Integral $\int_{0}^{1}{\frac{\exp(-rx)}{\sqrt{1-x}}dx}$ converges for $r\geq{1}$

Question

How do you prove that $\int_{0}^{1}{\frac{\exp(-rx)}{\sqrt{1-x}}dx}$ converges for $r\geq{1}$ ?

N.B. : I forgot everything about improper integrals, so please be very explicit :)

Answer 1

Very explicit: The integrand blows up at $x=1$, so you should be looking at $$\lim_{u\to1-}\int_{0}^{u}\frac{\exp(-rx)}{\sqrt{1-x}}dx.$$ The integrand is positive, so either this diverges to $\infty$ or it remains bounded. Let's show it is bounded. On the given interval, $\exp(-rx)\le1$ if $r>0$, while $\exp(-rx)\le e^{-r}$ if $r$ is negative. In either case just say $\exp(-rx)\le K$. So $$\int_0^u\frac{\exp(-rx)}{\sqrt{1-x}}dx\le\int_0^u\frac{K}{\sqrt{1-x}}dx=2K(1-\sqrt{1-u})<2K$$ for all $u<1$, so the integral is indeed bounded.

Answer 2

Let $x=1-u^2$. It suffices to show $\int_0^1 \exp(1-u^2)du$ converges. Note that $\exp(1-u^2)$ is controlled by a constant