So I'm trying to solve the integral $$\int_{-\infty}^\infty \frac{x^3e^{ix}}{x^4+16} \,dx$$
To do this, I'm integrating of a semicircle of radius $R$ in the positive imaginary axis. The line from $-\infty$ to $\infty$ will be my answer, I just need to show that the upper circular part goes to $0$. Then I can use the Residue Theorem to evaluate this integral.
My issue lies in showing that the upper part goes to $0$ as $R$ approaches $\infty$.
Here is the work I've done so far.
First I parameterize as $x = Re^{i\theta}$ and $dx = Rie^{i\theta}$. Plugging this into the integral (and changing the limits gives me
$$\int_{0}^\pi \frac{R^3e^{3i\theta}e^{iRe^{i\theta}}Ri}{R^4e^{4i\theta}+16} \,dx$$
If I take the modulus of this, $|R^3e^{3i\theta}| = R^3$, $|e^{iRe^{i\theta}}| = e^{-Rsin(\theta)}$, and $|R^4e^{4i\theta}+16| \geq |R^4-16|.$
So my integral is bounded by $$\int_{0}^\pi \frac{iR^4e^{-Rsin(\theta)}}{|R^4-16|} \,dx$$
Where do I go from here? I know that I'm supposed to let $R$ approach infinity, which will make the exponential approach $0$. However, I think I have to be more rigorous since I'm dealing with an integral, and there are points where $sin(\theta) = 0$, which would instead make the exponential become $1$, making this become $1$ as $R$ approaches infinity.