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# 1.7: Applications to Laplace transforms

• • Contributed by William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University
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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],

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