Given the integral $\int_{-\infty}^{\infty}\frac{x}{x^2+1}dx,$ we can clearly see this is the integral of an odd function with limits which are symmetric about the origin, and thus its integral is zero.
However, if I treat this as a curve along the real axis on the Riemann sphere (since the function is zero at infinity), then I can consider the interior of the curve to be the upper-half of the complex plane where it has a single singularity of order $1$ at $i$. Thus, applying the residue theorem, I obtain:
$$\int_{-\infty}^{\infty}\frac{x}{x^2+1}dx = 2\pi i\lim_{x\rightarrow i}(x-i)\frac{x}{(x-i)(x+i)} = 2\pi i\frac{i}{2i} = \pi i\neq 0.$$
Clearly I'm doing something wrong, can someone explain to me my misconception(s)?
Thanks
Best Answer
It is not true that the integral from -∞ to ∞ of odd function is always 0.
The upper half plane method is not applicable in this question because the integral along the circle with radius appoaching infinity is not 0.
So basicly both of your approaches are incorrect. This integral is the expected value of t(lamda)-distribution with lamda = 1, which is not defined.