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As other posters indicated, the Laplace transform is closely related to the Fourier transform. It is easier to explain the versatility of the Fourier transform.

If you are interested in differential equations, you wish that all functions were linear combinations of exponentials

$$e_\xi(x)= e^{ i \xi x}.$$

For example

The reason is the following simple identity

$$\frac{d}{dx}\left(\sum_k A(\xi_k) e_{\xi_k}(x)\right) = i\sum_k \xi_k A(\xi_k) e_{\xi_k}(x)$$

Thus, dor

which shows that for linear combinations of exponentials the transcendental operation of derivation is replaced with a much simpler algebraic operation. It is natural to ask if any function f(x) can be described as a linear combination of exponentials

$$f(x) = \sum_\xi A(\xi) e_\xi(x).$$

The answer is yes, if we allow for infinte superpositions of exponentials

$$f(x) "= " \sum_{\xi\in\mathbb{R}} A(\xi) e_\xi(x) :=\int_{\mathbb{R}} A(\xi) e_\xi(x) d\xi.$$

The

More precisely, the above function $\xi\mapsto A(\xi)$ above is the Fourier transfrom of $f(x)$

$$A(\xi)=\frac{1}{2\pi} \int_{\mathbb{R}} f(x) e_{-\xi}(x) dx.$$

1

If you are interested in differential equations, you wish that all functions were linear combinations of exponentials

$$e_\xi(x)= e^{ i \xi x}.$$

For example

$$\frac{d}{dx}\left(\sum_k A(\xi_k) e_{\xi_k}(x)\right) = i\sum_k \xi_k A(\xi_k) e_{\xi_k}(x)$$

Thus, dor linear combinations of exponentials the transcendental operation of derivation is replaced with a much simpler algebraic operation. It is natural to ask if any function f(x) can be described as a linear combination of exponentials

$$f(x) = \sum_\xi A(\xi) e_\xi(x).$$

The answer is yes, if we allow for infinte superpositions of exponentials

$$f(x) "= " \sum_{\xi\in\mathbb{R}} A(\xi) e_\xi(x) :=\int_{\mathbb{R}} A(\xi) e_\xi(x) d\xi$$

The function $\xi\mapsto A(\xi)$ above is the Fourier transfrom of $f(x)$

$$A(\xi)=\frac{1}{2\pi} \int_{\mathbb{R}} f(x) e_{-\xi}(x) dx.$$