Evaluate $$\int_{-\infty}^\infty \frac{dx}{(x^2+1)(x^2+9)}$$

Firsly I found the residues of this function:





I then closed the contour using the upper half semi circle $C$, parametrized by $Rz^{it}$ where $0\leq t\leq \pi$ and $R$ is radius, which I'll take to infinity.

Using partial fractions:


ML Lemma(I think) gives me that this integral is $\leq \pi R / (R^2-1)$ and $\leq \pi R / (R^2-9)$

But then the residue theorem:

$2\pi i (i/16+i/48)$ is negative. What is my mistake?

  • 4
    $\begingroup$ Look more closely at your own working!!!! The $i/16$ in the last line should be $-i/16$. $\endgroup$ – David Mar 23 '16 at 0:10
  • $\begingroup$ @David Arghh thanks a lot :) $\endgroup$ – GRS Mar 23 '16 at 0:14

Since complex integration and I do not get along, I would just do this with real integration. I know that this isn't what the OP asked for, but it can be useful to solve a problem in more than one way.

$\int_0^{\infty} \frac{dx}{x^2+a^2} = \frac1{a}\arctan(\frac{x}{a})|_0^{\infty} = \frac{\pi}{2a} $, so $\int_{-\infty}^{\infty} \frac{dx}{x^2+a^2} = \frac{\pi}{a} $.

Using your partial fraction decomposition of $\frac{1}{(z^2+1)(z^2+9)} =\frac1{8}(\frac{1}{z^2+1}-\frac{1}{z^2+9}) $, I get $\frac{\pi}{8}(1-\frac1{3}) =\frac{\pi}{12} $.

Note that $ \int \frac{dx}{x^2-a^2} = -\frac1{a}(\tanh^{-1}(x/a)) $.


$f(z) = \dfrac{1}{(z+i)(z-i)(z+3i)(z-3i)}$

$\dfrac{f(i)}{(z-i)} =\dfrac{1}{(2i)(4i)(-2i)} = \dfrac{1}{16i}$

$\dfrac{f(3i)}{(z-3i)} =\dfrac{1}{(2i)(4i)(6i)} = -\dfrac{1}{48i}$

$2\pi i(\dfrac{1}{16i} - \dfrac{1}{48i})= \dfrac{\pi}{12}$


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