how to integral $\int_{-\infty}^{\infty}{\frac{e^{iyx}}{1+y^2}}dy$ , $x\ge 0$ How to integral $$\int_{-\infty}^{\infty}{\frac{e^{iyx}}{1+y^2}}dy,x\ge 0$$ without using Fourier transform?
In fact, this integral is the fourier inverse transform of $\frac{1}{1+y^2}$. and $\frac{1}{1+y^2}$ is the fourier transform of $e^{-|x|}$. so the integral above is $\pi e^{-x}$. But I do not know how to integrate the above integral directly. 
 A: The suggestion of Friedrich is good, so take it. Once you got it, what you have to solve is
$$\int_{-\infty}^{+\infty} \frac{\cos(xy)}{y^2 + 1}\ \text{d}y$$
Using Residues calculus, we pass to the complex plane:
$$\int_{-\infty}^{+\infty} \frac{\cos(xz)}{z^2 + 1}\ \text{d}z$$
Poles of the denominator are for $z = \pm i$, indeed we have that the denominator can be written as $(z-i)(z+1)$.
When we integrate a function which is of the form $R(z)\cos(z)$, namely a rational function $R(z)$ multiplied by the Cosine or the Sine function, and when we have no pole in the real line (like in this case), we take into account only the upper half complex plane, thence for $z\to \infty$ we have $R(z) = O(z^{-1})$ so the solution $z_1 = i$ is acceptable, whilst the solution $z_2 = -i$ will be skipped. Applying residue calculus the solution is
$$2\pi i \text{Res}\left(\lim_{z\to i} (z-i)\frac{\cos(xi)}{(z+i)(z-i)}\right) = 2\pi i\frac{\cos(ix)}{2i} = \pi\cos(ix) = \pi\cosh(x)$$
Let's evaluate instead the whole integral
$$\int_{-\infty}^{+\infty}\frac{e^{i x y}}{y^2+1}\ \text{d}y$$
We run the problem in the same way as before, denoting
$$f(z) = \frac{e^{i x z}}{z^2+1} = \frac{e^{i x z}}{(z+i)(z-i)}$$
The singularity $z_1 = -i$ will be skipped for the same reasons as before, whilst with the other one, $z_2 = i$, we calculate the residue by substituting it, just as before:
$$\text{Res}(i, f(z)) = \frac{e^{i^2 x}}{i + i} = \frac{e^{-x}}{2i}$$
Thence the solution is
$$\int_{-\infty}^{+\infty}\frac{e^{i x y}}{y^2+1}\ \text{d}y = 2\pi i\ \text{Res}(i, f(z)) = 2\pi i \frac{e^{-|x|}}{2i} = \pi e^{-x}$$
