Using Laplace Transformation Calculate the given integral

Question

The question says that I should calculate the following integral using Laplace transformation.

I know the formula for the Laplace transform:

I've also added the -st to -x since the base is e

I am not sure how I would solve this, the ratio (sin(x))^2/x should be done with Taylor's expansion since this can't be solved with known integrals method..

Note: the (sin(x))^2/x ratio is not in the exponent (the power) it's rather in the exp line (multiplication)...

Answer 1

HINT:

Note that we have

$$F(s)=\int_0^\infty e^{-sx}\frac{\sin^2(x)}{x}\,dx$$

Therefore, we can write

$$\begin{align} F'(s)&=-\int_0^\infty e^{-sx}\sin^2(x)\,dx\\\\ &=-\frac{1}{2}\int_0^\infty e^{-sx}\left(1-\cos(2x)\right)\,dx \tag 1 \end{align}$$

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Recall that the Laplace Transforms of $1$ and $\cos(ax)$ are $\frac1s$ and $\frac{s}{s^2+a^2}$, respectively. Using this in $(1)$ reveals $$F'(s)=-\frac12\left(\frac{1}{s}-\frac{s}{s^2+4}\right)\tag 2$$Integrating $(2)$ and using the fact that $\lim_{s\to \infty }F(s)=0$ yields $$F(s)=\frac14 \log\left(1+\frac{4}{s^2}\right)$$from which we find $$F(1)=\int_0^\infty e^{-x}\frac{\sin^2(x)}{x}\,dx=\frac14 \log(5)$$And we are done!