78 Conclusion

$$\frac{e^{-xh}-1}{h}+x=h\left(\frac{e^{-xh}-1+hx}{h^2}\right)$$

where by comparing the coefficients in the power series, we see that

$$\left|\frac{e^{-xh}-1+hx}{h^2}\right|\le \frac{e^{\delta x}-1-\delta x}{{\delta }^2}$$

and $e^{\delta x}{e}^{-sx}f(x)$ is integrable. (This is justified by uniform convergence or dominated convergence theorems.)