Fourier transform is defined as $$F(jw)=\int_{-\infty}^{\infty}f(t)e^{-jwt}dt.$$ Inverse Fourier transform is defined as $$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(jw)e^{jwt}dw.$$

Let $f(t)=e^{-at}h(t),a>0$, where $$h(t)$$ is heaviside function and $$a$$ is real constant.

Fourier transform of this function is $$F(jw)=\int_{0}^{\infty}f(t)e^{-jwt}dt=\int_{0}^{\infty}e^{-at}e^{-jwt}dt=\frac{1}{a+jw}$$ How can I calculate inverse Fourier transform of $$\frac{1}{a+jw}$$, $$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{a+jw}e^{jwt}dw$$? Although $$\frac{1}{a+jw}$$ doesn't look complicated, there is no way I can calculate this integral. Generaly, I have problems calculating inverse FT. Any suggestion? Thanks in advance.

  • $\begingroup$ Because of heaviside function, f(t)=0, for every t<0. So limits of integrations are (0,infinity) $\endgroup$ – hari Nov 16 '15 at 1:21
  • $\begingroup$ The inverse transform extends over the entire real line as @Winther alluded. $\endgroup$ – Mark Viola Nov 16 '15 at 1:22
  • $\begingroup$ I definitely have problems with FT :). I will correct limits $\endgroup$ – hari Nov 16 '15 at 1:23
  • 1
    $\begingroup$ Have you learned the residue theorem? $\endgroup$ – Mark Viola Nov 16 '15 at 1:24
  • $\begingroup$ We didn't start with complex analysis yet... $\endgroup$ – hari Nov 16 '15 at 1:28


We can proceed to evaluate the integral if we invoke Generalized Functions. To that end, we write the inverse Fourier Transform representation for $f$ as

$$f(t)=\frac{1}{2\pi}\int_{- \infty}^{ \infty}\frac{e^{j\omega t}}{a+j\omega}\,d\omega \tag 1$$

Then, note that the derivative $f'$ is given by

$$f'(t)=\frac{1}{2\pi}\int_{- \infty}^{ \infty}\frac{j\omega\,e^{j\omega t}}{a+j\omega}\,d\omega \tag 2$$

Adding $(2)$ and $a$ times $(1)$ reveals that

$$\begin{align} f'(t) +af(t)&=\frac{1}{2\pi}\int_{- \infty}^{ \infty}e^{j\omega t}\,d\omega \\\\ &=\delta(t) \tag 3 \end{align}$$

where $\delta(t)$ is the Dirac Delta Distribution. Solution to the ODE expressed in $(3)$ is


as was to be shown!


We begin by writing the integral representation for $f(t)$ as the Cauchy Principal Value

$$f(t)=\frac{1}{2\pi}\lim_{R\to \infty}\int_{-R}^{ R}\frac{e^{j\omega t}}{a+j\omega}\,d\omega \tag 4$$

Note that we need not evaluate the integral in terms of its Cauchy Principal Value. Inasmuch as the improper integral converges, its Cauchy Principal Value converges to the same value.

Now, we define a new function $f_R(t)$ as

$$f_R(t)=\oint_C \frac{e^{jzt}}{a+jz}\,dz$$

where the closed contour $C$ is comprised of the real line segment from $z=-R$ to $z=+R$ and for $t>0$ ($t<0$) the semicircle $C_R$ of radius $R$ in the upper (lower) half plane.

Jordan's Lemma guarantees that as $R\to \infty$, the contribution of the integral from the integration over $C_R$ goes to zero. Therefore, we have $f(t)=\lim_{R\to \infty}f_R(t)$.

Now, we note that since $a>0$ that the only singularity of $\frac{e^{jzt}}{a+jz}$ is a $z=ja$. Thus, from the Residue Theorem we have

$$f(t)= \begin{cases} 2\pi i\text{Res}\left(\frac{e^{jzt}}{a+jz},z=ja\right)=e^{-at}&, t>0\\\\ 0&,t<0 \end{cases}$$

as expected!

  • $\begingroup$ Wow! Thanks a lot! $\endgroup$ – hari Nov 16 '15 at 14:26
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Nov 16 '15 at 14:55
  • $\begingroup$ Method 1 seems very nice! Thanks. I have a question about Eq. (3). How does the integral evaluate to Dirac Delta? $\endgroup$ – emrea Apr 9 '20 at 7:38
  • 1
    $\begingroup$ @emrea The Fourier Transform of the Dirac Delta is equal to $1$. So, the inverse Fourier Transform of $1$ is $2\pi \delta(t)$. $\endgroup$ – Mark Viola Apr 9 '20 at 15:18

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