Show that $\int_{- \infty}^{\infty} \frac{\sin x}{x+i} dx = \frac{\pi}{e}$

This is question 37 from Exercises 6.6.2 from Dennis Zill's An Introduction to Complex Analysis.

Show that $$\int_{- \infty}^{\infty} \frac{\sin x}{x+i} dx = \frac{\pi}{e}$$

I am not sure how to proceed. There is a hint to use $$\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}$$ for real $$x$$.

If I try to find the residue of $$\dfrac{\sin z}{z+i}$$ at $$z=-i$$, I get $$\text{Res}\left(\dfrac{\sin z}{z+i};-i\right) = -i \sinh 1.$$

But I am not sure what contour to even use, so I cannot relate these numbers to the integral in the question.

I tried to use a semicircular contour in the lower half plane. I see that $$|e^{iz}| = e^{-y} \leq e^{R}$$ for $$z=x+iy$$ on a semicircle of radius $$R$$. I am not sure how to proceed since this would mean the integral does not vanish. Could someone please offer some guidance? Thank you very much.

• You can use a different contour for the $e^{ix}$ piece vs. the $e^{-ix}$ piece.
– Ian
Dec 11, 2022 at 22:17

I split the integral into $$2i I_1 = \int_{\mathbb{R}} \dfrac{e^{ix}}{x+i} dx$$ and $$2i I_2 = \int_{\mathbb{R}} \dfrac{-e^{-ix}}{x+i} dx .$$
For the first integral, consider the contour $$\Gamma_R$$ consisting of the semicircle of radius $$R$$, in the upper half plane, denoted $$\gamma_R$$, and the line segment joining $$-R$$ to $$R$$. Note that the function is holomorphic on and inside this countour. Thus, using Cauchy's Integral formula, $$\int_{\Gamma_R} \dfrac{e^{iz}}{z+i} dz = \int_{-R}^R \dfrac{e^{ix}}{x+i} dx + \int_{\gamma_R} \dfrac{e^{iz}}{z+i} dz=0.$$
On $$\gamma_R$$, we have $$|e^{iz}| = e^{-R \sin \theta}$$. By the Estimation lemma and Jordan's inequality, the modulus of the integral will converge to $$0$$ as $$R \to \infty$$. Thus the integral converges to $$0$$ as $$R \to \infty$$. Thus we conclude that $$I_1 =0$$.
For $$I_2$$, we can substitute $$u = -x$$, and thus obtain get $$2iI_2 = \int_{\mathbb{R}} \dfrac{-e^{iu}}{-u+i} du$$. Note that the function has a residue of $$1/e$$ at the pole $$z=i$$. We apply Cauchy's Residue Theorem and the same method as in the earlier part to finally obtain that $$2iI_2 = 2 \pi i/e.$$ We can also use a contour in the lower half-plane, but the substitution we used transforms this into the first case which we already solved.