Differential Equation for improper integrals

Question

How do I use the definiton of the improper integral to find the Laplace transform $F(s)$ for the function $f(t)=e^{(t-1)^2}$

Answer 1

Well the formal definition of the Laplace transform is:

$$\mathcal{L}\left\{f(t)\right\}=\int_0^\infty e^{-st}f(t)dt$$

Here, your function is $f(t)=e^{(t-1)^2}$ so your Laplace transform is:

$$F(s)=\int_0^\infty e^{-st}e^{(t-1)^2}dt=\int_0^\infty e^{(t-1)^2-st}dt$$

Work this one out and you'll find what you want

Edit: As @anorton noted, this integral is divergent. Indeed, if you expand the exponent you get:

$$(t-1)^2-st=t^2-(s+2)t+1=\left(t-\frac{s+2}{2}\right)^2+\left(1-\left(\frac{s+2}{2}\right)^2\right)$$

so your integral becomes:

$$\large F(s)=e^{1-\left(\tfrac{s+2}{2}\right)^2}\int_0^\infty e^{\left(t-\tfrac{s+2}{2}\right)^2}dt$$

Now a variable change gives you:

$$\large F(s)=e^{1-\left(\tfrac{s+2}{2}\right)^2}\int_{-\tfrac{s+2}{2}}^\infty e^{x^2}dx$$

Which diverges.

Note:

Your function can be written as:

$$f(t)=e\cdot e^{-2t}\cdot e^{t^2}$$

$e$ is a constant so from the linearity of the Laplace transform, you have:

$$F(s)=\mathcal{L}\left\{f(t)\right\}=e\cdot\mathcal{L}\left\{e^{-2t}\cdot e^{t^2}\right\}$$

Now, you have another property that states that: If $G(s)$ is the Laplace transform of a function $g(t)$, then the Laplace transform of $e^{\lambda t}g(t)$ is $G(s-\lambda)$.

Here $g(t)=e^{t^2}$ and $\lambda=-2$ so you have:

$$F(s)=e\cdot G(s+2)$$

Where $G(s)$ is the Laplace transform of $e^{t^2}$. But $e^{t^2}$ doesn't have a Laplace transform, so neither does $f(t)$

Are you sure your function has to have a Laplace transform ? In which case maybe there's a mistake in it. Like for example $f(t)=\left(e^{(t-1)}\right)^2$ instead of $f(t)=e^{(t-1)^2}$

Answer 2

This Laplace transform does not exist. Why?

From Boyce and DiPrima's Elementary Differential Equations:

Theorem 6.1.2: Suppose that:

1. $f$ is piecewise continuous on the interval $0\le t\le A$ for any positive $A$
2. $|f(t)| \le Ke^{at}$ when $t\ge M$. In this inequality, $K$, $a$, and $M$ are real constants, $K$, $M$ necessarily positive.

Then the Laplace Transform $\mathcal{L}\{f(t)\} = F(s)$, defined by Eq. $(4)$, exists for $s \gt a$.

The problem we have is that $f(t) = e^{(t-1)^2}$ is not of exponential order--the $a$ and $K$ values above don't exist for this function.