Integrating over an Infinite Integral

Integrating a Discontinuous Integrand

Comparison Theorem

• Bigger and convergent
  • Smaller function is also
  • convergent
• Smaller and divergent
  • Bigger function is also divergent

<aside> 📎 Let $f(x)$ be a function that is continuous on $[a, \infty)$ for some real number $a$, we can definte the improper integral as $\int a ^\infty f(x) dx = \lim{t \to \infty} \int ^t _a f(x) dx$

If the limit exists, we say that the improper integral converges. If this limit does not exist, we say it diverges.

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• $\int _0 ^\infty e ^{-ax}dx$ is convergent if $a>0$ and divergent if $a ≤ 0$
• $\int ^\infty _0 \frac{1}{x^P}dx$ is convergent if $p > 1$ and divergent if $p ≤1$

Infinite Discontinuities at $x = a$

<aside> 📎 $\int_a ^b f(x) dx = \lim_{t \to a^+} \int ^b _t f(x) dx$

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Lets consider $\int _{-\infty} ^{\infty} \frac{1}{x^2}dx$

1. Unbounded domain as $x$ goes toward $- \infty$
2. Vertical asymptote as $x$ approaches 0 from the left
3. The same vertical asymptote, seen from the other side as we cross the x-axis
4. Unbounded domain as $x$ increases toward $+ \infty$