Integral of Laplace

first, we calculate the laplace transform: $\mathcal{L} \left \{\sum^{\infty}_{n=1} \dfrac{(-1)^{n-1} t^{2n-1}}{\left((2n-1)! \right)^2} \right\}=\sum^{\infty}_{n=1} \dfrac{(-1)^{n-1}}{\left((2n-1)! \right)^2}\mathcal{L}\{t^{2n-1}\}\\=\sum^{\infty}_{n=1} \dfrac{(-1)^{n-1}}{\left((2n-1)! \right)^2}\dfrac{(2n-1)!}{s^{2n}}=\dfrac{1}{s}\sin\left(\dfrac{1}{s}\right)$ In the last step the taylor series of sine was used. The integral becomes $2\int_0^\infty \dfrac{1}{s} \sin\left(\dfrac{1}{s}\right)\text{ds}=2\int_0^\infty \dfrac{\sin(x)}{x}\text{dx}$ A substitution of $x=t^{-1}$ was made. The integral is the famous trigonometric integral, which i will prove using laplace transform in one line $2\int_0^\infty\frac{sin(x)}{x} dx=2\int_0^\infty \mathcal{L}\{\sin(x)\}(s) ds=2\int_0^\infty \dfrac{1}{s^2+1} ds=2\arctan s\bigg|_{0}^{\infty}=\boxed{\pi}$