# Cauchy Principal Value calculation

I am self-studying the residue theorem and its applications and I tried solving a problem which involves finding the principal value for an improper integral but I am not sure if my approach/answer is correct.

I wish to find $$P.V. \int_{-\infty}^{\infty} \frac{dx}{x(x^2+1)}$$

I know that the integrand has a singularity at x = 0 on the real line, and this singularity is a simple pole since $$\frac{d}{dx} x(x^2 + 1) = 3x^2 + 1$$ which is not equal to zero at z = 0

I also know that the integrand has a simple pole at $$z = i$$ on the upper half plane and so I should have that

$$P.V. \int_{-\infty}^{\infty} \frac{dx}{x(x^2+1)} = 2 \pi i Res(f, i) + \pi i Res(f,0) = 2 \pi i (\frac{1}{-2}) + \pi i = 0$$

Does this look right, the answer being zero concerns me for some reason but I assume it seems true since the function may be symmetric

• The integrand is odd. – gammatester Dec 12 '18 at 19:09