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Advanced and Inverse Laplace Transforms

Laplace Transform

The Laplace Transform of a function f(t), where t ≥ 0, is defined as $\mathcal{L}(f(t)) = \int_{0}^{∞} f(t)e^{-st}dt = F(s)$.

One of the most important properties of the Laplace Transform is linearity, which states: $\mathcal{L}(af(t)+bg(t)) = a\mathcal{L}(f(t))+b\mathcal{L}(g(t))$

Table of Common Laplace Transforms

The following table summarizes the most frequent mappings used in system analysis.

Function	Laplace Transform
Unit Step, u(t)	$\mathcal{L}(u(t)) = \frac{1}{s}, s > 0$
Exponential, $e^{at}$	$\frac{1}{s - a}, s > a$
Complex Exp $e^{(a + bi)t}$	$\mathcal{L}(e^{(a + bi)t}) = \frac{1}{s - (a + bi)}, s > a$
Cosine $\cos(\omega t)$	$\mathcal{L}(\cos(\omega t)) = \frac{s}{s^2 + \omega^2}, s > 0$
Sine $\sin(\omega t)$	$\mathcal{L}(\sin(\omega t)) = \frac{\omega}{s^2 + \omega^2}, s > 0$
Power, $t^n$	$\mathcal{L}(t^n) = \frac{n!}{s^{n+1}}, s > 0$
Delayed Step, $u(t-a)$	$\mathcal{L}(u(t-a)) = \frac{e^{-as}}{s}, s > 0$
Dirac Delta $\delta(t-a)$	$\mathcal{L}(\delta(t-a)) = e^{-as}, a \geq 0$
$\frac{1}{t}$	$\mathcal{L}\left(\frac{1}{t}\right) = \text{not defined}$
$e^{-bt} \cos(\omega t)$	$\mathcal{L}(e^{-bt} \cos(\omega t)) = \frac{s + b}{(s + b)^2 + \omega^2}, s > -b$
$e^{-bt} \sin(\omega t)$	$\mathcal{L}(e^{-bt} \sin(\omega t)) = \frac{\omega}{(s + b)^2 + \omega^2}, s > -b$
$e^{at}f(t)$	$\mathcal{L}(e^{at}f(t)) = F(s-a)$

Laplace Transforms of Polynomials

This is a fundamental result in Laplace Transform theory and is widely used in solving differential equations and analyzing systems.

Recall that the Laplace Transform of a function $f(t)$ is defined as: $\mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} f(t)e^{-st} dt$ (where $s$ is a complex number, typically with its real part $Re(s) > 0$ to ensure the integral converges).

For $f(t) = t^n$, the Laplace Transform becomes: $\mathcal{L}\{t^n\} = \int_{0}^{\infty} t^n e^{-st} dt$

Below are two distinct methods for evaluating this integral.

Method 1: Recursive Integration by Parts

To compute this integral directly, we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.

💡 Strategy: Choose $u$ as a function that becomes simpler when differentiated (the polynomial $t^n$), and $dv$ as a function that remains manageable when integrated (the exponential $e^{-st}$). We will use this repeatedly to step the power of $t$ down to zero.

Step 1: Set up the parts. Let $u = t^n \implies du = nt^{n-1} dt$. Let $dv = e^{-st} dt \implies v = \frac{e^{-st}}{-s}$

Step 2: Apply the integration by parts formula (∫udv =[Part A] uv − [Part B] ∫vdu). $\int_{0}^{\infty} t^n e^{-st} dt = \left[ t^n \frac{e^{-st}}{-s} \right]_{0}^{\infty} - \int_{0}^{\infty} \left( \frac{e^{-st}}{-s} \right) nt^{n-1} dt$

Step 3: Evaluate the boundary term (the $uv$, [A] part).

Let’s look at the first term evaluated from $0$ to $\infty$: $\lim_{t \to \infty} \left( \frac{t^n e^{-st}}{-s} \right) - \left( \frac{0^n e^{-0}}{-s} \right)$

At $t \to \infty$: We evaluate $\lim_{t \to \infty} \frac{t^n}{-s e^{st}}$. Because $s > 0$, the exponential term $e^{st}$ grows significantly faster than the polynomial $t^n$. By applying L’Hôpital’s rule $n$ times, we can rigorously show that this limit goes to $0$.

At $t = 0$: $\left( \frac{0^n e^{-0}}{-s} \right) =[\text{n is a non-negative integer}] \frac{0\cdot 1}{-s} = 0$

Thus, the entire boundary term vanishes: $0 - 0 = 0$.

