Q3Advanced Engineering Mathematics I

Question

Prove $L {\frac{s i n ^{2} t}{t}} = \frac{1}{4} lo g (\frac{s ^{2} + 4}{s ^{2}})$ and hence deduce the integral $\int_{0}^{\infty} \frac{s i n ^{2} t}{t ^{2}} d t$ .

Answer 1

Using the property $L {\frac{f ( t )}{t}} = \int_{s}^{\infty} F (u) d u$ .

First, $L {sin^{2} t} = L {\frac{1 - c o s 2 t}{2}} = \frac{1}{2 s} - \frac{s}{2 ( s ^{2} + 4 )}$ .

Integrate this from $s$ to $\infty$ to get $\frac{1}{4} lo g (\frac{s ^{2} + 4}{s ^{2}})$ .

The integral $\int_{0}^{\infty} \frac{s i n ^{2} t}{t ^{2}} d t$ is evaluated by taking the limit as $s \to 0$ of the Laplace transform (final value theorem applications).

Q3Advanced Engineering Mathematics I

Question

Answer