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[SOLVED] improper complex integrals--residue

dwsmith

Well-known member
Feb 1, 2012
1,673
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
$$
\int_{-\infty}^{\infty}\frac{x^2}{x^4 + 1}dx = 2i\pi\sum_{z \ \text{upper half}}\text{Res}_{z}f = \frac{\pi\sqrt{2}}{2}
$$
 

Krizalid

Active member
Feb 9, 2012
118
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 learnt well about computing integrals by using complex analysis methods so I'll let another one which may confirm your procedure.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Yes because the integral is convergent and you're using the way back of the even function when having a symmetric interval.
I never learnt 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