Question

The Laplace transform of the multiple integral \( L \left\{ \int_{0}^{t} \int_{0}^{t} \int_{0}^{t} \int_{0}^{t} \cos au \, du \, du \, du \, du \right\} \) is:

• A. \( \frac{1}{s^2 + a^2} \)
• B. \( \frac{1}{s(s^2 + a^2)} \)
• C. \( \frac{1}{s^2(s^2 + a^2)} \)
• D. \( \frac{1}{s^3(s^2 + a^2)} \)

Hint:

Always reduce the power of \( s \) in the numerator by 1 for every level of integration when dealing with functions like \( \cos at \) or \( \sin at \).

Answer 1

The Correct Option is D

Step 1: Rewrite the repeated integral as a convolution.
Integrating a function four times from \( 0 \) to \( t \) is the same as convolving it four times with the constant function \( 1 \). Using \( L\{f * g\} = F(s)G(s) \) and \( L\{1\} = \dfrac{1}{s} \), each integration multiplies the transform by \( \dfrac{1}{s} \).

Step 2: Find the transform of the innermost function.
\[ L\{\cos at\} = \frac{s}{s^2+a^2} \]

Step 3: Apply the factor four times.
Four convolutions with \( 1 \) multiply the transform by \( \dfrac{1}{s^4} \), so \[ L\{\cdots\} = \frac{1}{s^4}\cdot\frac{s}{s^2+a^2} = \frac{1}{s^3(s^2+a^2)} \] \[ \boxed{\frac{1}{s^3(s^2+a^2)}} \]