Integrating Various Functions

Let the integral be $I$ . Then we have:

$\begin{aligned} I & = \int_0^\infty \frac{e^{-x}\sin^2 x}{x} \space dx \\ & = \int_0^\infty \frac{e^{-x}(1 - \color{#3D99F6}{\cos 2x})}{2x} \space dx \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{By Maclaurin series:}} \\ & = \int_0^\infty \frac{e^{-x}}{2x} \left( 1 - \color{#3D99F6}{\displaystyle \sum_{n=0}^\infty \frac{(-1)^n(2x)^{2n}}{(2n)!}}\right) \space dx \\ & = \int_0^\infty \frac{e^{-x}}{2x} \left(\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}(2x)^{2n}}{(2n)!} \right) \space dx \\ & = \int_0^\infty \displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}(2x)^{2n-1}e^{-x}}{(2n)!} \space dx \\ & = \sum_{n=1}^\infty \frac{(-1)^{n+1}\int_0^\infty (2x)^{2n-1}e^{-x} \space dx }{(2n)!} \\ & = \sum_{n=1}^\infty \frac{(-1)^{n+1} 2^{2n-1} \Gamma (2n)}{(2n)!} \\ & = \sum_{n=1}^\infty \frac{(-1)^{n+1} 2^{2n-1} (2n-1)!}{(2n)!} \\ & = \sum_{n=1}^\infty \frac{(-1)^{n+1} 2^{2n-1}}{2n} \\ & = \frac{1}{4} \sum_{n=1}^\infty \frac{(-1)^{n+1} 4^n}{n} \\ & = \frac{\ln (1+4)}{4} = \boxed{\dfrac{\ln 5}{4}} \end{aligned}$

now we know that :- So