How can $\int_a^b f(x)dx $ exist if either $f(a)$ or $f(b)$ does not exist?

Question

In class, I came across the integral: $$\int_0^1 \frac{dx }{\sqrt{1-x^2}}=\frac{\pi}{2}$$

This is easy enough to prove using a substitution or by recalling the derivative of $\arcsin x$.

However, how can the integral be evaluated to a finite value when the integrand evaluated at a bound does not exist? In my example, $\dfrac{1}{\sqrt{1-x^2}}$ is undefined at $x=1$.

My intuition tells me that this cannot be true, however, the above example easily says otherwise.

Put more simply, how can $\int_a^b f(x)dx $ exist if either $f(a)$ or $f(b)$ does not exist?

Answer 1

It has to do with the type of singularity.

• In the case you mention, the singularity as $x \to 1^-$, is in the form $$ \frac1{\sqrt{1-x^2}} \sim \frac1{\sqrt{2}}\frac1{\sqrt{1-x}} $$ and the latter integrand admits an antiderivative which is finite as $x \to 1^-$: $$ \int_a^{1-\epsilon}\frac{dx}{\sqrt{1-x}} = \left[-2\sqrt{1-x}\right]_a^{1-\epsilon}=2\sqrt{1-a}-2\sqrt{\epsilon} \to 2\sqrt{1-a}<\infty \quad ( \epsilon \to 0). $$
• Taking another integrand with a different singularity as $x \to 1^-$, say in the
  form $$ \frac1{(1-x^2)^{3/2}} \sim \frac1{2^{3/2}}\frac1{(1-x)^{3/2}} $$ then the latter integrand admits an antiderivative which is not finite as $x \to 1^-$: $$ \int_a^{1-\epsilon}\frac{dx}{(1-x)^{3/2}} = \left[\frac2{\sqrt{1-x}} \right]_a^{1-\epsilon}=\frac2{\sqrt{\epsilon}}-\frac2{\sqrt{1-a}} \to \infty \quad ( \epsilon \to 0). $$

Answer 2

We have

$$\int_0^1 \frac{dx}{\sqrt{1-x^2}} \stackrel{\text{def}}{=} \lim_{\epsilon \to 1^{-}} \int_0^\epsilon \frac{dx}{\sqrt{1-x^2}} $$ if the limit exists. Indeed, the limit equals $\frac{\pi}{2}$ as you claim.