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

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