Is $\int_{-\infty}^{\infty}\frac{1}{(x-a)}dx = \pi i $?

Question

Silly question I should know the answer to, but what is the integral of:

$$\int_{-\infty}^{\infty}\frac{1}{(x-a)}dx$$

I can use countour integrals to get one type of answer: $$\oint\frac{1}{(x-a)}dx = \int_{-\infty}^{\infty}\frac{1}{(x-a)}dx + \int_{arc}\frac{1}{(x-a)}dx$$

Where here I assume that I am integrating over a contour that contains the pole. Choosing a parameterized path:

$$\int_{arc}\frac{1}{(x(t)-a)}\frac{dt}{dx}dx$$ $$\int_{0}^{\pi}\frac{1}{(R e^{i t})} i R e^{i t}dx$$

$$2 \pi i = \int_{-\infty}^{\infty}\frac{1}{(x-a)}dx + \pi i \\ \int_{-\infty}^{\infty}\frac{1}{(x-a)}dx = i\pi $$

But if I solve the problem using u-substitution:

$$ u = ln(x-a) \\ du = \frac{1}{x-a}\\ \int_{-\infty}^{\infty}\frac{1}{(x-a)}dx = \int du = u = ln(x-a)|^{x\rightarrow\infty}_{x \rightarrow-\infty} = ln(\infty) - ln(-\infty)? $$

since $ln(-\infty)$ is undefined, so what is going on here?

EDIT: One of the answers points out that this integration only works when a isn't on the complex line. In the case where it is complex, (I think) we can do the following:

We breakdown our contour integral into more paths: $$\oint\frac{1}{(x-a)}dx = \int_{-\infty}^{a-\epsilon}\frac{1}{(x-a)}dx + \int_{\gamma_2(t)}\frac{1}{(x-a)}dx + \int_{a+\epsilon}^{\infty}\frac{1}{(x-a)}dx + \int_{arc}\frac{1}{(x-a)}dx$$

Were we've broken down the path along the real line into 3 pieces, which looks like the following picture:

In the limit where we take $\epsilon \rightarrow 0$ $\gamma_1$ and $\gamma_3$ combine to approach the integral alone the real line. And $\gamma_2$ can be expressed as the integral:

$$\int_{\pi}^{0}\frac{1}{(\epsilon e^{i t})} i \epsilon e^{i t}dx = -i \pi$$

which then gives us: $$\int_{-\infty}^{\infty}\frac{1}{(x-a)}dx -i \pi + i \pi = 2 \pi i\\ $$

So if it's on the complex plane the integral is $\pi i$ and its $2\pi i$ on the real plane

Answer 1

There are a few mistakes in your working.

1. Assuming $a \in \mathbb{R}$, the contour you chose is not suitable as $a$ lies on the border of the contour. This is not allowed - it must lie strictly inside the contour in order to apply Residue theorem.
2. The antiderivative of $\frac{1}{x - a}$ is $\ln|x - a|$. Thus, what you should obtain is instead "$\ln(\infty) - \ln(\infty)$", which is not well-defined.
3. The function is not Riemann integrable as it is not bounded in $[-1,1]$, which is one of the requirements of Riemann integrability.

Answer 2

Let $a = b+ic$ with $(b,c)\in\mathbb{R}^2$. So first let us remark that the integrand is not Lebesgue integrable since for example using the fact that $x≥ b+|c| \implies c^2+|x-b|^2 \leq 2\,|x-b|^2$ we have $$ \begin{align*} \int_{-\infty}^∞ \frac{\mathrm{d}x}{|x-a|} &= \int_{-\infty}^∞ \frac{\mathrm{d}x}{\sqrt{|x-b|^2+c^2}} \\ &≥ ∫_{b+|c|}^\infty \frac{\mathrm{d}x}{\sqrt{2}\,(x-b)} \\ &≥ \frac{1}{\sqrt{2}}\lim_{n\to\infty} \left(\ln(n) - \ln(|c|)\right) = +\infty \end{align*} $$ From this we deduce that the integral is also not Riemann integrable (as remarked by Clement Yung, if $a\in\mathbb{R}$, one also has that the integral is not Riemann integrable since it is unbounded, but more generally, Riemann integrability is only for bounded sets).

Since the integral is on an unbounded set and it is not Lebesgue integrable, the only way to give a meaning to the integral is by defining it as a limit, and different definitions might give different values. For example taking $a=0$, we have $$ \lim_{n\to\infty,\,\varepsilon\to 0 } \int_{-n}^{-\varepsilon}\frac{\mathrm{d}x}{x} + \int_{\varepsilon}^n \frac{\mathrm{d}x}{x} = 0 $$ by symmetry, but $$ \lim_{n\to\infty,\,\varepsilon\to 0 } \int_{-n}^{-\varepsilon}\frac{\mathrm{d}x}{x} + \int_{\varepsilon}^{2n} \frac{\mathrm{d}x}{x} = \lim_{n\to\infty} \ln(2n)-\ln(n) = \ln(2). $$ The same phenomenon occurs when you try to get the value of the integral as a limit of contour integrals. See also https://en.wikipedia.org/wiki/Riemann_integral#Generalizations.

Other remarks: one usually use the Residue Theorem on bounded sets and then take a limit. Of course, on bounded sets, analytic functions are continuous and so there is nothing to verify. What your example shows, is that one can not take directly an infinite contour in general if the function is not Lebesgue integrable.