Laplace transform for dummies

Maybe the heuristics below is for smarties rather than for dummies, but anyway here goes.
For the sake of rigor: assuming that all (improper) integrals exist and everything is real-valued. The Taylor series expansion of a function $ f(t+\tau) $ around $ t $ will be set up: $$ f(t+\tau) = \sum_{k=0}^\infty \frac{ \tau^k }{ k ! } f^{(k)}(t) = \left[ \sum_{k=0}^\infty \frac{1}{k!} \left( \tau \ \frac{d}{dt} \right)^k \right] f(t) $$ In the expression between square brackets the series expansion of $\,e^x\,$ is recognized. Therefore we can write , symbolically : $$ f(t+\tau) = e^{\Large \tau \frac{d}{dt} } f(t) \quad \Longrightarrow \quad f(t-\tau) = e^{\Large -\tau \frac{d}{dt} } f(t) $$ With the last formula in mind, consider an arbitrary convolution-integral: $$ \int_{-\infty}^{+\infty} h(\tau) f(t-\tau) \, d\tau $$ Convolution integrals do frequently occur. With a linear system, the response at a disturbance is the convolution-integral of the disturbance with the so-called (impulse) response. The unit response is the way in which the system reacts upon the simplest of all disturbances, that is a steep peak of very short duration at time zero, a Dirac delta. Our convolution integral can be rewritten with help of the expression for $ f(t-\tau) $ as follows: $$ = \int_{-\infty}^{+\infty} h(\tau) \left[e^{\Large - \tau \frac{d}{dt} } f(t) \right]\, d\tau = \int_{-\infty}^{+\infty} h(\tau) e^{\Large - \tau \frac{d}{dt} } \, d\tau \; \cdot \; f(t) $$ The integral on the right should be well known to us. Quite "incidentally" namely it is the (double-sided) Laplace transform: $$ H(p) = \int_{-\infty}^{+\infty} e^{\large - p \tau}\, h(\tau) \, d\tau $$ Thus it seems that Laplace's integral shows up quite spontaneously with elementary considerations about convolution-integrals in combination with Operational Calculus . The end-result is: $$ \int_{-\infty}^{+\infty} h(\tau) f(t-\tau) \, d\tau = H(\frac{d}{dt}) \, f(t) $$ The fact that Laplace transforms are a very powerful means for solving differential equations can now be understood without much effort. Suppose we have a linear inhomogeneous differential equation. In general it has the form: $$ D( \frac{d}{dt} ) \, \phi(t) = f(t) $$ Then with help of our Operator/Operational Calculus we can immediately write the solution as: $$ \phi(t) = \frac{1}{\large D( \frac{d}{dt} ) } f(t) $$ Put $\,H(d/dt) = 1/D(d/dt) $ , then the excercise becomes: find the inverse of the Laplace transform of $\, H(p) $ . Call this inverse function $\, h(t) $ . Finding the solution then follows the above pattern: $$ \phi(t) = \int_{-\infty}^{+\infty} h(\tau) f(t-\tau) \, d\tau $$ Example 1. Suppose that we have derived (for $p>\alpha$): $$ h(t) = e^{\large \alpha t}.u(t) \quad \Longrightarrow \quad H(p) = \int_{-\infty}^{+\infty} e^{\large -p\tau} e^{\large \alpha \tau}.u(\tau) d\tau =\\ \int_0^\infty e^{\large -p\tau} e^{\large \alpha \tau} d\tau = \left[\frac{-e^{\large -\tau(p-\alpha)}}{p-\alpha}\right]_{\tau=0}^\infty = \frac{1}{p-\alpha} $$ Where the Heaviside