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The Laplace transform of a function $ f(t) $ is defined by the improper integral

$ F(s)=\mathcal{L}\{f(t)\}=\int\limits_0^\infty e^{-st}f(t)dt $

where s is a complex number representing frequency

$ s=\sigma+i\omega $
The real part of s must be positive to ensure convergence.

The purpose of the Laplace transform is to take a real function of a variable $ t $ (often time, sometimes $ x $ is used for other properties) and transform it into a complex function of $ s $ , often representing frequency. One of the most common applications of the Laplace transform is in solving differential equations, as it converts an equation of varying degrees of differentiation into a polynomial of varying degrees of power.

Inverse Laplace transform

Given a $ F(s) $ , finding a function of $ t $ which satisfies $ \mathcal{L}\{f(t)\} $ is called taking the inverse Laplace transform. This process is often far more difficult than finding a Laplace transform.

For ordinary differential equations, it is usually sufficient to find an inverse Laplace transform by finding a similar Laplace transform (preferably one with a similar denominator) using a table. Often, this requires some algebraic manipulation. For example:

$ F(s)=\frac{3}{s+5}=3\frac{1}{s-(-5)}=\mathcal{L}\{3e^{-5t}\} $

By the complex inversion formula, the inverse Laplace transform is equal to the contour integral

$ f(t)=\mathcal{L}^{-1}\{F(t)\}=\frac{1}{2\pi i}\lim_{T\to\infty}\int\limits_{\alpha-iT}^{\alpha+iT}e^{st}F(s)ds. $

As the Laplace transform of any function goes to zero as r approaches infinity, and all residues can found to the left of some constant, by the residue theorem this is simply equal to

$ f(t)=\mathcal{L}^{-1}\{F(t)\}=\sum_{i=0}^k \text{Res}(f(a_i)). $

Using Laplace transforms to solve differential equations

Since the Laplace transform of any derivative will include $ F(s) $ , one can solve for this and take the inverse Laplace transform to find the original function $ f(t) $. For example, given the very simple second order differential equation

$ f''(t)=t^2,\ f(0)=1,\ f'(0)=1 $
$ s^2F(s)-sf(0)-f'(0)=\frac{2}{s^3} $
$ s^2F(s)-s-1=\frac{2}{s^3} $
$ F(s)=\frac{2}{s^5}+\frac{1}{s}+\frac{1}{s^2} $

Now by taking the inverse Laplace transform, we get

$ f(t)=\mathcal{L}^{-1}\{F(s)\}=\mathcal{L}^{-1}\left\{\frac{2}{s^5}+\frac{1}{s}+\frac{1}{s^2}\right\}=\frac{t^4}{12}+1+t $

Which does indeed satisfy the conditions set by the initial problem. This is a fairly elementary example which can easily solved by other methods, but the Laplace transform can be used in many situations where this is not the case, such as

$ f''(t)-4f(t)=7,\ f(0)=1,\ f'(0)=2 $
$ s^2F(s)-sf(0)-f'(0)-4F(s)=\frac{7}{s} $
$ F(s)=\frac{7}{s(s^2-4)}+\frac{s+2}{s^2-4}=\frac{s^2+2s+7}{s(s-2)(s+2)} $

We can expand this by partial fraction expansion to get

$ F(s)=\frac{-\tfrac{7}{4}}{s}+\frac{\tfrac{15}{8}}{s-2}+\frac{\tfrac{7}{8}}{s+2} $

Now by taking the inverse Laplace transform, we can find $ f(t) $ :

$ f(t)=\mathcal{L}^{-1}\left\{\frac{-\tfrac{7}{4}}{s}+\frac{\tfrac{15}{8}}{s-2}+\frac{\tfrac{7}{8}}{s+2}\right\}=\frac{15}{8}e^{2t}+\frac{7}{8}e^{-2t}-\frac{7}{4} $

See also

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