## Question 1

Determine whether the integrals are convergent or divergent. Evaluate those that are convergent.

This exercise is more to get us used to evaluating infinite integrals — none of these bar (iii) are necessarily related to Laplace transforms.

### (i) $\int_1^\infty \frac{1}{(3x+1)^2}\,dx$

#### Solution

We first try and evaluate $\int \frac{1}{(3x+1)^2}\,dx$.

Try the substitution $u=3x+1$: $\frac{du}{dx}=3\Rightarrow dx =\frac{du}{3}$. $I=\int\frac{1}{u^2}\cdot\frac{du}{3}=\frac{1}{3}\int u^{-2}\,du$ $=\frac{1}{3}\frac{u^{-1}}{-1}=-\frac{1}{3u}=-\frac{1}{3(3x+1)}$.

Now evaluating $I$ between the limits $x=1$ and $x=R$: $\left(-\frac{1}{3(3R+1)}\right)-\left(-\frac{1}{3(3+1)}\right)=\frac{1}{12}-\frac{1}{9R+3}$.

Now taking the limit as $R\rightarrow\infty$: $I=\lim_{R\rightarrow\infty}\left(\frac{1}{12}-\underbrace{\frac{1}{9R+3}}_{\rightarrow 0}\right)=\frac{1}{12}$.

The integral is convergent with value $1/12$.

### (ii) $\int_0^\infty\frac{x}{(x^2+2)^2}\,dx$

We will do this by evaluating the indefinite integral, then put in the limits ( $0$ and $R$), then then take limit as $R\rightarrow \infty$.

Consider $I= \int \frac{x}{(x^2+2)^2}\,dx$.

Note the function-derivative pattern $x^2+2$ – $x$ (or use LIATE), and hence let $u=x^2+2$: $\frac{du}{dx}=2x\Rightarrow dx=\frac{du}{2x}$ $I=\int \frac{x}{u^2}\cdot\frac{du}{2x}=\frac{1}{2}\int u^{-2}\,du$ $=\frac{1}{2}\frac{u^{-1}}{-1}=-\frac{1}{2}\frac{1}{x^2+2}$.

Now putting in the limits: $\left[-\frac{1}{2}\frac{1}{x^2+2}\right]_0^R=\left(-\frac{1}{2}\frac{1}{R^2+2}\right)-\left(-\frac{1}{2}\frac{1}{2}\right)$ $=\frac{1}{4}-\frac{1}{2R^2+4}$.

Now taking the limit as $R\rightarrow\infty$: $\lim_{R\rightarrow\infty}\left(\frac{1}{4}-\underbrace{\frac{1}{2R^2+4}}_{\rightarrow 0}\right)=\frac{1}{4}$;

i.e. the integral is convergent with value $1/4$.

### (iii) $\int_{2\pi}^\infty \sin\theta \,d\theta$

Consider $\int_{2\pi}^R\sin\theta\,d\theta=\left[-\cos \theta\right]_{2\pi}^R$ $=(-\cos R)-(-\cos 2\pi)$.

Now using a unit circle, the tables, a calculator, or otherwise, note that $\cos 2\pi=\cos 0=1$. That is the integral evaluates to: $1-\cos R$.

Now taking the limit as $R\rightarrow \infty$: $\lim_{R\rightarrow \infty}(1-\cos R)$

however this limit does not exist. $1-\cos R$ does not get closer and closer to a particular value as $R$ increases — instead it oscillates between $+2$ and $0$. (Nice Wolfram Alpha graph http://www.wolframalpha.com/input/?i=Plot%5B1-Cos%5Bx%5D%2C{x%2C0%2C100}%5D).

## Question 2

Find the Laplace transform of the zero function $f(t) = 0$.

### Solution

Using the definition: $\mathcal{L}\{f(t)\}=\int_0\infty (0)e^{-st}\,dt=\lim_{R\rightarrow\infty}\underbrace{\int_0^R(0)\,dt}_{=0}$ $=\lim_{R\rightarrow\infty} (0)=0$.

## Question 3

Find the Laplace transform of the following function $g(t)$: $g(t):=\left\{\begin{array}{cc}1 & \text{ if }t\in[0,1]\\ 0 & \text{ otherwise}\end{array}\right.$

I am under the impression that this would be a more Math 3.1 exercise — nothing like it will be on our exams.

### Solution

Again, using the definition: $\mathcal{L}\{g(t)\}=\int_0^\infty g(t)e^{-st}\,dt=\lim_{R\rightarrow 0}\int_0^Rg(t)e^{-st}\,dt$.

Hence splitting up the interval over which we integrate; $\int_0^Rg(t)e^{-st}\,dt=\int_0^1g(t)e^{-st}\,dt+\int_1^Rg(t)e^{-st}\,dt$ $=\int_0^1(1)e^{-st}\,dt+\underbrace{\int_1^R(0)e^{-st}\,dt}_{=0}=\left[\frac{e^{-st}}{-s}\right]_0^1$ $=\left(-\frac{e^{-s}}{s}\right)-\left(-\frac{e^0}{s}\right)=-\frac{e^{-s}}{s}+\frac{1}{s}=\frac{1}{s}(1-e^{-s})$.

Clearly taking the limit as $R\rightarrow \infty$ does not affect this so this is the answer.