Why the integral $\int_{-\infty}^\infty dx \frac{1}{(x-i)^2} \frac{1}{(x+i)^2}$ is not zero?

Question

This might be silly but I am almost pulling my hair.

Why the following integral (along the real axis) is not zero: $$\int_{-\infty}^\infty dx \frac{1}{(x-i)^2} \frac{1}{(x+i)^2}$$

The integral has no residues and can be completed by a semicircle either in the upper complex plane or the lower complex plane. According to the residue theorem, it should be zero. (Mathematica says it's $\pi/2$)

What's wrong with my logic here?

Edit: I was somehow given the wrong impression that high-order poles does not contribute to the residue.

Answer 1

Oh, but the residue theorem does give a nonzero value. There are second-order poles at $\pm i$, and the integrand's residue at such a pole is $$\lim_{x\to\pm i}\frac{d}{dx}\frac{1}{(x\pm i)^2}=\frac{-2}{(\pm 2i)^3}=\mp\frac{i}{4}.$$As I think your only mistake was in misunderstanding how higher-order poles provide residues, I'll leave it to you to understand why we can calculate the integral from the pole at $i$ alone, viz.$$\int_{\Bbb R}\frac{dx}{(x^2+1)^2}=2\pi i\times\frac{-i}{4}=\frac{\pi}{2},$$just as your software said. (It can also be proven with $x=\tan t$.)

Edit: let me motivate the way higher-order poles work. Suppose $f(c)$ is neither $0$ nor infinite. Then $$\oint\frac{f(z)}{(z-c)^n}dz=\oint\frac{f^{n-1}(c)}{(n-1)!(z-c)}dz=2\pi i\cdot\lim_{z\to c}\frac{f^{n-1}(z)}{(n-1)!}.$$

Answer 2

According to the residue theorem, it should be $\frac{\pi}{2}$ as it is. Let $\gamma(R)$ be the union of the segment from $-R$ to $R$ and the arc $Re^{i\theta}$ for $\theta\in[0,\pi]$, counter-clockwise oriented. Due to the ML lemma we have $$ \int_{-\infty}^{+\infty}\frac{dx}{(x^2+1)^2} = \lim_{R\to +\infty}\oint_{\gamma(R)}\frac{dz}{(z^2+1)^2}=2\pi i\cdot\operatorname*{Res}_{z=i}\frac{1}{(z^2+1)^2} $$ (notice that the pole at $-i$ is not enclosed by $\gamma(R)$) and by partial fraction decomposition $$ \frac{1}{(x^2+1)^2}=\left[\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right)\right]^2=\frac{-1/4}{(x-i)^2}+\frac{-i/4}{(x-i)}+\frac{i/4}{(x+i)}+\frac{-1/4}{(x+i)^2} $$ so $$ \int_{-\infty}^{+\infty}\frac{dx}{(x^2+1)^2}=2\pi i\cdot(-i/4)=\frac{\pi}{2}.$$ The real-analytic approach is even simpler: since for any $A>0$ we have $\int_{\mathbb{R}}\frac{dx}{A+x^2}=\frac{\pi}{\sqrt{A}}$, by differentiating both sides with respect to $A$ we have $\int_{\mathbb{R}}\frac{dx}{(A+x^2)^2}=\frac{\pi}{2A\sqrt{A}}$ and it is enough to evaluate at $A=1$.

Answer 3

Alternatively, i will compute explicitly the residue in $i$ by computing the Laurent expansion, we have using $h=x-i$ $$ \begin{aligned} (x-i)^{-2}\frac 1{(x+i)^2} &= h^{-2}\frac 1{(2i+h)^2} = h^{-2}\frac 1{4i^2}\frac 1{(1-\frac {ih}2)^2} \\ &= -\frac 14h^{-2} \left( 1 +2\left(\frac{ih}2\right) +3\left(\frac{ih}2\right)^2 +\dots\right) \ . \end{aligned} $$ So the residue in $i$ is $-i/4$.

Applying the residue theorem on that $D$-rotated contour, we get passing to the limit the value $2\pi i\cdot\left(-\frac i4\right)=\frac \pi 2$, which is matching the computer prediction:

sage: var( 'x' );
sage: f = 1/(x^2+1)^2
sage: integral( f, x, -oo, +oo )
1/2*pi
sage: f.residue( x==i )
-1/4*I
sage: 2*pi*i * f.residue( x==i )
1/2*pi