Find $\displaystyle\int_{-\infty}^{\infty}\frac{\cos x\,dx}{x^2+i}$
I noted $\displaystyle J=\int_{-\infty}^{\infty}\frac{\cos x\,dx}{x^2+i}$ and $\displaystyle I=\int_{-\infty}^{\infty}\frac{\sin x\,dx}{x^2+i}$
So $\displaystyle J+iI=\int_{-\infty}^{\infty}\frac{e^{ix}\,dx}{x^2+i},$ $\quad\displaystyle J-iI=\int_{-\infty}^{\infty}\frac{e^{-ix}\,dx}{x^2+i}$
I used the residues formula , but i did not find the value of the integral.
It's ok. Let's compute the general integral, meant as the principal value:
$$\int_{-\infty}^{+\infty} \frac{\cos(x)}{x^2+a}\ dx$$
at the end we will plug $a\to i$ (we can do it because what matters is that $\Re(a)>0$).
Passing in the complex plane $x\to z$, the denominator can be written as $(z + i\sqrt{a})(z-i\sqrt{a})$.
Also $\cos(z) = \Re(e^{iz})$ hence:
$$\text{res}[f(z), z_0] = \lim_{z\to z_0} (z-z_0)f(z)$$
$$\text{res}[f(z), z_0] = \lim_{z\to ia} (z-i\sqrt{a}) \frac{e^{iz}}{(z+i\sqrt{a})(z-i\sqrt{a})} = \frac{e^{i(i\sqrt{a})}}{2i\sqrt{a}} = \frac{e^{-\sqrt{a}}}{2i\sqrt{a}}$$
Whence
$$\int_{-\infty}^{+\infty} \frac{\cos(z)}{z^2+a}\ dx = 2\pi i \text{res} = 2\pi i\frac{e^{-\sqrt{a}}}{2i\sqrt{a}} = \color{red}{\pi \frac{e^{-\sqrt{a}}}{\sqrt{a}}}$$
Plugging now $a\to i$ gives you the final correct result:
$$\color{blue}{-(-1)^{3/4} e^{-\sqrt[4]{-1}} \pi}$$
