improper integrals and difficulty computing closed curve part.

Question

I am going to compute $\int_{-\infty}^{\infty} \frac{x^2}{(x^2+1)^2(x^2+2x+2)} \,dx$ I have difficulties doing the integration over semi circle contour using residues for $z=i$ and $z=-1+i$. I have tried this several times and the integral end up with an $i$ in its answer which I never had in my other examples. so I need help how to find the integration over whole semi circle contour. any help is appreciated.

Answer 1

Let $ R> 2 $. Integrating the function $ f:z\mapsto\frac{z^{2}}{\left(z^{2}+1\right)^{2}\left(z^{2}+2z+2\right)}$, over a contour $ \mathscr{C}_{R}=\left[-R,R\right]\cup\gamma_{R} $, where $ \gamma_{R}=\left\lbrace z:\left|z\right|=R,\ \mathcal{Im}\left(z\right)\geq 0\right\rbrace $, gives the following : $$ 2\pi\mathrm{i}\left(\mathrm{Res}\left(f,\mathrm{i}\right)+\mathrm{Res}\left(f,\mathrm{i}-1\right)\right)=\oint_{\mathscr{C}_{R}}{f\left(z\right)\mathrm{d}z}=\int_{-R}^{R}{f\left(x\right)\mathrm{d}x}+\int_{\gamma_{R}}{f\left(z\right)\mathrm{d}z} $$

Since : \begin{aligned}\small \left|\int_{\gamma_{R}}{f\left(z\right)\mathrm{d}z}\right|\leq\int_{\gamma_{R}}{\left|f\left(z\right)\right|\left|\mathrm{d}z\right|}&\small=\int_{\gamma_{R}}{\frac{\left|z\right|^{2}}{\left|z^{2}+1\right|^{2}\left|\left(z+1\right)^{2}+1\right|}\left|\mathrm{d}z\right|}\\ &\small\leq\frac{R^{2}}{\left(R^{2}-1\right)^{2}\left(\left(R-1\right)^{2}-1\right)}\int_{\gamma_{R}}{\left|\mathrm{d}z\right|}=\frac{R^{2}}{\left(R^{2}-1\right)^{2}\left(\left(R-1\right)^{2}-1\right)}\int_{0}^{\pi}{R\,\mathrm{d}\theta}\underset{R\to +\infty}{\longrightarrow}0 \end{aligned}

We get, by tending $ R $ to infinity : $$ \int_{-\infty}^{+\infty}{f\left(x\right)\mathrm{d}x}=2\pi\mathrm{i}\left(\mathrm{Res}\left(f,\mathrm{i}\right)+\mathrm{Res}\left(f,\mathrm{i}-1\right)\right) $$

Now : $$ \mathrm{Res}\left(f,\mathrm{i}-1\right)=\lim_{z\to\mathrm{i}-1}{\left(z-\mathrm{i}+1\right)f\left(z\right)}=\lim_{z\to \mathrm{i}-1}{\frac{z^{2}}{\left(z^{2}+1\right)^{2}\left(z+\mathrm{i}+1\right)}}=\frac{3}{25}-\frac{4\mathrm{i}}{25} $$

And : $$ \mathrm{Res}\left(f,\mathrm{i}\right)=\lim_{z\to\mathrm{i}}{\frac{\mathrm{d}}{\mathrm{d}z}\left(z-\mathrm{i}\right)^{2}f\left(z\right)}=\lim_{z\to \mathrm{i}}{-\frac{2z\left(z^{3}+z^{2}-\mathrm{i}z-2\mathrm{i}\right)}{\left(z+\mathrm{i}\right)^{3}\left(z^{2}+2z+2\right)^{2}}}=-\frac{3}{25}+\frac{9\mathrm{i}}{100} $$

Hence : $$ \int_{-\infty}^{+\infty}{f\left(x\right)\mathrm{d}x}=\frac{7\pi}{50} $$