integration - Integral over the real axis using residues

I have to evaluate the following integral
$$\int_{-\infty}^\infty \frac{x\sin{x}}{x^2-4x+8}dx$$
I know that for such integrals we can use residues and
$$\int_{-\infty}^\infty \frac{x\sin{x}}{x^2-4x+8}dx = 2\pi i \sum_{k=0}^n \text{res}[f(z),a_k] + \lim_{R\to\infty}\int_{\Gamma_R} f(z)dz$$
I see that $f(z) = \frac{x\sin{x}}{x^2-4x+8}$ has one singularity in the upper half of the plane which is $x = 2+2i$ and I have computed that res$[f(z),2+2i] = \left(\frac{1}{2}-\frac{i}{2}\right)\sin{(2+2i)}$, but I have problems evaluating the line integral. Usually we have shown that this goes to $0$ as $R$ gets large, but in this case it doesn't seem to be so. How can I go about evaluating the line integral? Thanks for any advice.