Step 4: Establish the recursive relation

We are left with just the integral term. Pulling the constants $n$ and $-s$ out of the integral yields: $\mathcal{L}\{t^n\} = \frac{n}{s} \int_{0}^{\infty} t^{n-1} e^{-st} dt$

Notice that the remaining integral is exactly the definition of the Laplace transform for $t^{n-1}$! This gives us our recursive relation: $\mathcal{L}\{t^n\} = \frac{n}{s} \mathcal{L}\{t^{n-1}\}$

Step 5: Unroll the recursion

We can apply (or unfold) this recursive formula repeatedly until we reach $t^0 = 1$: $\mathcal{L}\{t^n\} = \left(\frac{n}{s}\right) \left(\frac{n-1}{s}\right) \mathcal{L}\{t^{n-2}\}= \frac{n(n-1) \dots 1}{s^n} \mathcal{L}\{t^0\}$

Since $\mathcal{L}\{1\} = \frac{1}{s}$, we multiply it into the chain: $\mathcal{L}\{t^n\} = \frac{n!}{s^n} \cdot \frac{1}{s} = \boxed{\frac{n!}{s^{n+1}} \quad \text{for } s > 0}$

This result is fundamental and provides a direct way to compute the Laplace Transform of any non-negative integer power of t.

Method 2: Using the Gamma Function $\Gamma(z)$

An alternative, elegant way to solve this is by utilizing the Gamma function, which is the continuous generalization of the factorial function.

The Gamma function is defined as: $\Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} dt \quad \text{for } Re(z) > 0$

We aim to demonstrate that $\Gamma(n+1) = n!$

If we evaluate $\Gamma(n+1)$, we get: $\Gamma(n+1) = \int_{0}^{\infty} t^n e^{-t} dt$

Using integration by parts ($u = t^n, dv = e^{-t} dt$), we get: $\Gamma(n+1) = \left[ -t^n e^{-t} \right]_{0}^{\infty} + n \int_{0}^{\infty} t^{n-1} e^{-t} dt$

Just like before (Method 1), the boundary term vanishes so we are left with: $\Gamma(n+1) = n \Gamma(n)$

Applying this recursively down to $\Gamma(1) = 1$ proves that $\Gamma(n+1) = n!$.

Evaluating the Laplace Transform

Now, let’s look back at our Laplace Transform integral: $\mathcal{L}\{t^n\} = \int_{0}^{\infty} t^n e^{-st} dt$

We can use $u$-substitution to make this look exactly like the Gamma function.

Let $u = st$. This means $t = \frac{u}{s}$, then differentiating gives $dt = \frac{du}{s}$

Substitute these into the integral: $\int_{0}^{\infty} \left( \frac{u}{s} \right)^n e^{-u} \left( \frac{du}{s} \right)$

Factor out the constants (all the $s$ terms in the denominator): $= \frac{1}{s^{n+1}} \int_{0}^{\infty} u^n e^{-u} du$

Notice that the remaining integral is exactly the definition of $\Gamma(n+1)$! $= \frac{1}{s^{n+1}} \Gamma(n+1)$

Since we proved earlier that $\Gamma(n+1) = n!$, we arrive at the exact same, fundamental result: $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$

Growth condition for the Laplace Transform

This is the core issue: the Laplace integral is essentially a “race to infinity” between the function $f(t)$ trying to grow and the damping factor $e^{-st}$ trying to crush it to zero.

Recall that the Laplace Transform of a function $f(t)$ is defined as an improper integral: $\mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} f(t)e^{-st} dt$

For this transform to exist, the integral must converge to a finite value. Because the integration goes to infinity, $f(t)$ cannot grow so rapidly that the exponential decay factor $e^{-st}$ is unable to overpower it.

Piecewise Continuity.
The function $f(t)$ must be piecewise continuous on the interval $[0, \infty)$.
This means $f(t)$ is continuous everywhere except for possibly a finite number of jump discontinuities in any finite interval. Crucially, it means $f(t)$ cannot have vertical asymptotes (infinite discontinuities) within the domain of integration
Exponential Order $\alpha$
A function $f(t)$ is said to be of exponential order $\alpha$ if it does not grow faster than a standard exponential function as $t \to \infty$.
Formally, there must exist real constants $M > 0$, $\alpha$, and $t_0 > 0$ such that: $|f(t)| \le M e^{\alpha t} \quad \text{for all } t \ge t_0$
If $f(t)$ satisfies this condition, the damping factor $e^{-st}$ can successfully force the integrand to zero, provided we choose an $s$ that is large enough ($s > \alpha$).

Proof of Convergence

Let’s prove that if $f(t)$ is piecewise continuous and of exponential order $\alpha$, its Laplace Transform exists for $s > \alpha$.

