What does the sum of residues being zero mean for this integral?

[taylormade]

Homework Statement

Calculate
\int_0^{+\infty}\frac{\x^2}{1+x^6}dx

I found the poles/residues for this guy, and did the integral over the semicircular contour from -R to R, with R->infinity.

I get poles that contribute at:

\frac{1}{6}e^\frac{-4*pi*i}{6}.

\frac{1}{6}e^\frac{-12*pi*i}{6}.

\frac{1}{6}e^\frac{-20*pi*i}{6}.

Where I'm confused, it that when you sum the residues (I.e. 2*pi*i(Sum of Residues)), the sum of the residues is zero.

I also tried doing this integral over a smaller contour, from 0 to theta = 2pi/6 - and get similar weird answer.

My actual question is - What does this mean? I'm inclined to say that this integral equals zero - but looking at the function, that doesn't make sense to me.

Where have I gone off the rails?

I apologize - I'm still not very good with the Latex stuff.

 

[nrqed]

Homework Statement

Calculate
\int_0^{+\infty}\frac{\x^2}{1+x^6}dx

I found the poles/residues for this guy, and did the integral over the semicircular contour from -R to R, with R->infinity.

I get poles that contribute at:

\frac{1}{6}e^\frac{-4*pi*i}{6}.

\frac{1}{6}e^\frac{-12*pi*i}{6}.

\frac{1}{6}e^\frac{-20*pi*i}{6}.

Where I'm confused, it that when you sum the residues (I.e. 2*pi*i(Sum of Residues)), the sum of the residues is zero.

I also tried doing this integral over a smaller contour, from 0 to theta = 2pi/6 - and get similar weird answer.

My actual question is - What does this mean? I'm inclined to say that this integral equals zero - but looking at the function, that doesn't make sense to me.

Where have I gone off the rails?

I apologize - I'm still not very good with the Latex stuff.

You mean \int_0^{+\infty}\frac{x^2}{1+x^6}dx (don't put a back slash in front of the x in the numerator because the x does not show up!)

I don't see why youhave exponentials in your answer, furst of all.

Clearly, the integral is not zero. How did you find the poles and residues? You have to be careful if the poles are not simple poles!

 

[taylormade]

You mean \int_0^{+\infty}\frac{x^2}{1+x^6}dx (don't put a back slash in front of the x in the numerator because the x does not show up!)

I don't see why youhave exponentials in your answer, furst of all.

Clearly, the integral is not zero. How did you find the poles and residues? You have to be careful if the poles are not simple poles!

-sigh- yes - that is what I meant. Thanks!

OK - for x^6+1, there would be 6 poles, all at x^6 = -1, and they would be simple poles (right?). (well, z^6 = -1, after converting to the contour).

I took those poles to be z=e^pi*i/6, e^3*pi*i/6 and e^5*pi*i/6 as the only ones contained in the semi-circular contour.

I did the residue calculation - (z-zo*f(z)), then took the limit as z->those poles.

And I get those residues in my first post (I messed up and called them poles) :( .

\frac{1}{6}e^\frac{-4*pi*i}{6}.

\frac{1}{6}e^\frac{-12*pi*i}{6}.

\frac{1}{6}e^\frac{-20*pi*i}{6}.

The sum of those residues = 0, Hence my confusion.

I also tried a different approach and used a contour that only extends to 2*pi/6, and only contains the first pole, at z=e^pi*i/6.

When I work through that I get another result that points to the integral = 0, which just baffles me.

 

[Gib Z]

I'm wondering why you are bothering to use complex analysis when this is an elementary integral!

\int^{\infty}_0 \frac{x^2}{1+x^6} dx

Let u=x^3, then du= 3x^2. Hence the integral becomes
\frac{1}{3}\int^{\infty}_0 \frac{1}{1+u^2} du = \frac{1}{3} \left( \arctan \infty - \arctan 0 \right) = \frac{\pi}{12}

 

[taylormade]

I'm wondering why you are bothering to use complex analysis when this is an elementary integral!

\int^{\infty}_0 \frac{x^2}{1+x^6} dx

Let u=x^3, then du= 3x^2. Hence the integral becomes
\frac{1}{3}\int^{\infty}_0 \frac{1}{1+u^2} du = \frac{1}{3} \left( \arctan \infty - \arctan 0 \right) = \frac{\pi}{12}

I know - because the Prof says: "Solve Using Calculus of Residues".

I think I figured it out. I'm just beat from working on these - easy stuff is giving me fits.