Note that the definite integral starting from 1 is not defined for negative values of a, since it passes through a singularity, though since 1/x is an odd function, one can base the definite integral for negative powers at −1. If one is willing to use improper integrals and compute the Cauchy principal value, one obtains ∫ − c c 1 x d x = 0 , {\displaystyle \int _{-c}^{c}{\frac {1}{x}}\,dx=0,} which can also be argued by symmetry (since the logarithm is odd), so ∫ − 1 1 1 x d x = 0 , {\displaystyle \int _{-1}^{1}{\frac {1}{x}}\,dx=0,} so it makes no difference if the definite integral is based at 1 or −1. As with the indefinite integral, this is special to the real numbers, and does not extend over the complex numbers.