Show that $$I = \int_{-\infty}^{\infty}\frac{dx}{(x^2+1)^2} = \frac{\pi}{2}$$
Using the same method from Calculating a real integral using complex integration ,
*Except for the use of the Residue theorem because it's not covered*
I arrived at:
$$\lim_{R \to \infty}\int_\psi \frac{dz}{(z^2+1)^2} = \int_{-\infty}^\infty \frac{dx}{(x^2 + 1)^2}$$
Where $\psi$ is the same contour taken in the linked question.
Now if we observe that $$(z^2+1)^2 = (z-i)^2(z+i)^2$$ And since we have a simple and closed contour, we can use Cauchy's Line Integral Theorem for derivatives with $$z_0 = -i$$$$f(z) = \frac{1}{(z-i)^2}$$$$n = 1$$
I think this is what I did wrong, since $z_0$ is not in the interior of the path but I'm not entirely sure
We can show that $$I = \frac{i\pi}{-2}$$
Of course, the answers do not match. Is what I think went wrong what actually went wrong?
$-i$ is not in the interior of the contour, rather $i$ is. Now, we apply Cauchy's Integral Formula. We have that $g(z) = \frac{1}{(z+i)^2}$ is analytic in a neighborhood of $i$. Then, by CIF (and its derivative formula) we have $$g'(i) = \frac{1}{2\pi i} \int \frac{g(z)}{(z-i)^2} = \frac{-2}{(2i)^3} = \frac{-i}{4}.$$ Then, the integral evaluates as $\frac{-i(2\pi i)}{4} = \frac{\pi}{2}$.