1.48 An integral from the theory of differential equations

Now we rearrange, and get $(1+\frac{a^{2}}{s^{2}})I=-\frac{s\sin ax+a\cos ax}{s^{2}}e^{-sx}+c$, so

$$I=-\frac{s\sin ax+a\cos ax}{a^{2}+s^{2}}e^{-sx}+c^{\prime}.$$

Thus

$$\int_{0}^{R}\sin ax\,e^{-sx}\,dx={-\left[\frac{s\sin ax+a\cos ax}{a^{2}+s^{2}}e^{-sx}\right]_{0}^{R}}$$

$${=\frac{a-e^{-Rs}(s\sin aR+a\cos aR)}{a^{2}+s^{2}}.}$$

Finally, $|s\sin aR+a\cos aR|\leq|a|+s$, so

$${e^{-Rs}(s\sin aR+a\cos aR)\rightarrow 0\;\;\mbox{as}\;\;R\rightarrow\infty.}$$

Thus the integral converges to $\frac{a}{a^{2}+s^{2}}$ as $R\rightarrow\infty$.