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

[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