[SOLVED]improper complex integrals--residue

dwsmith

Well-known member
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
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
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