Evaluate \[ \int_{-\infty}^{\infty}\frac{x^6}{(1+x^4)^2}\,dx. \]
Solution
The value of the integral is \[ \boxed{\frac{3\pi\sqrt2}{8}}. \]
1. Let\[ F(z)=\frac{z^6}{(1+z^4)^2}. \]On the upper semicircle \(|z|=R\), the integral of \(F\) tends to zero as\(R\to\infty\).
For \(R\) sufficiently large, \[ |1+z^4|\ge |z|^4-1=R^4-1 \] on \(|z|=R\). Hence \[ |F(z)|\le \frac{R^6}{(R^4-1)^2}=O(R^{-2}). \] The semicircle has length \(\pi R\), so the \(ML\) estimate gives an \(O(R^{-1})\) bound for its contribution, which tends to zero.
2. The poles of \(F\) in the open upper half-plane are the two doublepoles\[ a_1=e^{i\pi/4}, \qquad a_2=e^{3i\pi/4}. \]
The poles are the roots of \[ z^4=-1=e^{i(\pi+2\pi k)}, \] namely \[ e^{i(\pi/4+k\pi/2)},\qquad k=0,1,2,3. \] Exactly the first two have positive imaginary part. Since every root of \(1+z^4\) is simple and the denominator is squared, these are double poles of \(F\).
3. If \(a^4=-1\), then\[ \operatorname{Res}(F,a)=\frac{3}{16a}. \]
Factor \[ 1+z^4=(z-a)q(z). \] Then \[ F(z)=\frac{1}{(z-a)^2}\frac{z^6}{q(z)^2}, \] so, because the pole is double, \[ \operatorname{Res}(F,a) =\left.\frac{d}{dz}\left(\frac{z^6}{q(z)^2}\right)\right|_{z=a}. \] From \[ (1+z^4)'=4z^3 \] we get \[ q(a)=4a^3. \] Differentiating \[ 1+z^4=(z-a)q(z) \] twice and evaluating at \(z=a\) gives \[ 12a^2=2q'(a), \qquad q'(a)=6a^2. \] Put \[ h(z)=\frac{z^6}{q(z)^2}. \] Then \[ h(a)=\frac{a^6}{(4a^3)^2}=\frac1{16} \] and logarithmic differentiation gives \[ \frac{h'(a)}{h(a)} =\frac6a-2\frac{q'(a)}{q(a)} =\frac6a-2\frac{6a^2}{4a^3} =\frac3a. \] Therefore \[ \operatorname{Res}(F,a)=h'(a)=\frac{3}{16a}. \]
4. The sum of the upper-half-plane residues is\[ -\frac{3\sqrt2}{16}i. \]
By 3, \[ \begin{aligned} \operatorname{Res}(F,a_1)+\operatorname{Res}(F,a_2) &=\frac3{16}\left(e^{-i\pi/4}+e^{-3i\pi/4}\right)\\ &=\frac3{16}\left( \frac{1-i}{\sqrt2}+\frac{-1-i}{\sqrt2} \right)\\ &=-\frac{3\sqrt2}{16}i. \end{aligned} \]
5. Hence\[ \int_{-\infty}^{\infty}\frac{x^6}{(1+x^4)^2}\,dx =\frac{3\pi\sqrt2}{8}. \]
Integrate \(F\) over the positively oriented upper-half-plane semicircle of radius \(R\). By the residue theorem, \[ \int_{-R}^{R}F(x)\,dx +\int_{\text{semicircle}}F(z)\,dz =2\pi i\left(-\frac{3\sqrt2}{16}i\right). \] By 1 the semicircle contribution tends to zero. Letting \(R\to\infty\) therefore gives \[ \int_{-\infty}^{\infty}\frac{x^6}{(1+x^4)^2}\,dx =2\pi i\left(-\frac{3\sqrt2}{16}i\right) =\frac{3\pi\sqrt2}{8}. \]