# Fourier transform of a function

I am tasked with finding the fourier transform of: $$f(t)=\frac{1}{1+jt}$$ where $j=\sqrt{-1}$

The fourier transform is given by: $$\int_{-\infty}^{\infty} \frac{e^{-j\omega t}}{1+jt}\,dt$$ substituting $1+jt=x \Rightarrow dt=dx/j$, I get: $$\frac{e^{\omega}}{j}\int_{1-j\infty}^{1+j\infty} \frac{e^{-\omega x}}{x}\,dx$$ I am stumped here. I think above can be solved using complex analysis but I am not sure about which counter to use.

Please avoid the use of duality property because I am more interested in solving this problem from the definition of fourier transform.

Any help is appreciated. Thanks!

• u can apply jordans lemma and integrate around a bic semicircle closed in the uhp or lhp depending on the sign of $\omega$ – tired Oct 24 '15 at 16:09
• @tired: if you don't mind, can you please provide a complete solution? I am a bit rusty on my complex analysis skills as its been months I used those tools. Thanks! :) – Pranav Arora Oct 24 '15 at 16:10
• i have no time at the moment, but have a look at example 2 here: en.wikipedia.org/wiki/Methods_of_contour_integration to get an idea what i mean. if more help is needed, i can offer it tomorrow, – tired Oct 24 '15 at 16:19
• Do you still need help? – tired Oct 24 '15 at 20:55

Here comes the standard solution for calculating this integral. However, I suggest to have a look at Jordan's lemma, since what is below will merely be a proof of that in our specific case.

I suggest that you use the following contour for $\omega<0$:

and the similar one in the lower half plane for $\omega>0$.

Let me hint on the calculation for $\omega<0$. The one for $\omega>0$ is similar, but simpler since, there is no pole. By the Residue theorem, you have $$\int_{-R}^R\frac{e^{-i\omega t}}{1+it}\,dt+\int_{\gamma_R}\frac{e^{-i\omega t}}{1+it}\,dt=2\pi i \,\text{Res}\,\Bigl(\frac{e^{-i\omega t}}{1+it},i\Bigr)$$ As $R\to+\infty$, the first integral converges to what we want. The residue is easy to calculate, and I leave that to you. Let us concentrate on showing that the second integral tends to $0$ as $R\to+\infty$.

Parametrize $\gamma_R$ in the figure as $$t=Re^{i\theta},\quad 0<\theta<\pi.$$ Then, since $|1+it|\geq |it|-1=R-1$, $dt=iRe^{i\theta}\,d\theta$, and $$|e^{-i\omega t}|=e^{\omega R\sin\theta},$$ we find that, by the triangle inequality, $$\begin{split} \biggl|\int_{\gamma_R}\frac{e^{-i\omega t}}{1+it}\,dt\biggr|&\leq \frac{R}{R-1}\int_0^{\pi}e^{\omega R\sin\theta}\,d\theta\\ &=\frac{2R}{R-1}\int_0^{\pi/2}e^{\omega R\sin\theta}\,d\theta \end{split}$$ On the interval $0<\theta<\pi/2$, we have $\sin\theta>2\theta/\pi$, so the right-hand side above is bounded by $$\frac{2R}{R-1}\int_0^{\pi/2}e^{\omega R2\theta/\pi}\,d\theta=\frac{\pi}{(R-1)\omega}\bigl(e^{R\omega}-1\bigr).$$ Since $\omega<0$, this expression tends to zero as $R\to+\infty$.

I don't know what you consider OK to use, but I would recommend to write $$\frac{1}{1+jt}=\frac{1}{1+t^2}-jt\frac{1}{1+t^2},$$ And $$\mathcal F\Bigl(\frac{1}{1+t^2}\Bigr)(\omega)=\pi e^{-|\omega|}$$ is probably known (and asked for many times on this site). Moreover, since $$\mathcal F(jt f(t))(\omega)=-\frac{d}{d\omega}\mathcal F(f(t))(\omega)$$ we find that $$\mathcal F\Bigl(jt\frac{1}{1+t^2}\Bigr)(\omega)=-\frac{d}{d\omega}\pi e^{-|\omega|}=\pi e^{-\omega}H(\omega)-\pi e^{\omega}H(-\omega),$$ where $H$ denotes the Heaviside step function. Finally,

$$\mathcal F\Bigl(\frac{1}{1+jt}\Bigr)(\omega)=\pi e^{-|\omega|}-\pi e^{-\omega}H(\omega)+\pi e^{\omega}H(-\omega)=2\pi e^{\omega}H(-\omega).$$

• I know about this way but I would like to see a solution for the integral I end up with. :) – Pranav Arora Oct 24 '15 at 16:25
• OK, I was not sure, and thought this might be the simplest (except the dual) way. – mickep Oct 24 '15 at 16:27

$$\begin{eqnarray*}\int_{-\infty}^{+\infty}\frac{e^{-i\omega t}}{1+it}\,dt &=& \int_{0}^{+\infty}\left(\frac{e^{-i\omega t}}{1+it}+\frac{e^{i\omega t}}{1-it}\right)\,dt\\&=&2\int_{0}^{+\infty}\frac{\cos(\omega t)- t\sin(\omega t)}{1+t^2}\,dt\\&=&\int_{-\infty}^{+\infty}\frac{\cos(\omega t)- t\sin(\omega t)}{1+t^2}\,dt\\&\color{red}{=}&\pi\, e^{-|\omega|}-\pi\,\text{Sign}(\omega)\, e^{-|\omega|}\\&=&2\pi\, e^{\omega}\cdot\mathbb{1}_{\omega\leq 0}.\end{eqnarray*}$$ The critical step $\color{red}{=}$ can be achieved by using the Laplace transform or other techniques.