my problem in inverse laplace $\frac{1}{s}\arctan\frac{1}{s}$

Question

I want to find inverse laplace $$\dfrac{1}{s}\arctan\dfrac{1}{s}$$ My approach: $$F(s)=\dfrac{1}{s}\arctan\dfrac{1}{s}$$ $$(sF)'=\dfrac{1}{s^2+1}$$ $$L(-ty')=\dfrac{1}{s^2+1}$$ $$-ty'=\sin t+C$$ $$y=-\int\dfrac{\sin t+C}{t}dt$$ how can i continue from here?

Answer 1

At this point, the integral $$\int \frac{\sin(t)}{t}dt$$ has no closed form: this is called the sine integral