I am asked to evaluate
$$\int_{-\infty}^{\infty}\frac{x\sin(mx)}{x^4 + 1}\ dx$$
So I know it's a D shaped contour, and I've calculated that the poles that I need are \begin{align}e^{\frac{1+i}{\sqrt 2}}\\ e^{\frac{-1+i}{\sqrt2}}\end{align}
and I know I have to calculate the residue at this point etc.
But the issue I have is that the answer I'm getting is different to the one in the solution. And I don't think the answer in the solution can be right as it says to take Real parts at the end but surely I want to take imaginary parts as it's sin function.
The answer in the solution is $$\pi e^{\frac{−m}{\sqrt2}}\sin\left(\frac{m}{\sqrt2}\right)$$
Let $$f(z)=\frac{ze^{imz}}{1+z^4}.$$ Let $\gamma_r$ be the D-shaped contour running across the real line from $-r$ to $+r$ then counter-clockwise through the upper half-plane via the semi-circle of radius $r$. Trivial estimates show that as $r\to +\infty$ the contour integral of $f(z)$ over the arc tends to $0$. From this it follows that $$\int_{-\infty}^{+\infty}\frac{x\sin(mx)}{1+x^4}\ dx=\Im\lim_{r\to+\infty}\oint_{\gamma_r}f(z)\ dz.$$
For $r$ sufficiently large, $\gamma_r$ will contain all the poles with positive imaginary parts and the integral of $f(z)$ over $\gamma_r$ will be constant. Thus to evaluate the integral in the limit $r\to +\infty$ it suffices to evaluate the residues in the poles with positive imaginary part, namely in the poles $$z_\pm=\frac{\pm1+i}{\sqrt{2}}.$$ Some algebra will show that $$\textrm{Res}_{z=z_-}\frac{1}{1+z^4}=\sqrt{2}\frac{1-i}{8}\\ \textrm{Res}_{z=z_+}\frac{1}{1+z^4}=\sqrt{2}\frac{-1-i}{8}$$ hence $$\textrm{Res}_{z=z_-}f(z)=z_-e^{imz_-}\sqrt{2}\frac{1-i}{8}=\frac{1}{4}e^{-\frac{m}{\sqrt{2}}}\left(\sin{\frac{m}{\sqrt{2}}+i\cos{\frac{m}{\sqrt{2}}}}\right)\\ \textrm{Res}_{z=z_+}f(z)=z_+e^{imz_+}\sqrt{2}\frac{-1-i}{8}=\frac{1}{4}e^{-\frac{m}{\sqrt{2}}}\left(\sin{\frac{m}{\sqrt{2}}-i\cos{\frac{m}{\sqrt{2}}}}\right).$$ Now we sum the residues and multiply by $2\pi i$ to find that for sufficiently large $r$: $$\oint_{\gamma_r}f(z)\ dz=\pi i e^{-\frac{m}{\sqrt{2}}}\sin{\frac{m}{\sqrt{2}}}$$ hence the desired result:$$\int_{-\infty}^{+\infty}\frac{x\sin(mx)}{1+x^4}\ dx=\pi e^{-\frac{m}{\sqrt{2}}}\sin{\frac{m}{\sqrt{2}}}.$$