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
 
I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...

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