Question

Using Laplace transformation, the value of $ \int_{0}^{\infty} \frac{\sin 2t}{t} dt $ is

• A. \(0 \)
• B. \( \frac{3}{s^2+4} \)
• C. \( \frac{2}{\pi} \)
• D. \( \frac{\pi}{2} \)

Hint:

Remember the Laplace transform property for division by \( t \): \( L\left\{\frac{f(t)}{t}\right\} = \int_{s}^{\infty} F(u) du \). To evaluate integrals of the form \( \int_{0}^{\infty} \frac{f(t)}{t} dt \), find \( L\left\{\frac{f(t)}{t}\right\} \) and then set \( s=0 \).

Answer 1

The Correct Option is D

Step 1: Understand the Laplace Transform of \( \frac{f(t)}{t} \).
The Laplace transform of a function \( f(t) \) is defined as \( L\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt \).
One important property of Laplace transforms is:
\( L\left\{\frac{f(t)}{t}\right\} = \int_{s}^{\infty} F(u) du \), where \( F(u) = L\{f(t)\} \).
Step 2: Find the Laplace transform of \( \sin 2t \).
Let \( f(t) = \sin 2t \).
The Laplace transform of \( \sin at \) is \( L\{\sin at\} = \frac{a}{s^2 + a^2} \).
For \( a=2 \), we have:
\( F(s) = L\{\sin 2t\} = \frac{2}{s^2 + 2^2} = \frac{2}{s^2 + 4} \). 
Step 3: Apply the property for \( \frac{f(t)}{t} \).
Now, we need to find \( L\left\{\frac{\sin 2t}{t}\right\} \):
\[ L\left\{\frac{\sin 2t}{t}\right\} = \int_{s}^{\infty} \frac{2}{u^2 + 4} du \] \[ = 2 \int_{s}^{\infty} \frac{1}{u^2 + 2^2} du \] We know that \( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) \). \[ = 2 \left[ \frac{1}{2} \tan^{-1}\left(\frac{u}{2}\right) \right]_{s}^{\infty} \] \[ = \left[ \tan^{-1}\left(\frac{u}{2}\right) \right]_{s}^{\infty} \] \[ = \tan^{-1}(\infty) - \tan^{-1}\left(\frac{s}{2}\right) \] Since \( \tan^{-1}(\infty) = \frac{\pi}{2} \): \[ L\left\{\frac{\sin 2t}{t}\right\} = \frac{\pi}{2} - \tan^{-1}\left(\frac{s}{2}\right) \] 
Step 4: Evaluate the integral by setting \( s=0 \).
The integral \( \int_{0}^{\infty} \frac{\sin 2t}{t} dt \) is equivalent to \( L\left\{\frac{\sin 2t}{t}\right\} \) evaluated at \( s=0 \).
\[ \int_{0}^{\infty} \frac{\sin 2t}{t} dt = \left[ L\left\{\frac{\sin 2t}{t}\right\} \right]_{s=0} \] \[ = \frac{\pi}{2} - \tan^{-1}\left(\frac{0}{2}\right) \] \[ = \frac{\pi}{2} - \tan^{-1}(0) \] Since \( \tan^{-1}(0) = 0 \): \[ = \frac{\pi}{2} - 0 \] \[ = \frac{\pi}{2} \] The final answer is $\boxed{\text{4}}$.