# Fourier Transform of $\displaystyle \frac{x}{x+ic}$

$$\int_{-\infty} ^\infty \frac{x}{x+ic}e^{ikx}dx$$ where $c$ is some positive real constant.

I've tried substituton $u=x+ic$ giving:

$$\int_{-\infty}^\infty \frac{u-ic}{u}e^{ik(u-ic)}du \\ = \int_{-\infty} ^\infty \frac{u-ic}{u}e^{iku}e^{kc}du \\ = e^{kc}\left(\int_{-\infty} ^\infty e^{iku}du-ic\int_{-\infty} ^\infty\frac{e^{iku}}{u}du\right) \\ = e^{kc}\left(\left.\frac{e^{iku}}{ik}\right|_{-\infty} ^\infty -ic\int_{-\infty} ^\infty\frac{e^{iku}}{u}du\right)$$

I don't know how to evaluate the second integral, and I'm not sure if there's just an easier way to do the whole calculation that I'm missing.

• Do you know about complex contour integration? This function has a simple pole at $z=-ic$ that can be easily computed. Commented Nov 20, 2017 at 4:16
• Note that $\frac{x}{x+ic} = 1 - \frac{ic}{x+ic}$ and that $\int_{-\infty}^{+\infty} e^{i k x}\, dx = 2\pi\,\delta(k)$. Commented Nov 20, 2017 at 4:43
• Never crossed my mind to use a contour integral! Thanks Dylan Commented Nov 20, 2017 at 8:46

Using $\frac{x}{x+ic} = 1 - \frac{ic}{x+ic}$, we have \begin{align} I(k,c) &= \int_{-\infty}^{+\infty}dx\, \frac{x}{x+ic} e^{i k x}\\ &= \int_{-\infty}^{+\infty}dx\,e^{i k x} \;-\; ic \int_{-\infty}^{+\infty}dx\,\frac{e^{i k x}}{x+ic}\\ &=2\pi\,\delta(k)\;-\; ic \int_{-\infty}^{+\infty}dx\,\frac{e^{i k x}}{x+ic}\, ,\qquad\qquad\qquad (1) \end{align} where we have used the integral representation of the Dirac delta function, $$\delta(k) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}dx\,e^{i k x}\, .$$
The latter integral in equation (1) above can be evaluated by contour integration using a semicircular contour. Since $c>0$, the pole at $-ic$ is in the lower half-plane. When $k>0$ we must close the contour in the upper-half-plane, so that the $e^{ikx}$ term will decay properly. Since there are no poles in the upper-half-plane, this yields $0$.
When $k<0$, we close the contour in the lower-half-plane, enclosing the pole. Thus: $$-\; ic \int_{-\infty}^{+\infty}dx\,\frac{e^{i k x}}{x+ic} \;=\; -ic \cdot (-2\pi i) \cdot e^{i k (-ic)} \;=\; -2\pi\, c\, e^{-c |k|} \qquad (k < 0)$$
When $k = 0$, the integral in Eqn. (1) becomes \begin{align} -ic \int_{-\infty}^{+\infty}dx\,\frac{1}{x+ic} &= -ic \int_{-\infty}^{+\infty}dx\,\frac{x-ic}{x^2+c^2}\\ &= -ic \int_{-\infty}^{+\infty}dx\,\frac{x}{x^2+c^2} \;-\; c^2 \int_{-\infty}^{+\infty}dx\,\frac{1}{x^2+c^2}\\ &= -ic \times 0 \;-\; \pi c\, , \end{align} where the first integral becomes zero by symmetry.
Putting all of this together, we have $$I(k,c) \;=\; 2\pi\,\delta(k)\;-\;\pi\, c\, e^{-c |k|}\times \begin{cases} 0 & k >0\\ 1 & k = 0\\ 2 & k < 0 \end{cases}\, .$$