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.

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