Evaluate the following integral using Laplace transform
$$\int_0^\infty\frac{e^{-ax}\sin bx}{x}\,dx$$
I obtained this partial result
$$=\int_0^\infty \frac{1}{p+1} \frac{b}{p^2+b^2}\,dp$$
and I am stuck here. I know that the final answer is $$\arctan\frac{b}{a}.$$ I would appreciate if someone could help me finish to attain the final answer.
Best Answer
Your partial result does not hold. According to Laplace transform properties $$\int_0^\infty\frac{e^{-ax}\sin bx}{x}\,dx=\cal{L}\left(\frac{\sin bx}{x}\right)(a)=\int_a^{+\infty} \cal{L}\left(\sin bx\right)(p)dp=\int_a^{+\infty} \frac{b}{p^2+b^2}\, dp.$$ Can you take it form here?