We analyze the absolute value of the integral to test for absolute convergence: $|F(s)| = \left| \int_{0}^{\infty} e^{-st}f(t) dt \right| \le \int_{0}^{\infty} \left| e^{-st}f(t) \right| dt$

We split this integral into two parts, namely before $t_0$ and after $t_0$: $= \int_{0}^{t_0} |e^{-st}f(t)| dt + \int_{t_0}^{\infty} |e^{-st}f(t)| dt$

The First Integral (Finite Interval): Because $f(t)$ is piecewise continuous and does not go to infinity on $[0, t_0]$, and $e^{-st}$ is strictly bounded, the integral over this finite interval evaluates to a finite number.

On a compact interval $[0,t_0]$, a piecewise continuous function has only finitely many jump discontinuities and is continuous on each subinterval. Each continuous piece is bounded, and a finite union of bounded pieces is bounded. The exponential factor is bounded on $[0,t_0]$. Product of bounded functions is bounded.
The Second Integral (Infinite Interval): Here, we apply our exponential order condition ($|f(t)| \le Me^{\alpha t}$):
$$\int_{t_0}^{\infty} |e^{-st}f(t)| dt \le \int_{t_0}^{\infty} e^{-st} M e^{\alpha t} dt = M \int_{t_0}^{\infty} e^{-(s-\alpha)t} dt$$
Now, evaluate this upper bound: $= M \left[ \frac{e^{-(s-\alpha)t}}{-(s-\alpha)} \right]_{t_0}^{\infty}$

If $s > \alpha$: The exponent $-(s-\alpha)t$ is negative. As $t \to \infty$, the term $e^{-(s-\alpha)t} \to 0$. The upper bound evaluates to a finite number: $M \frac{e^{-(s-\alpha)t_0}}{s-\alpha}$.
If $s \le \alpha$: The exponent is positive or zero, causing the term to blow up or stagnate as $t \to \infty$, and the integral diverges.

Conclusion: The Laplace Transform exists and is absolutely convergent for any $s > \alpha$.

Examples and Counterexamples of Functions of Exponential Type

Bounded Functions: $\sin(t)$

The sine function oscillates between $-1$ and $1$, meaning it is strictly bounded: $|\sin(t)| \le 1$.

We can rewrite this as $|\sin(t)| \le 1 \cdot e^{0 \cdot t}$.

In this case, the sine function is of exponential type with $\alpha=0$ and $M=1$, indicating that it does not grow exponentially at all, so its Laplace Transform exists for all $s > 0$

Polynomials: $t^n$

For any positive $\alpha$, an exponential function will always eventually outgrow a polynomial. We can prove that $t^n$ is of exponential order by taking the limit of their ratio as $t \to \infty$: $\lim_{t \to \infty} \frac{t^n}{e^{\alpha t}}$

By applying L’Hôpital’s rule $n$ times, each differentiation reduces the degree of the numerator, it becomes a constant n! and the denominator remains an exponential function multiplied by a polynomial term, which still grows exponentially ($\alpha^n e^{\alpha t}$). $\lim_{t \to \infty} \frac{n!}{\alpha^n e^{\alpha t}} = 0$

Since the ratio goes to zero, $t^n$ is strictly bounded by $M e^{\alpha t}$ for sufficiently large $t$. Therefore, polynomials have a Laplace Transform for $s > 0$.

Piecewise Continuity: $\frac{1}{t}$

The function $f(t) = \frac{1}{t}$ does not fail the exponential order test as $t \to \infty$ (in fact, it shrinks to zero). It fails the piecewise continuity test at $t = 0$.

Because there is an infinite discontinuity (a vertical asymptote) right at the starting bound of the integral, the area under the curve near $t=0$ blows up to infinity. Therefore, the Laplace Transform of $\frac{1}{t}$, $\mathcal{L}\left\{\frac{1}{t}\right\}$, does not exist.

Grows Too Fast: $e^{t^2}$

This is the classic example of a function that breaks the exponential order rule. Let’s look at the integrand: $f(t)e^{-st} = e^{t^2}e^{-st} = e^{t^2 - st} = e^{t(t-s)}$

As $t \to \infty, t(t-s) \gt 0$, so $f(t)e^{-st}$ grows with bond. The integrand blows up, the area is infinite, and thus, $\mathcal{L}\{e^{t^2}\}$ does not exist.

In practice, the exponential growth condition ensures that the Laplace Transform can handle a wide variety of useful functions, including polynomials, exponentials, and trigonometric functions, all of which appear frequently in solving differential equations and modeling physical systems.