I was wondering if we can use complex contour integration to evaluate the integral $$ \int_0^\infty \frac {\sin x}{1+x^3}dx. $$ Since the integrand is not even, we cannot extend the integration domain to $\mathbb R$ and use the upper semicircular contour $Re^{it}$, $0\le t\le \pi$. We cannot use the keyhole contour either, since the integral of $\frac{e^{iz}}{1+z^3}$ over the lower semicircle $Re^{it}$, $\pi\le t\le 2\pi$ does not converge to $0$ as $R\to \infty$.
Using Wolframalpha to evaluate the above improper integral, it does not give a closed form answer. Therefore I was wondering if complex integration is not applicable in this case.
Without complex integration.
Write $$1+x^3=(x+1)(x-a)(x-b)$$ where $$a=\frac{1+i \sqrt{3}}{2}\qquad \text{and} \qquad b=\frac{1-i \sqrt{3}}{2}$$ $$\frac 1{1+x^3}=\frac{1}{(a+1) (a-b) (x-a)}+\frac{1}{(b+1) (b-a) (x-b)}+$$ $$\frac{1}{(a+1) (b+1) (x+1)}$$ to face three integrals looking like $$I_k=\int \frac {\sin(x)}{x+k}\,dx$$ Let $x+k=t$, expand the sine to face the sine and cosine integrals. Back to $x$ $$I_k=\cos (k)\, \text{Si}(k+x)-\sin (k)\, \text{Ci}(k+x)$$
So, if $\Re(k)>0\lor k\notin \mathbb{R}$ $$J_k=\int_0^\infty \frac {\sin(x)}{x+k}\,dx=\text{Ci}(k) \sin (k)+\frac{1}{2} (\pi -2 \text{Si}(k)) \cos (k)$$
The formula is a bit too long to type here but you have the formula.