1

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?

gt6989b
  • 54,422
Threnody
  • 898

1 Answers1

2

$-i$ 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 $\frac{-i(2\pi i)}{4} = \frac{\pi}{2}$.

  • What confuses me is that here we pick $R$ to be arbitrary. So given the fact that we can choose $R$ to be bigger than 1 is what allows us to use CIF, correct? – Threnody Dec 27 '19 at 16:42
  • 1
    CIF may be used for any value of $R$ that you pick, provided that the path along with you are integrating your function has no poles. For example, we cannot make sense of the integral when $R = 1$. When $R < 1$ you can see that the integral will always be 0 since the integrand will always be analytic in that interior. However, we wish to take $R$ to $\infty$ in order to evaluate the integral along the real line, so eventually $R > 1$ and that is where this technique comes into play. –  Dec 27 '19 at 16:48
  • I have one question, since $-i$ is not in the counter, and $i$ is, this is why you chose your function $g$ to be the 'about $z=-i$ portion and not about the $z=i$, so you could apply CIF to the "rest" if you will. I am just clarifying for myself, I just finished complex analysis and want to better understand all of it. – homosapien Dec 28 '19 at 15:14
  • Yes, that is pretty much what is going on here. I could have also mirrored the contour so that $-i$ was inside and we use the "other" portion of the function for analyticity. With the residue theorem and Laurent series, this trick of applying CIF to the "rest" actually always works. What we did was actually expanded our function as a Laurent series here and then calculated the residue. –  Dec 30 '19 at 16:44