Just Stick To The Basics...Or Don't

Using $\sin^3x=\frac{3\sin x-\sin 3x}{4}$ , we can rewrite the integral as $\frac{3}{4}\int_0^\infty \frac{e^{-x}\sin x}{x}\,dx-\frac{1}{4}\int_0^\infty \frac{e^{-x}\sin 3x}{x}\,dx$ Now, consider $I_a(b)=\int_0^\infty \frac{e^{-ax}\sin bx}{x}\,dx\qquad,\qquad\mbox{where}\,\,a>0$ so that $I_a(0)=0$ . Differentiating w.r.t. $b$ and then integrating back, we have $\begin{aligned} I_a'(b)&=\int_0^\infty e^{-ax}\cos bx\,dx\\ &=\frac{a}{a^2+b^2}\\ I_a(b)&=\int\frac{a}{a^2+b^2}\,db\\ &=\arctan\left(\frac{b}{a}\right)+C \end{aligned}$ Since $I_a(0)=0$ , then $C=0$ . Our original integral is $\begin{aligned} \int_0^\infty \frac{e^{-x}\sin^ 3x}{x}\,dx&=\frac{3}{4}I_1(1)-\frac{1}{4}I_1(3)\\ &=\frac{3}{4}\arctan(1)-\frac{1}{4}\arctan(3)\\ &=\frac{3\pi}{16}-\frac{\arctan(3)}{4} \end{aligned}$

Anastasiya Romanova $\quad\color{#D61F06}{\Rightarrow}\qquad$ Romanov's Integral