# Definite integral $1/(x^2-b^2)$ over the real axis

I'm interested in the definite integral

\begin{align} I\equiv\int_{-\infty}^{\infty} \frac{1}{x^2-b^2}=\int_{-\infty}^{\infty} \frac{1}{(x+b) (x-b)}.\tag{1} \end{align}

Obviously, it has two poles ($x=b, x=-b$) on the real axes and is thus singular. I tried to apply the contour integration methods mentioned here, where they discuss the integral \begin{align} \int_{-\infty}^{\infty}\frac{e^{iax}}{x^2 - b^2}dx = -\frac{\pi}{b}\sin(ab),\tag{2} \end{align} where the r.h.s. is the solution derivable in multiple ways as shown in the above thread (e.g. circumventing the poles with infinitesimal arcs).

However, since in the seemingly more simple case (1) the nominator is symmetric in constrast to the situation in (2), I obtain $$I=0,$$ as the residues equal up to different signs. E.g. consider the limit $a\rightarrow 0$ in (2) which gives $\sin(ab)\rightarrow 0$.

Based on some literature (in the context in which the integral is appearing) it seems that one should obtain

\begin{align} I=-\frac{i\pi}{b}. \end{align}

Of course, this can be realized by considering the modified integral \begin{align} I_{mod}\equiv\lim_{\eta\rightarrow 0^+} \int_{-\infty}^{\infty} dx \frac{1}{x^2-b^2+i\eta}, \end{align} and closing the contour (e.g. a box closed at infinity) in the lower half plane.

However, in this approach one seems to have some freedom (sign of the infinitesimal contribution, why shift one pole upwards and another pole downwards and not e.g. both upwards?)

So let me explicitly phrase my questions:

1. Is the value of the definite integral in (1) well-defined?
2. Is it equal to zero?
3. In any case, why would I include an infinitesimal shift as in (2) and not in another way?

Thank you very much in advance!

Notice, $$\int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx\iff f(-x)=f(x)$$ $$\int_{-\infty}^{\infty} \frac{1}{x^2-b^2}dx=2$$ $$=\int_{0}^{\infty} \frac{1}{x^2-b^2}dx$$ $$=\frac{1}{2b}\left[\ln\left|\frac{x-b}{x+b}\right|\right]_{0}^{\infty}$$$$=\lim _{x\to \infty}\frac{1}{2b}\ln\left|\frac{x-b}{x+b}\right|-\ln\left|\frac{0-b}{0+b}\right|$$

$$=\lim _{x\to \infty}\frac{1}{2b}\ln\left|\frac{x-b}{x+b}\right|-0$$ Let, $x=\frac{1}{t}\implies t\to 0\ as\ x\to \infty$, we get $$=\lim _{t\to 0}\frac{1}{2b}\ln\left|\frac{\frac{1}{t}-b}{\frac{1}{t}+b}\right|$$ $$=\lim _{t\to 0}\frac{1}{2b}\ln\left|\frac{1-bt}{1+bt}\right|=\frac{1}{2b}\ln|1|=0$$

• How do you get $\ln (-1) = 0$ in the evaluation of the integral at the lower limit? – Mark Bennet Aug 6 '15 at 8:49
• Where do you find this in the answer? Log is not defined for negative numbers. – Harish Chandra Rajpoot Aug 6 '15 at 8:54
• Sorry, I see you put absolute value bars on. But if you don't, and you get the factors of $2$ right, you can see where the value $\frac{(1\pm 2n)\pi i}b$ for the value of the integral might come from. Also with reference to the original question a clue as to what might happen if you compute the integral in different ways. – Mark Bennet Aug 6 '15 at 9:02

By replacing $x$ with $by$ it is trivial that we just need to deal with the $b=1$ case.

Now the principal value $$I=PV\int_{-\infty}^{+\infty}\frac{dx}{x^2-1}$$ is well defined: $$I= PV\int_{0}^{+\infty}\frac{2\,dx}{x^2-1}=PV\int_{0}^{+\infty}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)\,dx = PV\int_{-1}^{1}\frac{dx}{x}=0.$$

On the other hand, the principal value is just one way to regularize divergent integrals.

If we consider: $$I_n = \int_{-\infty}^{+\infty}\frac{dx}{x^2-1+\frac{i}{n}} = \pi\sqrt{\frac{n}{i-n}}$$ we are just approaching the real line, as integration path, in a different way, so there is no wonder in: $$\lim_{n\to +\infty} I_n = -i\pi \neq 0.$$