Divergent Integral?

Question

Why is $$\int_{-\infty}^{\infty} \frac{2x}{1+x^2}dx$$

divergent, when the function being described is clearly an odd function and $$\int_{-a}^{a} \frac{2x}{1+x^2}dx = 0$$ for any finite a?

Answer 1

Its Cauchy principle value is indeed $0$: $$\lim_{\ell\to\infty}\int_{-\ell}^\ell\frac{2x}{1+x^2}dx=0,$$ as the integrand is odd and hence the integral is zero for all values of $L$, as you noted. However, if one of the limits of integration goes to infinity faster than the other, the integral may not be finite. For example: $$\lim_{\ell\to\infty}\int_{-\ell}^{\exp\ell}\frac{2x}{1+x^2}dx=\lim_{\ell\to\infty}\log(\exp2\ell+1)-\log(\ell^2+1),$$ which doesn't exist. Since the integral's convergence depends on the speed at which your bounds go to infinity, we say the integral diverges. Contrast this with the following integral: $$\int_{-\infty}^\infty\frac{dx}{1+x^2}$$ Let $f$ and $g$ be functions which approach $\infty$ as $x\to\infty$. Then: $$\lim_{x\to\infty}\int_{-g(x)}^{f(x)}\frac{dt}{1+t^2}=\lim_{x\to\infty}\tan^{-1}\big(f(x)\big)-\tan^{-1}\big(-g(x)\big)=\pi/2+\pi/2=\pi$$ for any such $f$ and $g$ (probably under some assumptions that they are nice enough).