Vertical Asymptotes
Sometimes the function itself goes to infinity at a point within the limits of integration. We handle this by taking the limit as we approach the discontinuity.
Worked Examples
Example 1: Approaching Zero
Evaluate \(\int_0^1 \frac{1}{\sqrt{x}} dx\).
- The function is undefined at \(x=0\).
- \(\lim_{a \to 0^+} \int_a^1 x^{-1/2} dx = \lim_{a \to 0^+} [2\sqrt{x}]_a^1 = 2 - 0 = 2\).
- Result: 2. (Even though the function is infinitely tall, its area is finite!)
The Bridge to Quantum Mechanics
Quantum particles often interact with "Point Sources" (like the nucleus of an atom). The potential energy near a point charge goes to infinity (\(V \propto 1/r\)). To find the average energy of an electron, we have to integrate over this singularity. Improper integrals allow us to show that even though the potential is infinitely deep, the total energy of the atom remains finite. This is the only reason atoms don't collapse in on themselves.