What is the integral of x^2 over 1+x^10 from 0 to infinity?

[Ackbach]

Here is this week's POTW:

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Evaluate $\displaystyle \int_0^{\infty}\frac{x^2}{1+x^{10}} \, dx.$

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[Ackbach]

Congratulations to kiwi for his correct answer! It is posted below:

Spoiler

Consider the related closed contour integral in the complex plane (with r > 1):

\oint_R \frac{z^2}{1+z^{10}}dz=\int^{(r,0)}_{(-r,0)} \frac {z^2}{1+z^{10}}dz+\int _{arc} \frac {z^2}{1+z^{10}}dz

where "arc" is the semi circle in the upper half plane, centred on the origin and terminating on (r,0) and (-r,0) and the first integral is in the complex plane along the real axis.

The curve R contains five simple poles, located at:
p_1=e^{\frac{\pi i}{10}}, p_2=e^{\frac{3 \pi i}{10}}, p_3=e^{5 \frac{\pi i}{10}}, p_4=e^{\frac{7 \pi i}{10}}, p_5=e^{\frac{9 \pi i}{10}}.

The residues of these poles are given by:
Res(z_i)=\frac{x^2}{\frac{d}{dz}(1+x^{10})}=\frac 1{10}z^{-7}
So
Res(z_1)=-\frac {1}{10}e^{\frac{3 \pi i}{10}}

Res(z_2)=\frac {1}{10}e^{\frac{- \pi i}{10}}

Res(z_3)=-\frac {i}{10}

Res(z_4)=-\frac {1}{10}e^{\frac{ \pi i}{10}}

Res(z_5)=\frac {1}{10}e^{\frac{-3 \pi i}{10}}

The sum of the real part of these residues is zero, so:

SumResidues=\frac {1}{10}\large(-2i \sin(\frac{3 \pi}{10})-2i \sin(\frac{\pi}{10})+i\large)

\oint_R \frac{z^2}{1+z^{10}}dz=\int^{(r,0)}_{(-r,0)} \frac {z^2}{1+z^{10}}dz+\int _{arc} \frac {z^2}{1+z^{10}}dz=2 \pi i \times SumResidues

\therefore \int^{(r,0)}_{(-r,0)} \frac {z^2}{1+z^{10}}dz+\int _{arc} \frac {z^2}{1+z^{10}}dz=\frac {2 \pi i}{10}\large(-2i \sin(\frac{3 \pi}{10})-2i \sin(\frac{\pi}{10})+i\large)

\therefore \int^{(r,0)}_{(-r,0)} \frac {z^2}{1+z^{10}}dz+\int _{arc} \frac {z^2}{1+z^{10}}dz=\frac {\pi }{5}\large(2 \sin(\frac{3 \pi}{10})+2 \sin(\frac{\pi}{10})-1\large)

Now taking the limit as r goes to infinity:

\therefore \int^{\infty}_{-\infty} \frac {x^2}{1+x^{10}}dx+\int _{arc} \frac {1}{z^{8}}=\frac {\pi }{5}\large(2 \sin(\frac{3 \pi}{10})+2 \sin(\frac{\pi}{10})-1\large)

\therefore \int^{\infty}_{-\infty} \frac {x^2}{1+x^{10}}dx=\frac {\pi }{5}\large(2 \sin(\frac{3 \pi}{10})+2 \sin(\frac{\pi}{10})-1\large)

And finally using the fact that the integrand is an even function:

\therefore \int^{\infty}_{0} \frac {x^2}{1+x^{10}}dx=\frac {\pi }{10}\large(2 \sin(\frac{3 \pi}{10})+2 \sin(\frac{\pi}{10})-1\large)