Discussion
Laplace transform of f(x)=tanx where tanx has limit of infinity is
Answer: D
A function has a Laplace transform only if the integral
\[ \int_{0}^{\infty} e^{-st}f(t)\,dt \]
converges. That requires \( f \) to be piecewise continuous and of exponential order.
\( \tan t \) has infinite discontinuities at \( t = \tfrac{\pi}{2}, \tfrac{3\pi}{2}, \ldots \), where it blows up faster than any exponential can damp. The integral diverges, so no Laplace transform exists.
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