Improper Integrals

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Improper Integrals

Type 1 - Improper Integrals with Infinite Intervals of Integration
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Type 1 - Improper Integrals with Infinite Intervals of Integration

An improper integral of type 1 is an integral whose interval of integration is infinite. This means the limits of integration include $\infty$ or $-\infty$ or both. Remember that $\infty$ is a process (keep going and never stop), not a number. Therefore, we cannot use $\infty$ as an actual limit of integration as in the FTC II. We make the limit of integration a number, and take the limit as that goes to infinity or negative infinity. We define

$\displaystyle\int_a^\infty f(x) \,dx= \lim_{t \to \infty} \int_a^t f(x)\, dx,$

$\displaystyle\int_{-\infty}^b f(x)\, dx = \lim_{t \to -\infty} \int_t^b f(x)\, dx,$ and

$\displaystyle\int_{-\infty}^\infty f(x)\, dx = \int_{-\infty}^a f(x) \,dx + \int_a^\infty f(x) \,dx,$ where we can choose any number for the break point $a$ . (Zero is often convenient.)

On Convergence

$\displaystyle\int_a^\infty f(x)\,dx$ converges if $\displaystyle\lim_{t\to\infty}\int_a^t f(x)\,dx$ exists, and diverges if the limit doesn't exist, including when it is infinite.
$\displaystyle\int_{-\infty}^b f(x)\,dx$ converges if $\displaystyle\lim_{t \to -\infty} \int_t^b f(x)\, dx$ exists, and diverges if the limit doesn't exist, including when it is infinite.
$\displaystyle\int_{-\infty}^\infty f(x)\,dx$ converges if both $\displaystyle\int_a^\infty f(x)\,dx$ and $\displaystyle\int_{-\infty}^b f(x)\,dx$ converge.

We'll be talking a lot more about convergence and divergence when we get to sequences and series.

The following video explains improper integrals with infinite intervals of integration (type 1) and works out a number of examples.