Evaluate using complex integration: $int_{-infty}^infty frac{dx}{(x^2+1)(x^2+9)}$

Evaluate

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

Firsly I found the residues of this function:

$$Res(i)=-i/16$$

$$Res(-i)=i/16$$

$$Res(3i)=i/48$$

$$Res(-3i)=-i/48$$

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:

$$\frac{1}{(z^2+1)(z^2+9)}=\frac{1}{8(z^2+1)}-\frac{1}{8(z^2+9)}$$

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?

Answer

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