I'm having difficulty evaluating the following integral using residue theory, and would love some advice on proceeding. Below I develop my approach to the problem:
$$\int_{0}^{\infty} \frac{x^3sin(kx)}{x^4+a^4} dx; \quad k \in \!R,\ a^4>0$$
For simplicity, let's refer to $\frac{x^3}{x^4+a^4}$ as $f(x)$, where we note that $f(x)$ is an odd function, which makes this integral a bit tricky. In the textbook I use, integrals of this type call for a contour that takes the shape of a 'partial arc of a circle' composed of three pieces: $C_R$, $C_L$, and $C_x$, where $C_R:Re^{i\theta}:0\leq \theta \leq \phi$, $C_x$: is along the x-axis, and $C_L$ connects $C_R$ and $C_x$. This results in the following:
$$\int_{C}^{} \frac{z^3e^{ikz}}{z^4+a^4} dz = (\int_{C_L}^{}+\int_{C_x}^{}+\int_{C_R}^{})\frac{z^3e^{ikz}}{z^4+a^4}dz = 2\pi i\sum_{j}^{} Res(f;z_j)$$
Above I have replaced $sin(kz)$ with $e^{ikz}$ since we can take the imaginary part of the resulting integral to get our respective solution. Also note, we specifically select $\phi$ such that $C_L$ will simplify down based on 'symmetry properties' - one of the difficulties I am facing. Furthermore, using Jordan's Lemma, $C_R$ will go to $0$ as R approaches $\infty$. We are then left with evaluating $C_L$ and $C_x$. What choice of $\phi$ should I make to simplify $C_L$? Once I know this, I can easily calculate the relevant residues and solve the problem.
Note: After further thought, I'm unsure that a proper selection of $\phi$ will simplify $C_L$ due to the pesky $e^{ikz}$ term.
We may get rid of a parameter by assuming $a,k>0$ and by setting $x=az, m=ak$.
$$I(a,k)=\int_{0}^{+\infty}\frac{x^3}{x^4+a^4}\sin(kx)\,dx = \int_{0}^{+\infty}\frac{z^3}{z^4+1}\sin(mz)\,dz =J(m).$$ By the residue theorem: $$\begin{eqnarray*} \frac{z^3}{z^4+1} &=& \frac{1}{4}\cdot\frac{1}{z-\frac{1+i}{\sqrt{2}}}+\frac{1}{4}\cdot\frac{1}{z-\frac{-1+i}{\sqrt{2}}}+\frac{1}{4}\cdot\frac{1}{z-\frac{-1-i}{\sqrt{2}}}+\frac{1}{4}\cdot\frac{1}{z-\frac{1-i}{\sqrt{2}}}\\ \end{eqnarray*}$$ and: $$ J(m) = \frac{1}{2}\int_{\mathbb{R}}\frac{z^3\sin(mz)}{z^4+1}\,dz = \frac{1}{2}\text{Im}\int_{\mathbb{R}}\frac{z^3 e^{miz}}{z^4+1}\,dz$$ Now we may consider a contour in the upper half-plane, in such a way that the last integrand function fulfills Jordan's lemma. We get: $$ J(m) = \frac{\pi}{2} e^{-m/\sqrt{2}}\cos\left(m/\sqrt{2}\right).$$