cover of the 4th edition?

16. Jan 31, 2005

### msmith12

it is in the 6th edition...

17. Jan 31, 2005

### arildno

18. Jan 31, 2005

### dextercioby

The interesting part is that the ones that spotted the mistake couldn't /wouldn't come up with the result... :tongue2: Else it would have been there to check it...

Daniel.

P.S.Of all those thousands of integrals,the OP ran into the most devious one... :tongue:

19. Jan 31, 2005

### msmith12

the interesting part, is that the one that spotted the mistake is the professor that im working with...

small world ain't it....

~matt

20. Jan 31, 2005

### arildno

He hadn't forgotten that in the meantime, had he??

Last edited: Jan 31, 2005
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