integrate $\frac 1{x^2 + i}$ from $-\infty$ to $\infty$

Question

I have a question regarding this complex integral. How do I calculate $$\int_{-\infty}^{\infty} \frac{1}{x^2 + i} \,{\rm d}x?$$ I keep getting the result $0$ and most probably doing it the wrong way.

Answer 1

Hint. Alternatively, one may observe that $$ \int_{-\infty}^\infty\frac{1}{1+x^4}\:dx=\int_{-\infty}^\infty\frac{1}{\left(x-\frac1x\right)^2+2}\:\frac{dx}{x^2}=\int_{-\infty}^\infty\frac{1}{x^2+2}\:dx=\frac{\pi }{\sqrt{2}} $$similarly $$ \int_{-\infty}^\infty\frac{x^2}{1+x^4}\:dx=\int_{-\infty}^\infty\frac{1}{\left(x-\frac1x\right)^2+2}\:dx=\int_{-\infty}^\infty\frac{1}{x^2+2}\:dx=\frac{\pi }{\sqrt{2}} $$ then, by writing $$ \int_{-\infty}^\infty\frac{dx}{x^2+i}=\int_{-\infty}^\infty\frac{x^2-i}{x^4-i^2}\:dx=\int_{-\infty}^\infty\frac{x^2\:dx}{1+x^4}-i\int_{-\infty}^\infty\frac{dx}{1+x^4} $$ one may conclude with the preceding identities.

Answer 2

An efficient approach is to use the Residue Theorem. Proceeding, we have immediately

$$\begin{align} \int_{-\infty}^\infty \frac{1}{x^2+i}\,dx7&=2\pi i\text{Res}\left(\frac{1}{z^2+i}, z=e^{i3\pi/4}\right)\\\\ &=2\pi i \frac{1}{-2e^{-i\pi/4}}\\\\ &=-\pi e^{i3\pi/4}\\\\ &=\frac{\pi}{\sqrt{2}}(1-i) \end{align}$$

And we are done!