Laplace transform of $\frac{\sin at}{t}$

Question

Laplace transform of $\displaystyle \frac{\sin at}{t}$

My Attempt:

Rule used: $\displaystyle L[\frac{1}{t}f(t)]=\int_{s}^{\infty}\bar f(s)ds$ So,

$\displaystyle L[\frac{1}{t}\sin at]=\int_{s}^{\infty}\frac{a}{s^2+a^2}ds= \left. \tan^{-1}\frac{s}{a} \right|_{s}^{\infty}$

This is equal to $\displaystyle \frac{\pi}{2}-\tan^{-1}\frac{s}{a}$

The given answer is $\displaystyle \cot^{-1}\frac{s}{a}$

Where did I go wrong? Please help.

Answer 1

Hint:

$\tan^{-1} x+\cot^{-1} x=\frac{\pi}{2}$