What is the crazy integral that the professor and student are struggling with?

[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...

<br /> \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}))}<br />

the answer that the book gives is
<br /> \frac{\pi}{2\sqrt6}<br />

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...


<br /> \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}))} = <br /> \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}))}<br />

also...

<br /> \frac{\pi}{2\sqrt6} = \frac{\sqrt(\sum_{n=1}^\infty \frac{1}{n^2})}{2}<br />

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

any help or comments would be greatly appreciated...

cheers
~matt

 

[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!

 

[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...

 

[msmith12]

yeah... that is the integral...


~matt

 

[dextercioby]

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

Awful object...

Daniel.

 

[msmith12]

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

~matt

 

[dextercioby]

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

Daniel.

 

[NateTG]

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

 

[dextercioby]

I made the obvious one
x=\sinh u

And and got to a dead end...

Daniel.

 

[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
<br /> \frac {\pi}{2\sqrt6}<br />
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

 

[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...

 

[msmith12]

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

hope that's the info that u needed

~matt

 

[Integral]

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 infinite 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...

 

[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..)
EDIT:
Possibly, I made some mistake in identifying the singularities; I must check that as well..sigh

 

[Integral]

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?

 

[msmith12]

it is in the 6th edition...

 

[arildno]

Hi msmith12:
Here's the errata list for the 6th edition of G&R:
http://www.mathtable.com/errata/gr6_errata.pdf

And guess what?
I quote in full:
"Integral 3.248.5, page 321, is incorrect.
(Thanks to..blah-blah-blah)"

 

[dextercioby]

The interesting part is that the ones that spotted the mistake couldn't /wouldn't come up with the result... 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...

 

[msmith12]

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

small world ain't it...

~matt

 

[arildno]

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

small world ain't it...

~matt

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

 

[msmith12]

no, he just knew that the answer was wrong... and he asked me to try to figure out both the answer to the integral, and find an integral that equals the answer that they put in...

 

[arildno]

Well, what do you know of complex analysis and use of residues?
I got stuck..

 

[msmith12]

not very much actually, I asked my professor, and he said that he would work on them with me...

it seemed like he has tried this method on this problem, because when i first asked him about them, he fairly quickly replied that he did not believe that this method would work for this particular problem... but either way, I get to learn about them...

~matt

 

[saltydog]

Now that is a problem to work with complex contours! Thanks arildno for presenting it precisely (well you too dextercioby, didn't notice it at first). I'll look into too. May take a while for me.

 

[mathwonk]

may i ask why anyone would care about this integral?

 

[NateTG]

may i ask why anyone would care about this integral?

...Because it is there?

Actually, it seems like going through various attempts to crack an integral would be a pretty good excercise.

 

[mathwonk]

i guess i have less energy than you guys, i need more motivation than that to tackle something of this nature.

presumably it arises in engineering for some reason?

 

[NateTG]

presumably it arises in engineering for some reason?

But then numerical approaches are just fine.

 

[saltydog]

...Because it is there?

Works for me!

Salty

 

[StatusX]

Where'd you get stuck arildno? It seems pretty straightforward from there, though extremely tedious. Just find the residues and add them up.

 

[arildno]

Well, tediousness makes me resign faster..

Besides, there's been a few years since I did this, and I wondered if the fact that we need to pick an analytical branch of the complex logarithms (masquerading as roots) could lead to some problems.
Probably not, but I ended up feeling disinclined to proceed further.

 

[saltydog]

Well, tediousness makes me resign faster..

Besides, there's been a few years since I did this, and I wondered if the fact that we need to pick an analytical branch of the complex logarithms (masquerading as roots) could lead to some problems.
Probably not, but I ended up feeling disinclined to proceed further.

How about this: How can the problem be "simplified" so that it can be solved reasonably well with residues? How about just one radical with that quotient in it (or modified so that the residue theorem works)? Can someone demonstrate how to do that or no?

Salty

 

[StatusX]

One way to simplify it (that might be the integral the answer corresponds to, as the op was looking for) is to make it an infinitely nested radical:

y = \sqrt(1+\frac{4x^2}{3(1+x^2)^2}+\sqrt(1+\frac{4x^2}{3(1+x^2)^2}+\sqrt(...)))

y = \sqrt{1+\frac{4x^2}{3(1+x^2)^2}+y}

y^2 = 1+\frac{4x^2}{3(1+x^2)^2}+y

y = \frac{1}{2} \pm \frac{1}{2} \sqrt{5+\frac{16x^2}{3(1+x^2)^2}

I'm not sure which sign corresponds to the radical, but for now I'll take the plus sign, and plugging back in:

\int_0^{\infty} \frac{2 dx}{(1+x^2)^{\frac{3}{2}} (5 + \sqrt{1+\frac{16x^2}{3(1+x^2)^2}} )}

That should be somewhat easier to work with.

