Using Complex contour integration calculate $int_{-infty}^{infty} frac{sin x}{x+i}$

Using Complex contour integration calculate $$\int_{-\infty}^{\infty}

\frac{\sin x}{x+i} dx$$ . Use the form $\int_{-\infty}^{\infty} f(x) \sin
(\alpha x) dx$

Now I used the form $\int_{-\infty}^{\infty} f(x) \sin (\alpha x) dx$ and converted the integral to $Img(\oint_{-\infty}^{\infty} \frac{e^{ix}}{x+i}dx)$ where the contour is the positive semi-circle around the origin from $[-R,R]$ as $R \to \infty$

But then the only pole of the above integral is $x=-i$, which is not in the above contour hence the value of the integral in the above contour is zero thus the value of the integral is zero . But the answer given in the text is not so

Could someone please calculate this integral

Answer

Hint: $$\int_{-\infty}^{\infty}\frac{\sin x}{x+i}\,dx= \int_{-\infty}^{\infty}\frac{x\sin x}{x^2+1}\,dx-i\int_{-\infty}^{\infty}\frac{\sin x}{x^2+1}\,dx$$

and you can easily apply Jordan's lemma on these guys and calculate the integrals from residue theorem.