>>10834855fine, i'll do it properly
first note that the integral is improper of both types, so in fact the bounds are m and M and we'll send each back to where it needs to be. now make the substitution u^3 = x. then 3u^2 = dx/du, so the integral becomes
It is likely that one may solve this through more elementary means, but that sounds annoying, so I'm going to use residue calculus. Merely consider the contour C given by moving R along the real axis, then along the circle of radius R ccw until reaching Re^(i*2pi/3), then back to the origin along that ray. Within this contour for R > 1 we have one simple pole, namely e^(i*pi/3), and the residue of z/(1 + z^3) here is just e^(i*pi/3) / (3e^(i*2pi/3)) = (1/3)e^(-i*pi/3). So we get that
Now, let's break down the contour integral. On the circular path we have the integral
as R goes to infinity.
Then, the other parts of the path are moving from 0 to R, in which we now take R to infinity, as well as moving from Re^(i*2pi/3) to 0, which is the integral
which is just the other integral times -e^(i*4pi/3). So now we have (1 - e^(i*4pi/3)) * integral = 2pi/3 * e^(i*pi/6), and of course we also have to multiply by 3. Note 1 - e^(i * 4pi/3) = 1 - ( - 1/2 - sqrt(3)/2 i) = 3/2 + sqrt(3)/2 i = sqrt(3)e^(i*pi/6). so dividing, we find the final value of .