 

[saltydog]

Thanks for that derivation StatusX. Actually I was thinking of a completely separate, more simple problem to attempt first (unrelated to the one above).

I've been looking at the residue method and decided I need to spend time figuring out how to calculate residues of functions. Actually, I might just post a question on the board: "I have this expression, (a quotient with a radical) , can someone help me figure out what the residues are?". But I'll attempt to work out the problems first before I do that.

It's like one of the first post I made here: If you can't solve a difficult problem, put it up and work on simple ones that look like it and then just gradually build up to the more complex one.

Works for me,
Salty

 

[arildno]

I've been looking at the residue method and decided I need to spend time figuring out how to calculate residues of functions. Actually, I might just post a question on the board: "I have this expression, (a quotient with a radical) , can someone help me figure out what the residues are?". But I'll attempt to work out the problems first before I do that.

That was also a disheartening experience for me.
I was thinking of finding the Laurent expansion (or just the coefficient of the first negative power, anyways..), but I then thought that since the expansion is to be defined on an annular region containing the singularity, wouldn't the construction of such an annulus be invalidated since we need to restrict ourselves to an analytical branch of the complex logarithm?

I was suddenly unsure if I could do this stuff; I've only had an intro course in complex analysis some years back, so I felt the ice beneath my feet was getting too thin to proceed further upon.

 

[saltydog]

Some progress

Daniel Lichtblau of Wolfram Research (Mathematica) responded to an e-mail I sent him regarding the problem. He tells me the integrand cannot be represented as a "proper Laurent" series but rather is more amendable to a Puiseux series (one involving fractional powers) thus the problem as written cannot be solved through the methods of residues. Think I might still spend some time on the theory though.

 

[arildno]

Daniel Lichtblau of Wolfram Research (Mathematica) responded to an e-mail I sent him regarding the problem. He tells me the integrand cannot be represented as a "proper Laurent" series but rather is more amendable to a Puiseux series (one involving fractional powers) thus the problem as written cannot be solved through the methods of residues. Think I might still spend some time on the theory though.

I am glad you consulted someone more competent than myself in this matter!
Apparently, some of my concerns were justified..

 

[StatusX]

I looked it up and a puiseux series is just a laurent series in the variable x^1/d (instead of x) for some integer d. I don't know why this would be necessary, or how to find a d that would work, but maybe someone else has some ideas.

 

[arildno]

Well, since the standard version of the residue theorem requires a function with a regular Laurent expansion, similar results as the residue theorem for series in fractional powers ought to be somewhat different, when existing.

 

[saltydog]

I looked it up and a puiseux series is just a laurent series in the variable x^1/d (instead of x) for some integer d. I don't know why this would be necessary, or how to find a d that would work, but maybe someone else has some ideas.

I'm thinking maybe the problem can be solved by including the following in the Residue Theorem:

\oint_C (z-z_0)^m dz

with m=(2n+1)/2 since the series expansion of the nested radical is in terms of these powers.

Note the Residue Theorem as applied to definite integrals, requires the series expansion of f(z) to be in terms of:

(z-z_0)^m

with m an integer. The integral reduces to Sines and Cosines which behave nicely for (2n+1)t/2 for limits of 0 and 2pi (which are the limits you get when you convert it). I'm still working on it when I have time.

 

[msmith12]

I haven't checked back in in a while--sorry about that, it has been Mardi Gras time down here, so everything has been hectic...

thanks so much for all of the help and ideas...

~matt

 

[msmith12]

after playing around with the integral some more, there are a few more interesting things that might help.

first, if you make the substitution of x=1/x,

the new integral is

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

first of all, this integral is now odd, which means that

\int_{-\infty}^{+\infty} \frac{xdx}{(1+x^{2})^{\frac{3}{2}}\sqrt{1+\frac{4x^ {2}}{(1+x^{2})^{2}}+\sqrt{1+\frac{4x^{2}}{(1+x^{2} )^{2}}}}} = 0

This might be of use to someone...

also, if you substitute

u = \frac{dx}{\sqrt{1+x^2}}

you get the new integral of

-\int_{0}^{\sqrt{2}} \frac{dx}{\sqrt{1+\frac{12(1-x^2)}{x^6}+\sqrt{1+\frac{12(1-x^2)}{x^6}}}

--so, I can't seem to find the error in the coding for this, but the link has the basic gist of the integral--

(if i did my algebra right--i did it during an econ class, so I might be a little off)

hope that some of this might help