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.


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


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


Now evaluating I between the limits x=1 and x=R:


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.



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.


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


Now putting in the limits:



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.


\int_{2\pi}^\infty \sin\theta \,d\theta


\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{x%2C0%2C100}%5D).


Question 2

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


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.


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;




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