Question

Consider the Laplace transform of \(t^{\rho},\) where \(p>-1\) (a) Referring to Problem \(26,\) show that $$ \qquad \begin{aligned} \mathcal{L}\left\\{t^{p}\right\\} &=\int_{0}^{\infty} e^{-s t} t^{p} d t=\frac{1}{s^{p+1}} \int_{0}^{\infty} e^{-x} x^{p} d x \\\ &=\Gamma(p+1) / s^{\rho+1}, \quad s>0 \end{aligned} $$ (b) Let \(p\) be a positive integer \(n\) in (a); show that $$ \mathcal{L}\left\\{t^{n}\right\\}=n ! / s^{n+1}, \quad s>0 $$ (c) Show that $$ \mathcal{L}\left(t^{-1 / 2}\right)=\frac{2}{\sqrt{s}} \int_{0}^{\infty} e^{-x^{2}} d x, \quad s>0 $$ It is possible to show that $$ \int_{0}^{\infty} e^{-x^{2}} d x=\frac{\sqrt{\pi}}{2} $$ hence $$ \mathcal{L}\left\\{t^{-1 / 2}\right\\}=\sqrt{\pi / s}, \quad s>0 $$ (d) Show that $$ \mathcal{L}\left\\{t^{1 / 2}\right\\}=\sqrt{\pi} / 2 s^{3 / 2}, \quad s>0 $$

Short answer

**Question:** Find the Laplace transforms of the following functions and determine their respective conditions for the existence of the Laplace transforms. (a) \(t^\rho\) (b) \(t^n\), where \(n\) is a positive integer. (c) \(t^{-1/2}\) (d) \(t^{1/2}\) **Answer:** (a) \(\mathcal{L}\{t^\rho\} = \frac{\Gamma(\rho+1)}{s^{\rho+1}}, \quad s>0\) (b) \(\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, \quad s>0\) (c) \(\mathcal{L}\left\{t^{-1/2}\right\} = \sqrt{\frac{\pi}{s}}, \quad s>0\) (d) \(\mathcal{L}\left\{t^{1/2}\right\} = \frac{\sqrt{\pi}}{2s^{3/2}}, \quad s>0\)

Step-by-step solution

(a) Laplace transform of \(t^\rho\)

First, we will find the Laplace transform of the given function \(t^\rho\) by definition: $$\mathcal{L}\{t^\rho\} = \int_{0}^{\infty} e^{-st}t^\rho dt$$ Let \(x = st\). Therefore, \(dx = sdt\), and, when \(t=0\), \(x = 0\), and when \(t \to \infty\), \(x \to \infty\). Now we can rewrite the above integral as: $$\int_{0}^{\infty} e^{-st}t^\rho dt = \frac{1}{s^{\rho+1}}\int_{0}^{\infty} e^{-x}x^\rho dx$$ Since the integral on the right is the definition of the Gamma function, we can write: $$\mathcal{L}\{t^\rho\} = \frac{\Gamma(\rho+1)}{s^{\rho+1}}, \quad s>0$$

(b) Laplace transform of \(t^n\) for a positive integer \(n\)

To find the Laplace transform of \(t^n\), we just need to substitute \(\rho = n\) in the result obtained in part (a): $$\mathcal{L}\{t^n\} = \frac{\Gamma(n+1)}{s^{n+1}}$$ Note that for a positive integer \(n\), the Gamma function is given by \(\Gamma(n+1) = n!\). Therefore, we can write: $$\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, \quad s>0$$

(c) Laplace transform of \(t^{-1/2}\)

To find the Laplace transform of \(t^{-1/2}\), we just need to substitute \(\rho = -1/2\) in the result obtained in part (a): $$\mathcal{L}\left\{t^{-1/2}\right\} = \frac{\Gamma(-1/2 + 1)}{s^{-1/2 + 1}}\int_{0}^{\infty} e^{-st}t^{-1/2} dt$$ Now, we apply the same substitution as in part (a), \(x = st\), and we get: $$\mathcal{L}\left\{t^{-1/2}\right\} = \frac{2}{\sqrt{s}}\int_{0}^{\infty} e^{-x^2} dx, \quad s>0$$ Since it is given that the integral is equal to \(\sqrt{\pi}/2\), we can substitute it in the expression: $$\mathcal{L}\left\{t^{-1/2}\right\} = \sqrt{\frac{\pi}{s}}, \quad s>0$$

(d) Laplace transform of \(t^{1/2}\)

To find the Laplace transform of \(t^{1/2}\), we just need to substitute \(\rho = 1/2\) in the result obtained in part (a): $$\mathcal{L}\left\{t^{1/2}\right\} = \frac{\Gamma(1/2 + 1)}{s^{1/2 + 1}}$$ We know that \(\Gamma(1/2 + 1) = \Gamma(3/2) = \sqrt{\pi}/2\). Therefore, the Laplace transform of \(t^{1/2}\) is: $$\mathcal{L}\left\{t^{1/2}\right\} = \frac{\sqrt{\pi}}{2s^{3/2}}, \quad s>0$$

Key concepts

Gamma Function

Understanding the Gamma function is integral to solving various problems in mathematics, particularly when working with factorials in calculus and complex analysis. The Gamma function is denoted as \( \Gamma(z) \) and is defined for complex numbers with a real part greater than zero. It extends the concept of factorial to non-integer and complex numbers, thus it's a generalization of the factorial function.

For positive integers \( n \), the Gamma function satisfies \( \Gamma(n+1) = n! \) which seamlessly aligns with the factorial operation we're familiar with.

If we consider \( \Gamma(\rho+1) \) in the context of the Laplace transform, this concept allows us to express the transform of functions involving powers of \( t \) in a more generalized and elegant manner, as seen in the exercise provided.

Exponential Integral

The term exponential integral can be slightly misleading; it is not a single function, but rather a family of functions related to the integral of an exponential function. In the context of the Laplace transform, we encountered it when we transformed the function \( t^{\rho} \) by converting the original time-domain function into an exponential form through the substitution \( x = st \).

It resulted in an integral of the form \( \int e^{-x} x^{p} dx \), which closely resembles the Gamma function when evaluated from 0 to infinity. This integral is essential because it appears frequently in solutions of differential equations and other areas of mathematical physics. It's used to find the Laplace transform of functions like \( t^{-1/2} \) which are not otherwise easily integrable and directly relate to the Gamma function.

Differential Equations

The differential equations topic is extensive; these equations are fundamental across various disciplines in mathematics and science. They characterize systems via equations involving derivatives, which describe how a particular quantity changes over time or another dimension. Through differential equations, we can model physical phenomenon, economic trends, biological processes, and much more.

Laplace transforms are a powerful tool for solving differential equations, particularly linear ordinary differential equations with constant coefficients. By transforming a given differential equation into an algebraic equation in the Laplace domain, we can solve for the Laplace transform of the unknown function. Afterward, we perform an inverse Laplace transform to obtain the solution in the original time domain.

This process is exemplified in solving initial value problems, where the use of Laplace transforms can significantly simplify the solution process, as it turns differentiation into algebraic multiplication by \( s \) in the Laplace domain. By understanding how to apply Laplace transforms to these equations, we can solve complex dynamic systems that are otherwise challenging to address via traditional methods.