I'm reviewing lecture notes on Laplace Transform and there's one step that I don't understand:
Find the solution to: $$x y'' + y' + xy = 0, y(0) = 1, y'(0) \mbox{ finite}$$ Taking the Laplace Transform, we get: \begin{align} 0 &= (-s^2 - 1) Y'(s) + (-2s + s) Y(s) \\ (s^2 + 1) Y'(s) &= -s Y(s) \\ \frac{Y'(s)}{Y(s)} &= -\frac{s}{s^2 + 1} \\ \log(Y(s)) &= - \int \frac{s}{s^2 + 1} ds \\ &= -\frac{1}{2} \log(s^2 + 1) + C \\ Y(s) &= \frac{D}{\sqrt{s^2 + 1}} \end{align}
To find D, let: $$\frac{D}{\sqrt{s^2 + 1}} = \int_0^{\infty} e^{-xs} y(x) dx \rightarrow \frac{y(0)}{s} \mbox{as $s \rightarrow \infty$}$$ which then yields $D = 1$
So this is where I don't understand; how can we approximate the integral to $\frac{y(0)}{s}$ as $s \rightarrow \infty$?