Ngô Quốc Anh

September 1, 2010

The inverse of the Laplace transform by contour integration

Filed under: Giải tích 7 (MA4247) — Ngô Quốc Anh @ 17:23

Usually, we can find the inverse of the Laplace transform \mathcal L[\cdot](s) by looking it up in a table. In this entry, we show an alternative method that inverts Laplace transforms through the powerful method of contour integration.

Consider the piece-wise differentiable function f(x) that vanishes for x < 0. We can express the function e^{-cx}f(x) by the complex Fourier representation of

\displaystyle f(x){e^{ - cx}} = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{e^{i\omega x}}\left[ {\int_0^\infty {{e^{ - ct}}f(t){e^{ - i\omega t}}dt} } \right]d\omega }

for any value of the real constant c, where the integral

\displaystyle I = \int_0^\infty {{e^{ - ct}}|f(t)|dt}

exists. By multiplying both sides of first equation by e^{cx} and bringing it inside the first integral

\displaystyle f(x) = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{e^{(c + i\omega )x}}\left[ {\int_0^\infty {f(t){e^{ - (c + i\omega )t}}dt} } \right]d\omega } .

With the substitution z = c+\omega i, where z is a new, complex variable of integration,

\displaystyle f(x) = \frac{1}{{2\pi }}\int_{c - \infty i}^{c + \infty i} {{e^{zx}}\left[ {\int_0^\infty {f(t){e^{ - zt}}dt} } \right]d\omega }.

The quantity inside the square brackets is the Laplace transform \mathcal L[f](z). Therefore, we can express f(t) in terms of its transform by the complex contour integral

Theorem (Bromwich’s integral). The following line integral runs along the line x = c parallel to the imaginary axis and c units to the right of it, the so-called Bromwich contour

\displaystyle f(t) = \frac{1}{{2\pi i}}\int_{c - \infty i}^{c + \infty i} {\mathcal L[f](z){e^{tz}}dz} .

Example. Let us invert

\displaystyle L(s) = \frac{{{e^{ - 3s}}}}{{{s^2}(s - 1)}}.

Solution. From Bromwich’s integral,

\displaystyle\begin{gathered} f(t) = \frac{1}{{2\pi i}}\int_{c - \infty i}^{c + \infty i} {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}dz} \hfill \\ \qquad= \frac{1}{{2\pi i}}\oint_C {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}dz} - \frac{1}{{2\pi i}}\int_{{C_R}} {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}dz} \hfill \\ \end{gathered}

where C_R is a semicircle of infinite radius in either the right or left half of the z-plane and C is the closed contour that includes C_R and Bromwich’s contour.

Our first task is to choose an appropriate contour so that the integral along C_R vanishes. By Jordan’s lemma, this requires a semicircle in the right half-plane if t-3 < 0 and a semicircle in the left half-plane if t- 3>0. Consequently, by considering these two separate cases, we force the second integral to zero and the inversion simply equals the closed contour.

Consider the case t < 3 first. Because Bromwich’s contour lies to the right of any singularities, there are no singularities within the closed contour and f(t) = 0.

Consider now the case t > 3. Within the closed contour in the left half-plane, there is a second-order pole at z = 0 and a simple pole at z = 1. Therefore,

\displaystyle f(t) = \mathop {\rm Res}\limits_{z = 0} \left[ {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}} \right] + \mathop {\rm Res}\limits_{z = 1} \left[ {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}} \right]


\displaystyle\mathop {\rm Res}\limits_{z = 0} \left[ {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}} \right] = \mathop {\lim }\limits_{z \to 0} \frac{d}{{dz}}\left[ {{z^2}\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}} \right] = 2 - t


\displaystyle\mathop {\rm Res}\limits_{z = 1} \left[ {\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}} \right] = \mathop {\lim }\limits_{z \to 1} \frac{d}{{dz}}\left[ {(z - 1)\frac{{{e^{(t - 3)s}}}}{{{z^2}(z - 1)}}} \right] = {e^{t - 3}}.

Taking our earlier results into account, the inverse equals

\displaystyle f(t) = \left[ {{e^{t - 3}} - (t - 3) - 1} \right]H(t - 3).

Source: Green’s Functions with Applications by Dean G. Duffy



  1. Excuse me, I have a question.
    the first equality on the solution of the example, I think z have to be put instead of s.
    like this
    \displaystyle f(t) = \frac{1}{2\pi i}\int_{c-\infty i}^{c+\infty i}\frac{e^{(t-3)z}}{z^2 (z-1)}dz
    Am I right?

    Comment by zariski — September 10, 2010 @ 19:41

    • Hi zariski.

      You are right, it also appears that s in the second line need to be changed, i.e., the whole stuff reads as follows

      \displaystyle\begin{gathered} f(t) = \frac{1}{{2\pi i}}\int_{c - \infty i}^{c + \infty i} {\frac{{{e^{(t - 3)z}}}}{{{z^2}(z - 1)}}dz} \hfill \\ \qquad= \frac{1}{{2\pi i}}\oint_C {\frac{{{e^{(t - 3)z}}}}{{{z^2}(z - 1)}}dz} - \frac{1}{{2\pi i}}\int_{{C_R}} {\frac{{{e^{(t - 3)z}}}}{{{z^2}(z - 1)}}dz}. \hfill \\ \end{gathered}

      Thanks for visiting my blog and also pointing out the misprint.

      Comment by Ngô Quốc Anh — September 10, 2010 @ 21:27

  2. Thanks for this post. It was a useful overview.

    Comment by Sunil — August 12, 2012 @ 9:10

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