Q10Advanced Engineering Mathematics

Question

Find L (sin t) and hence obtain L (\frac{cos t}{t}) .

Answer 1

The problem requires finding the Laplace transform of a non-standard function, $f (t) = sin (t)$ . Because of the square root inside the trigonometric argument, standard integration tables and elementary Laplace transform properties (like shifting or scaling) cannot be directly applied. The most robust mathematical strategy is to expand the sine function into its infinite Maclaurin power series and then meticulously apply the Laplace transform to each individual term in the series.

1. Maclaurin Series Expansion

The well-known Maclaurin series expansion for $sin (x)$ is $sin (x) = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \frac{x ^{7}}{7 !} + \dots$ . Substituting $x = t = t^{1/2}$ into this infinite series, we obtain:

sin (t) = t^{1/2} - \frac{( t ^{1/2} ) ^{3}}{3 !} + \frac{( t ^{1/2} ) ^{5}}{5 !} - \dots = t^{1/2} - \frac{t ^{3/2}}{3 !} + \frac{t ^{5/2}}{5 !} - \dots

2. Term-by-Term Laplace Transform

We now take the Laplace transform of both sides. Because the Laplace transform is a linear operator, we can apply it term-by-term. We utilize the generalized formula for the Laplace transform of a power of $t$ , which is $L {t^{n}} = \frac{Γ ( n + 1 )}{s ^{n + 1}}$ , where $Γ$ represents the Gamma function.

L {sin (t)} = L {t^{1/2}} - \frac{1}{3 !} L {t^{3/2}} + \frac{1}{5 !} L {t^{5/2}} - \dots

L {sin (t)} = \frac{Γ ( 3/2 )}{s ^{3/2}} - \frac{Γ ( 5/2 )}{3 ! s ^{5/2}} + \frac{Γ ( 7/2 )}{5 ! s ^{7/2}} - \dots

We calculate the Gamma function values using the property $Γ (n + 1) = n Γ (n)$ and knowing that $Γ (1/2) = π$ :

$Γ (3/2) = \frac{1}{2} Γ (1/2) = \frac{π}{2}$
$Γ (5/2) = \frac{3}{2} Γ (3/2) = \frac{3 π}{4}$
$Γ (7/2) = \frac{5}{2} Γ (5/2) = \frac{15 π}{8}$

Substituting these values back into the infinite series:

= \frac{π}{2 s ^{3/2}} - \frac{3 π}{4 \cdot 6 s ^{5/2}} + \frac{15 π}{8 \cdot 120 s ^{7/2}} - \dots = \frac{π}{2 s ^{3/2}} (1 - \frac{1}{4 s} + \frac{1}{32 s ^{2}} - \dots)

We recognize the expression inside the parentheses as the Maclaurin series expansion for the exponential function $e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \dots$ , where $x = - \frac{1}{4 s}$ . Therefore, the expression simplifies elegantly to a closed form:

L {sin (t)} = \frac{π}{2 s ^{3/2}} e^{- \frac{1}{4 s}}

3. Deducing the Related Transform

The problem also asks to deduce $L {\frac{c o s ( t )}{t}}$ . Let $f (t) = sin (t)$ . Its derivative using the chain rule is $f^{'} (t) = \frac{c o s ( t )}{2 t}$ . Using the Laplace transform property of derivatives $L {f^{'} (t)} = s L {f (t)} - f (0)$ (where $f (0) = 0$ ):

L {\frac{cos ( t )}{2 t}} = s (\frac{π}{2 s ^{3/2}} e^{- \frac{1}{4 s}}) = \frac{π}{2 s ^{1/2}} e^{- \frac{1}{4 s}}

Multiplying both sides by 2 yields the final desired deduction: $L {\frac{c o s ( t )}{t}} = \frac{π}{s} e^{- \frac{1}{4 s}}$ .

Q10Advanced Engineering Mathematics

Question

Answer

1. Maclaurin Series Expansion

2. Term-by-Term Laplace Transform

3. Deducing the Related Transform