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I have been asked to solve this integral with residue theory.

\begin{equation} \int_{-\infty}^{\infty} \frac{1}{(2+2x+2x^2)^2} dx \end{equation}

What I have done is to consider the curve formed by the upper half-circle of radius R. If this is $\gamma$, then $\gamma_1$ is the line in the x-axis from $x=-R$ to $x=R$ ($y=0$) and $\gamma_2$ is the half-circle of radius R. Firstly, I compute the integral over $\gamma$ (a closed curve) using Residue Theory, and I obtain that it is equal to $\pi$/$2$. Then I compute the integrals over $\gamma_1$ and $\gamma_2$, in such a way that the first one is just the integral we want to compute and I would like to estimate the second one to conclude that it goes to zero as R goes yo $\infty$, the problem is thta I don't know how to estimate the integral over $\gamma_2$.

Could anybody help me, please?

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3 Answers 3

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\begin{align}\left|\int_{\gamma_2}\frac1{(2x^2+2x+2)^2}\,\mathrm dx\right|&\leqslant\pi R\sup_{|z|=R,\operatorname{Im}z\geqslant 0}\frac1{|2x^2+2x+2|^2}\\&\leqslant\frac{\pi R}{4R^4-8R^3-12R^2-8R-4}\end{align}and$$\lim_{R\to+\infty}\frac{\pi R}{4R^4-8R^3-12R^2-8R-4}=0.$$

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  • $\begingroup$ Why is true the second inequality? $\endgroup$
    – MGF01
    Mar 17, 2018 at 11:10
  • $\begingroup$ \begin{align}|2x^2+2x+2|^2&=|4x^2+8x^3+12x^2+8x+4|\\&\geqslant|4x^4|-|8x^3+12x^2+8x+4|\\&\geqslant4|x|^4-8|x|^3-12|x|^2-8|x|-4\\&=4R^4-8R^3-12R^2-8R-4.\end{align} $\endgroup$ Mar 17, 2018 at 11:13
  • $\begingroup$ Oh I see, perfect. Thank you so much. $\endgroup$
    – MGF01
    Mar 17, 2018 at 11:14
  • $\begingroup$ My approach is fine, right? I computed the integral with wolphram alpha and the result was not $\pi$/$2$ $\endgroup$
    – MGF01
    Mar 17, 2018 at 11:15
  • $\begingroup$ @IreneGil Your approach is fine, but your computations are wrong:$$\operatorname{Res}_{z=-\frac12+\frac{\sqrt3}2i}\left(\frac1{(2z^2+2z+2)^2}\right)=-\frac i{6\sqrt3}.$$ $\endgroup$ Mar 17, 2018 at 11:24
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Some preliminary simplification: $$ \int_{\mathbb{R}}\frac{dx}{(2+2x+2x^2)^2}=\frac{1}{4}\int_{\mathbb{R}}\frac{dx}{\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right)^2}=\frac{1}{4}\int_{\mathbb{R}}\frac{dx}{\left(x^2+\frac{3}{4}\right)^2}=\frac{1}{3\sqrt{3}}\int_{\mathbb{R}}\frac{dx}{(x^2+1)^2} $$ now you may consider a large semicircle in the upper half-plane and evaluate the residue at $x=i$,
or simply notice that by differentiating $\int_{\mathbb{R}}\frac{dx}{x^2+a}=\frac{\pi}{\sqrt{a}}$ with respect to $a$, then evaluating at $a=1$, the RHS of the previous line equals $\frac{\pi}{6\sqrt{3}}$.

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Let $q(z) = \sum_{\nu=0}^{m}a_{\nu}z^{\nu} $. Then, assuming $ q\neq 0$, $$ \frac{1}{q(z)}=z^{-m}\frac{1}{\frac{a_0}{z^m}+...+a_m}.$$ For $ \lvert z\rvert \rightarrow \infty $ the second factor goes to $ a_m $, hence there is $ M>0 $ and $ C>0 $ s.t. for $ M\leq\lvert z\rvert $ $$ \lvert \frac{1}{q(z)}\rvert\leq \lvert z \rvert^{-m}C.$$ If $\gamma_2$ is the half-circle of radius $R\geq M$, we have $$ \lvert \int_{\gamma_2}\frac{1}{q(z)}\rvert = \lvert \int_{0}^{\pi}\frac{Rie^{it}}{q(Re^{it})}dt\rvert \leq R\pi \frac{C}{\lvert Re^{it} \rvert^m}.$$ If $ deg(q)=m \geq 2 $ this goes to zero for $ R\rightarrow \infty $.

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