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

[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