I am trying to calculate $\int_{0}^{\infty}\frac{x^2}{(x^2+9)^2(x^2+4)}dx$. I want to use the upper half-circle as the contour (since the integrand is even), yet the function as poles in $2i , 3i$, but the pole $3i$ is an order $2$ pole, I couldn't find a way to calculate the residue of $3i$ in order to calculate the integral. Is there a way to split the integrand to $\frac{A}{x^2+9}$ + $\frac{B}{x^2+9}$ + $\frac{C}{x^2+4}$? or simple way to calculate the residue to $x=3i$?
Contour integral with pure imaginary pole of order two : $\int_{0}^{\infty}\frac{x^2}{(x^2+9)^2(x^2+4)}dx$
complex-analysiscontour-integrationresidue-calculus
Related Solutions
You could also use the following contour.
$$\int_{-\infty}^0 \frac{2x^2-1}{x^4+1}\operatorname dx =\int_{0}^{+\infty} \frac{2x^2-1}{x^4+1}\operatorname dx $$
Has an analytic continuation as $$\int_{\Gamma} \frac{2z^2-1}{z^4+1}\operatorname dz $$ with 4 poles, but just on pole inside of the contour.
$$\operatorname*{res}_{z=e^{i\frac{\pi}{4}}} f(z) = \frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8} $$
I think you made a calculation error here (?)
Using $$\int_{\Gamma} \frac{2z^2-1}{z^4+1}\operatorname dz = \color{blue}{\int_{\Gamma_1}\frac{2z^2-1}{z^4+1}\operatorname dz} + \int_{\Gamma_2}\frac{2z^2-1}{z^4+1}\operatorname dz + {\color{red}{\int_{\Gamma_3}\frac{2z^2-1}{z^4+1}\operatorname dz}}$$
- Now $\color{blue}{\int_{\Gamma_1} \to 0}$ as $R\to +\infty$ which can be proven using the triangle inequality.
- Use $\Gamma_2 \leftrightarrow z(x) = x$ and $x:0\to R$
- Use $\color{red}{\Gamma_3 \leftrightarrow z(y) = iy}$ and $y: R \to 0$
Which finally results in (as $R \to +\infty$) $$2\pi i \left(\frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8}\right) = \color{blue}{0}+\int_{0}^{+\infty}\frac{2x^2-1}{x^4+1}\operatorname dx + \color{red}{ i \int_{+\infty}^0\frac{-2y^2-1}{y^4+1}\operatorname dy}$$
Here you can read the real parts which results in:
$$\int_{0}^{+\infty}\frac{2x^2-1}{x^4+1}\operatorname dx = \frac{\pi\sqrt{2}}{4}$$
Hint
Consider $$f(z)=\frac{ze^{iz}}{1+z^2}$$ and integrate on $$\{[-r,r]\mid r>a\}\cup\{re^{i\theta}\mid \theta\in[0,\pi]\}.$$
Best Answer
Note that$$\frac{z^2}{(z^2+9)^2(z^2+4)}=\frac{z^2}{(z-3i)^2(z+3i)^2(z^2+4)}.$$Now, let $g(z)=\frac{z^2}{(z+3i)^2(z^2+4)}$. Then $g(3i)=-\frac1{20}$ and $g'(3i)=-\frac{13i}{300}$. Therefore,\begin{align}\frac{z^2}{(z-3i)^2(z+3i)^2(z^2+4)}&=\frac{-\frac1{20}-\frac{13i}{300}(z-3i)+\cdots}{(z-3i)^2}\\&=-\frac1{20(z-3i)^2}-\frac{13i}{300(z-3i)}+\cdots,\end{align}and so$$\operatorname{res}_{z=3i}\frac{z^2}{(z^2+9)^2(z^2+4)}=-\frac{13i}{300}.$$