MHB Improper complex integrals--residue

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Dustinsfl
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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
$$
\int_{-\infty}^{\infty}\frac{x^2}{x^4 + 1}dx = 2i\pi\sum_{z \ \text{upper half}}\text{Res}_{z}f = \frac{\pi\sqrt{2}}{2}
$$
 
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dwsmith said:
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.
 
Krizalid said:
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
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.

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