1.7: Applications to Laplace transforms

The Laplace transform of a function $f$ locally integrable on $[0,\infty)$ is

$$F(s)=\int_{0}^{\infty}e^{-sx}f(x)\,dx$$

for all $s$ such that integral converges. Laplace transforms are widely applied in mathematics, particularly in solving differential equations.

We leave it to you to prove the following theorem (Exercise [exer:26]).

[theorem:12] Suppose $f$ is locally integrable on $[0,\infty)$ and $|f(x)|\le M e^{s_{0}x}$ for sufficiently large $x$. Then the Laplace transform of $F$ converges uniformly on $[s_{1},\infty)$ if $s_{1}>s_{0}$.

[theorem:13] If $f$ is continuous on $[0,\infty)$ and $H(x)=\int_{0}^{\infty}e^{-s_{0}u}f(u)\,du$ is bounded on $[0,\infty),$ then the Laplace transform of $f$ converges uniformly on $[s_{1},\infty)$ if $s_{1}>s_{0}.$

If $0\le r\le r_{1}$,

$$\int_{r}^{r_{1}}e^{-sx}f(x)\,dx =\int_{r}^{r_{1}}e^{-(s-s_{0})x}e^{-s_{0}x}f(x)\,dt =\int_{r}^{r_{1}}e^{-(s-s_{0})t}H'(x)\,dt.$$

Integration by parts yields

$$\int_{r}^{r_{1}}e^{-sx}f(x)\,dt=e^{-(s-s_{0})x}H(x)\biggr|_{r}^{r_{1}} +(s-s_{0})\int_{r}^{r_{1}}e^{-(s-s_{0})x} H(x)\,dx.$$

Therefore, if $|H(x)|\le M$, then

$$\begin{aligned} \left|\int_{r}^{r_{1}}e^{-sx}f(x)\,dx\right|&\le& M\left|e^{-(s-s_{0})r_{1}} +e^{-(s-s_{0})r} +(s-s_{0})\int_{r}^{r_{1}}e^{-(s-s_{0})x}\,dx\right|\\ &\le &3Me^{-(s-s_{0})r}\le 3Me^{-(s_{1}-s_{0})r},\quad s\ge s_{1}.\end{aligned}$$

Now Theorem [theorem:4] implies that $F(s)$ converges uniformly on $[s_{1},\infty)$.

The following theorem draws a considerably stonger conclusion from the same assumptions.

[theorem:14] If $f$ is continuous on $[0,\infty)$ and

$$H(x)=\int_{0}^{x}e^{-s_{0}u}f(u)\,du$$

is bounded on $[0,\infty),$ then the Laplace transform of $f$ is infinitely differentiable on $(s_{0},\infty),$ with

$$\label{eq:30} F^{(n)}(s)=(-1)^{n}\int_{0}^{\infty} e^{-sx} x^{n}f(x)\,dx;$$

that is, the $n$-th derivative of the Laplace transform of $f(x)$ is the Laplace transform of $(-1)^{n}x^{n}f(x)$.

First we will show that the integrals

$$I_{n}(s)=\int_{0}^{\infty}e^{-sx}x^{n}f(x)\,dx,\quad n=0,1,2, \dots$$

all converge uniformly on $[s_{1},\infty)$ if $s_{1}>s_{0}$. If $0<r<r_{1}$, then

$$\int_{r}^{r_{1}}e^{-sx}x^{n}f(x)\,dx= \int_{r}^{r_{1}}e^{-(s-s_{0})x}e^{-s_{0}x}x^{n}f(x)\,dx =\int_{r}^{r_{1}}e^{-(s-s_{0})x}x^{n}H'(x)\,dx.$$

Integrating by parts yields

$$\begin{aligned} \int_{r}^{r_{1}}e^{-sx}x^{n}f(x)\,dx &=&r_{1}^{n}e^{-(s-s_{0})r_{1}}H(r)-r^{n}e^{-(s-s_{0})r}H(r)\\ &&-\int_{r}^{r_{1}}H(x)\left(e^{-(s-s_{0})x}x^{n}\right)'\,dx,\end{aligned}$$

where $'$ indicates differentiation with respect to $x$. Therefore, if $|H(x)|\le M\le \infty$ on $[0,\infty)$, then

$$\left|\int_{r}^{r_{1}}e^{-sx}x^{n}f(x)\,dx\right|\le M\left(e^{-(s-s_{0})r}r^{n}+e^{-(s-s_{0})r}r^{n} +\int_{r}^{\infty}|(e^{-(s-s_{0})x})x^{n})'|\,dx\right).$$

Therefore, since $e^{-(s-s_{0})r}r^{n}$ decreases monotonically on $(n,\infty)$ if $s>s_{0}$ (check!),

$$\left|\int_{r}^{r_{1}}e^{-sx}x^{n}f(x)\,dx\right|<3Me^{-(s-s_{0})r}r^{n},\quad n<r<r_{1},$$

so Theorem [theorem:4] implies that $I_{n}(s)$ converges uniformly $[s_{1},\infty)$ if $s_{1}>s_{0}$. Now Theorem [theorem:11] implies that $F_{n+1}=-F_{n}'$, and an easy induction proof yields [eq:30] (Exercise [exer:25]).

[example:13] Here we apply Theorem [theorem:12] with $f(x)=\cos ax$ ($a\ne0$) and $s_{0}=0$. Since

$$\int_{0}^{x}\cos au\,du=\frac{\sin ax}{a}$$

is bounded on $(0,\infty)$, Theorem [theorem:12] implies that

$$F(s)=\int_{0}^{\infty}e^{-sx}\cos ax\,dx$$

converges and

$$\label{eq:31} F^{(n)}(s)=(-1)^{n}\int_{0}^{\infty}e^{-sx}x^{n}\cos ax\,dx, \quad s>0.$$

(Note that this is also true if $a=0$.) Elementary integration yields

$$F(s)=\frac{s}{s^{2}+a^{2}}.$$

Hence, from [eq:31],

$$\int_{0}^{\infty}e^{-sx}x^{n}\cos ax=(-1)^{n}\frac{d^n}{ds^n} \frac{s}{s^{2}+a^{2}}, \quad n=0,1, \dots.$$