Improper integral using Contour

Question

What is the value of $$\int_{-\infty}^ \infty \frac{1}{x} dx$$

I calculated it to be zero using contour integration . Is it correct ? What is the easiest/shortest way to find this ? Thanks.

Answer 1

The integral of an odd integrable function over a symmetric interval with respect to the origin is always zero. The issue here is that $\frac{1}{x}$ is not integrable in a neighbourhood of zero, since for any $\varepsilon>0$ we have $\int_{\varepsilon}^{1}\frac{dx}{x}=-\log\varepsilon$ and $\lim_{\varepsilon\to 0^+}(-\log\varepsilon)=+\infty$. We are just allowed to say that the principal value of the integral is zero, since for any $\varepsilon,M>0$ we have $$ \int_{-M}^{-\varepsilon}\frac{dx}{x}+\int_{\varepsilon}^{M}\frac{dx}{x}=0 $$ but just as written the given integral is not converging.

Answer 2

One interpretation of an integral from $a$ to $b$ is an integral along a contour from $a$ to $b$. Since you mentioned contour integration, I thought I'd give this answer a go.

For holomorphic functions with no singularities, the particular contour does not matter. One can likewise consider a contour that does not cross $z=0$ to evaluate your divergent integral. Consider $C_R:Re^{i\theta},\theta\in[0,\pi]$. Then,

$$\begin{align}-\int_{-\infty}^{+\infty}\frac1z~\mathrm dz&=\lim_{R\to\infty}\int_{+R}^{-R}\frac1z~\mathrm dz\\&=\lim_{R\to\infty}\int_C\frac1z~\mathrm dz\\&=\pi i\end{align}$$

In fact, any curve from $-R$ to $+R$ will result in $(2k+1)\pi i,k\in\mathbb Z$.