Q.Evaluate $\displaystyle\int_0^1 \dfrac{1}{\sqrt{1-x^2}}\,dx$.

Improper Integral Evaluation

The Intuition First

You already know how to integrate over a finite interval: $\int_1^3 f(x)\,dx$ is the area under the curve from $x=1$ to $x=3$. But what if the region stretches to infinity, or the function shoots up to infinity somewhere in the interval?

That is what improper integrals handle — two situations that break the ordinary rules:

1. Infinite limits — integrating up to $\infty$ or down to $-\infty$.
2. Infinite discontinuities — the function blows up at some point of the interval.

The core idea is the same in both cases: replace the "bad" point with a limit. Integrate up to a finite value, then let that value approach the trouble spot. If the result approaches a finite number, the integral converges; if it grows without bound, it diverges.

The Precise Definitions

Type 1: Infinite limits

For a doubly-infinite integral, split at a convenient point $c$ and require both pieces to converge:

Type 2: Infinite discontinuities

If $f$ has a vertical asymptote at an endpoint, approach it from inside the interval:

If the blow-up is at an interior point $c$, split at $c$ and treat each side separately.

Worked Examples

Infinite limit. Evaluate $\int_1^\infty \frac{1}{x^2}\,dx$:

So it converges to $1$. Contrast $\int_1^\infty \frac{1}{x}\,dx = \lim_{b\to\infty}\log b = \infty$, which diverges.

Infinite discontinuity. Evaluate $\int_0^1 \frac{1}{\sqrt{x}}\,dx$, where the integrand blows up at $x=0$:

Converges to $2$.