step function $u(t)$ is defined by: $$ u(t) = \begin{cases} 0 & \mbox{for} & t < 0\\ 1 & \mbox{for} & t > 0\end{cases} $$ Now consider the differential equation: $$ \frac{d\phi}{dt} + \phi(t) = 0 \quad \mbox{with} \quad \phi(0)=1 $$ Which safely can be replaced by finding a Green's function in the time domain: $$ \frac{d\phi}{dt} + \phi(t) = \delta(t) \quad \mbox{with} \quad \phi(-\infty)=0 $$ It follows that: $$ \phi(t) = \frac{1}{\large \frac{d}{dt} + 1 } = \int_{-\infty}^{+\infty} e^{\large - \tau}.u(\tau) \delta(t-\tau) \, d\tau = e^{\large - t}.u(t) $$
Example 2. Still with us? Then let's investigate the Laplace transform of $\,\exp(-\mu t^2)$ : $$ \int_{-\infty}^{+\infty} e^{-pt} e^{-\mu t^2} \, dt = \int_{-\infty}^{+\infty} e^{-\mu t^2-pt}\, dt $$ Completing the square $\;\mu t^2 + pt= \mu\left[t^2+p/\mu.t+p^2/(2\mu)^2\right]-p^2/4\mu = x^2-p^2/4\mu\;$ with $\,x = t + p/2\mu\,$ results in: $$ = \int_{-\infty}^{+\infty} e^{-\mu x^2} \, dx \,.\, e^{\,p^2/4\mu } = \sqrt{ \frac{\pi}{\mu} } e^{\,p^2 / 4\mu } $$ The last move by using a well-known result for the integral of the Gaussian probability distribution.
Laplace transform $H$ and inverse Laplace transform $h$ are thus mutually related as follows, after having replaced $1/4\mu$ by $1/2\sigma^2$ : $$ H(p) = e^{\, \frac{1}{2} \sigma^2 p^2 } \quad \Longleftrightarrow \quad h(t) = \frac{1}{ \sigma \sqrt{2\pi} } e^{-t^2 / 2\sigma^2 } $$ A convolution integral with the normal distribution $h(t)$ as the kernel can thus be re-written as: $$ \int_{- \infty}^{+ \infty} \! h(\xi) \phi(x-\xi) \, d\xi = e^{\frac{1}{2} \sigma^2 \frac{d^2}{dx^2} } \phi(x) $$ The physical meaning of this is that the (Gaussian blur) operator $\,\exp(\frac{1}{2} \sigma^2 \large \frac{d^2}{dx^2})\,$ "spreads out" the function $\,\phi(x)\,$ over a domain with size of the order $\,\sigma $ .
The above outcome is immediately applicable to the following problem. Let's consider the (partial differential) equation for diffusion of heat in one-dimensional space and time: $$ \frac{\partial T}{\partial t} = a \frac{\partial^2 T}{\partial x^2} $$ Here $x=$ space, $t=$ time, $T=$ temperature, $a=$ constant. Rewrite in the first place as follows: $$ \lambda \frac{\partial}{\partial t} T = \lambda a \frac{\partial^2}{\partial x^2} T $$ As a next step we exponentiate at both sides the operators in place: $$ e^{\lambda \partial/\partial t } \, T = e^{\lambda a \partial^2 / \partial x^2} \, T $$ The resulting operator-expressions can be converted into classical mathematics with the acquired knowledge: $$ T(x,t+\lambda) = \int_{- \infty}^{+ \infty} \! h(\xi) T(x-\xi,t) \, d\xi $$ Where $ \frac{1}{2} \sigma^2 = \lambda a $. Therefore: $$ h(t) = \frac{1}{ \sigma \sqrt{2\pi} } \, e^{-t^2/2\sigma^2 } \quad \to \quad h(\xi) = \frac{1}{ \sqrt{4\pi \lambda a} } \, e^{-\xi^2/(4\lambda a) } $$ At last exchange $t$ and $\lambda$, and substitute $\lambda = 0$. Then we quickly find the solution of our PDE: $$ T(x,t) = \int_{- \infty}^{+ \infty} \! \frac{1}{\sqrt{4\pi a t}}\, e^{- \xi^2/(4 a t) }\, T(x-\xi,0) \, d\xi $$