calculate the integral $ \int_{-\infty}^{+\infty} \frac{e^{-itx}}{1 + t^2} dt $

Question

It is possible to calculate the integral $ \int_{-\infty}^{+\infty} \frac{e^{-itx}}{1 + t^2} dt $

without using the residue theorem, nor fourier transforms

Answer 1

Without Fourier transform, consider $$\frac{e^{-itx}}{1 + t^2}=\frac{i \,e^{-itx}}2 \left(\frac{1}{ (t+i)}-\frac{1}{ (t-i)} \right)$$ making the antiderivative to be $$\int \frac{e^{-itx}}{1 + t^2}\,dt=\frac{1}{2} i e^{-x} \text{Ei}(x(1-i t))-\frac{1}{2} i e^x \text{Ei}(-(x+i t))$$ where appears the exponential integral function.

Using the bounds and provided $x\in \mathbb{R}$ $$\int_{-\infty}^{+\infty} \frac{e^{-itx}}{1 + t^2}\, dt=\pi e^{-|x|}$$