<s>[INST] Calculate $\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}$ The question:\, 

Calculate $$\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}.$$

Book's final solution: $\dfrac\pi 2$.
My mistaken solution: I don't see where is my mistake because my final solution is $-\dfrac{\pi i}2$:
$$\begin{align}
\text{A}:\int_{-\infty}^\infty {x^2\over (1+x^2)^2}dx &= 2\int_{0}^\infty {x^2\over (1+x^2)^2}dx \quad\text{as the function is even}\\&=
-2\left(\text{Res}\left({z^2\over (1+z^2)^2}\cdot \ln(z) ,i\right)+\text{Res}\left({z^2\over (1+z^2)^2}\cdot \ln(z) ,-i\right)\right)\end{align}
$$
Calculate the residues: Since $\pm i$ are poles of order $2$, then
$$\begin{align}\text{B}:\text{Res}\left({z^2\ln z\over (1+z^2)^2} ,i\right)&={(2z\ln z+z^2\cdot z^{-1})(z+i)^2-z^2\ln z\cdot 2(z+i)\over (z+i)^4} \\&= \frac{(2i\ln i+i)(-4)+\ln i\cdot 4i }{16} \\ &= {1\over 16} (-8i\ln i -4i+4i\ln i)={-1\over 4}i(\ln i+1)\end{align}
$$
whereas
$$\begin{align}\text{C}:\text{Res}\left({z^2\ln z\over (1+z^2)^2} ,-i\right)&={(2z\ln z+z^2\cdot z^{-1})(z-i)^2-z^2\ln z\cdot 2(z-i)\over (z-i)^4} \\ &=\frac{(-2i\ln(- i)-i)(-4)+\ln (-i)\cdot (-4i)}{16} \\ &= {1\over 16} (8i\ln (-i) +4i-4i\ln(- i))={1\over 4}i(\ln (-i)+1)\end{align}
$$
Combining $\text A$, $\text B$ and $\text C$, we get
$$
\int_{-\infty}^\infty {x^2\over (1+x^2)^2}\,dx=-2\left({i\over 4}(\ln(-i)+1)-{i\over 4}(\ln i+1)\right)={-i\over 2}\left({3\pi\over 2}-{\pi\over 2}\right)={-\pi i \over 2}
$$
Where is my mistake? Thanks in advance!
 [/INST] If we take the contour $\sf C$ to be the upper half-plane (with $\sf{\Im z>0}$), then we can write the integrand as $$\sf{\frac{z^2}{(z^2+1)^2}=\frac{\left(\frac{z}{z+i}\right)^2}{(z-i)^2}}$$ since the singularity in $\sf C$ is at $\sf{z=i}$ (of order two). Thus, using Cauchy's Integral Formula, we obtain $$\sf{\int_{-\infty}^\infty{x^2\over (1+x^2)^2}\,dx=\oint_C\frac{\left(\frac{z}{z+i}\right)^2}{(z-i)^2}\,dz=2\pi i\frac d{dz}\left[\left(\frac{z}{z+i}\right)^2\right]_{z=i}}=2\pi i\cdot\frac{2i\cdot i}{(i+i)^3}=\frac\pi2$$ as required. In your solution, there is no reason to introduce the natural logarithm, as it is not even part of the integrand!
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