Is this integral done correctly $\int_{-\infty}^{\infty} \frac{x^2+3}{(x^2+4)^2} dx $

Question

$\int_{-\infty}^{\infty} \frac{x^2+3}{(x^2+4)^2} dx = 2 \pi i * res f(x_0)$
Where $x_0$ is residue where $x\geq 0$ for our function $x_0=2i$
$$resf(x_0)= \lim_{x \rightarrow 2i} \frac{d}{dx}[(x-2i)^2 \frac{x^2+3}{(x-2i)^2(x+2i)^2}=\lim_{x \rightarrow 2i} \frac{-x^4+x^3(2-4i)+x^2(8i+7)-x(8+12i)+12}{(x+2i)^4}=-1 $$ So then the integral $$\int_{-\infty}^{\infty} \frac{x^2+3}{(x^2+4)^2} dx = 2 \pi i *(-1)=-2\pi i$$

Answer 1

It looks like you took the limit as $x\to\infty$ instead of evaluating it at $x=2i$. Also you should note the following from the quotient rule when you have a squared denominator:

$\dfrac{d}{dx}\dfrac{f(x)}{[g(x)]^2}=\dfrac{f'g^2-2fgg'}{g^4}=\dfrac{f'g-2fg'}{g^3},$

canceling out a factor of $g(x)$. This gives a simpler form to calculate later on.

Answer 2

A real integral which gives an imaginary solution? That's sounds strange. Taking $$f(z)=\frac{z^2+3}{(z^2+4)^2},$$ and the contour $$\Gamma_R=\gamma_R\cup [-R,R],$$ for $R>2$, where $\gamma_R(t)=Re^{it},$ for $0\leq t\leq \pi,$ gives us that $$\int_{-\infty}^\infty\frac{x^2+3}{(x^2+4)^2}dx=\lim_{R\to\infty}\left(\int_{\Gamma_R}\frac{z^2+3}{(z^2+4)^2}dz-\int_{\gamma_R}\frac{z^2+3}{(z^2+4)^2}dz\right).$$ By the residue's theorem gives us, since the unique pole inside the contour is $2i$, open pole of order two,we have that $$\int_{\Gamma_R}\frac{z^2+3}{(z^2+4)^2}dz=2\pi i\operatorname{Res}(f,z=2i)=2\pi i\frac{d}{dz}\bigg|_{z=2i}\frac{z^2+3}{(z+2i)^2}=2\pi i\frac{2(2iz-3)}{(z+2i)^3}\bigg|_{z=2i}=2\pi i\frac{-14}{-64i}=\frac{7\pi}{16}.$$To conclude we have to check that $$\int_{\gamma_R}\frac{z^2+3}{(z^2+4)^2}dz\to 0,$$ as $R\to \infty$. Using the parametrization we get that the integral converts to $$\int_0^\pi\frac{R^2e^{2it}+3}{(R^2e^{2it}+4)^2}Rie^{it}dt,$$ since $|e^{2it}|=1,$ and we can write $$R^2-a=|R^2-a|=||R^2e^{2it}|-|-a||\leq |R^2e^{2it}+a|,$$ for $a=3,4$, we get that $$\left|\int_{\gamma_R}\frac{z^2+3}{(z^2+4)^2}dz\right|\leq \int_{\gamma_R}\left|\frac{z^2+3}{(z^2+4)^2}\right|dz\leq \pi\frac{R(R^2-3)}{(R^2-4)^2}\to 0,$$ as $R\to\infty$. So we conclude that $$\int_{-\infty}^\infty\frac{x^2+3}{(x^2+4)^2}dx=\lim_{R\to\infty}\left(\int_{\Gamma_R}\frac{z^2+3}{(z^2+4)^2}dz-\int_{\gamma_R}\frac{z^2+3}{(z^2+4)^2}dz\right)=\frac{7\pi}{16}.$$

Answer 3

Real method:Integration by Parts $$ \begin{aligned} \int_{-\infty}^{\infty} \frac{x^2+3}{\left(x^2+4\right)^2} d x & =-\int_{-\infty}^{\infty} \frac{x^2+3}{2 x} d\left(\frac{1}{x^2+4}\right) \\ & =\frac{1}{2} \int_{-\infty}^{\infty}\left(1-\frac{3}{x^2}\right) \frac{1}{x^2+4} d x \\ & =\frac{1}{4}\left[\tan ^{-1}\left(\frac{x}{2}\right)\right]_{-\infty}^{\infty}+\frac{3}{16} \int_{-\infty}^{\infty}\left(\frac{1}{x^2+4}-\frac{1}{x^2}\right) d x \\ & =\frac{\pi}{4}+\frac{3}{16}\left[\tan ^{-1}\left(\frac{x}{2}\right)+\frac{1}{x}\right]_{-\infty}^{\infty} \\ & =\frac{7 \pi}{16} \\ & \end{aligned} $$