# 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