Compute $\int_{\mathbb R}\frac{dx}{1+2x^2+x^4}$

Question

I want to compute Compute $$\int_\mathbb R\frac{1}{1+2x^2+x^4}dx$$ using the residue theorem and I've already applied the standard method of integrating along the upper half of the unit circle. But I have encountered a little problem when trying to show that one of the contour integral vanishes:

So first I parameterize the upper arc via $$\gamma:=\begin{cases}[0,\pi]\to\mathbb C\\ t\mapsto Re^{it}\end{cases}$$ and I want to show that with $f(z):=\frac{1}{z^4+2z^2+1}$ we have $$\lim_{R\to\infty}\left|\int_\gamma f(z)dz\right|=0$$ because I can't use the standard estimation. The denominator has 3 summands so I'm unsure how to apply a certain inequality to it. How can I estimate $|z^4+2z^2+1|$? Essentially I just want to say that $|z^4+2z^2+1|\geq |z|^4-2|z|^2-1$ and maybe that's even true but as of now I don't see why.

Answer 1

$|z^{4}+2z^{2}+1|=|z^{2}+1|^{2} \geq (R^{2}-1)^{2}$.

Answer 2

Just use the fact that\begin{align}\lvert z^4+2z^2+1\rvert&=\bigl\lvert z^4-(-2z^2-1)\bigr\rvert\\&\geqslant\lvert z^4\rvert-\lvert-2z^2-1\rvert\\&\geqslant\lvert z\rvert^4-\bigl(\lvert-2z^2\rvert+1\bigr)\\&\geqslant\lvert z\rvert^4-2\lvert z\rvert^2-1.\end{align}

Answer 3

Here's a real-variable way if you're interested: $$I=\int_{-\infty}^{\infty}\frac{dx}{x^4+2x^2+1}=2\int_0^{\infty}\frac{dx}{(x^2+1)^2}$$

Now let $x=\frac{1}{y}\implies dx=\frac{-dy}{y^2}$

So we have $$I=-2\int^0_{\infty}\frac{1}{y^2}\frac{dy}{\left(\frac{1}{y^2}+1\right)^2}=2\int_0^{\infty}\frac{1}{y^2}\frac{dy}{\left(\frac{y^2+1}{y^2}\right)^2}=2\int_0^{\infty}\frac{y^2}{(y^2+1)^2}dy$$ Then write $$I=2\int_0^{\infty}\frac{\left(y^2+1\right)-1}{(y^2+1)^2}dy=2\int_0^\infty\left(\frac{1}{y^2+1}-\frac{1}{(y^2+1)^2}\right)dy$$ $$=2\int_0^{\infty}\frac{dy}{y^2+1}-2\int_0^\infty\frac{dy}{(y^2+1)^2}=2\arctan(y)\big|^{\infty}_0-I$$

Thus we have $$I=2\arctan(\infty)-2\arctan(0)-I\implies2I=2\left(\frac{\pi}{2}\right)-0\implies \boxed{I=\frac{\pi}{2}}$$

Answer 4

Another solution without residue theorem: Substitute $$ x=\tan u, ~~ dx=\frac{du}{\cos^2 u}, ~~ 1+x^2=\frac1{\cos^2 u} $$ so that $$ I=\int_{\Bbb R}\frac{dx}{(1+x^2)^2} =\int_{-\frac\pi2}^{\frac\pi2}\cos^2u\,du =\int_0^{\frac\pi2}(1+\cos(2u))\,du =\frac\pi2 $$

Answer 5

While we're apparently solving this without the residue theorem:

For large enough $s$, let $$I(s) = \int_{-\infty}^{\infty} (x^2 + 1)^{-s} \, \mathrm{d}x.$$ Then $$\Gamma(s) I(s) = \int_0^{\infty} \int_{-\infty}^{\infty} y^{s-1} (x^2 + 1)^{-s} e^{-y} \, \mathrm{d}x \, \mathrm{d}y.$$ Making the change of variables $y \mapsto y(x^2+1)$ turns this into \begin{align*} \Gamma(s) I(s) &= \int_0^{\infty} y^{s-1} e^{-y} \int_{-\infty}^{\infty} e^{-yx^2} \, \mathrm{d}x \, \mathrm{d}y \\ &= \sqrt{\pi} \int_0^{\infty} y^{s-3/2} e^{-y} \, \mathrm{d}y \\ &= \Gamma(s-1/2)\sqrt{\pi} , \end{align*}

so $\int_{-\infty}^{\infty} (x^2 + 1)^{-s} \, \mathrm{d}x = \frac{\Gamma(s-1/2)\sqrt{\pi}}{\Gamma(s)}.$ At $s=2$ you obtain $\Gamma(3/2)\sqrt{\pi} = \pi/2.$