Improper complex integrals--residue

[Dustinsfl]

Given this:
$$
\int_0^{\infty}\frac{x^2}{x^6 + 1}dx = \frac{\pi}{6}
$$
Can I do this:
$$
\int_0^{\infty}\frac{x^2}{x^6 + 1}dx = \frac{1}{2}\int_{-\infty}^{\infty}\frac{x^2}{x^6 + 1}dx
$$
and solve the integral like this
$$
\int_{-\infty}^{\infty}\frac{x^2}{x^4 + 1}dx = 2i\pi\sum_{z \ \text{upper half}}\text{Res}_{z}f = \frac{\pi\sqrt{2}}{2}
$$

 

[Krizalid1]

Given this:
$$
\int_0^{\infty}\frac{x^2}{x^6 + 1}dx = \frac{\pi}{6}
$$
Can I do this:
$$
\int_0^{\infty}\frac{x^2}{x^6 + 1}dx = \frac{1}{2}\int_{-\infty}^{\infty}\frac{x^2}{x^6 + 1}dx
$$

Yes because the integral is convergent and you're using the way back of the even function when having a symmetric interval.
I never learned well about computing integrals by using complex analysis methods so I'll let another one which may confirm your procedure.

 

[Dustinsfl]

Yes because the integral is convergent and you're using the way back of the even function when having a symmetric interval.
I never learned well about computing integrals by using complex analysis methods so I'll let another one which may confirm your procedure.

It works because it is an even function. I actually just shown the integral is pi/6