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## Definition: Improper Integral

The improper integral is defined as the limit of a convergent sequence of Riemann integrals for the following three cases. We also say that the improper integral converges (or is convergent).

### Case 1: Unbounded Intervals

Let $f:[a,\infty)$ be a real-valued function such that

Then we set the improper interval to this limit:

$$\int_a^\infty f(x)dx:=\lim_{b\to\infty}\int_a^bf(x)dx$$

Analogously, if $f:(\infty,b]$ is a real-valued function such that

• $f$ is Riemann-integrable for all closed intervals $[a,b]$ for $-\infty < a < b$ and
• the limit $\lim_{a\to-\infty}\int_a^bf(x)dx$ exists,

then we set

$$\int_{-\infty}^b f(x)dx:=\lim_{a\to-\infty}\int_a^bf(x)dx.$$

### Case 2: The Riemann integral is undefined on one or even two integration endpoints

If $f:[a,b)\to\mathbb R$ is a function such that

• $f$ is Riemann-integrable for all closed intervals $[a,b-\epsilon]$ for all $\epsilon > 0$ and
• the limit $\lim_{\epsilon\searrow 0}\int_a^{b-\epsilon}f(x)dx$ exists,

then we set

$$\int_a^b f(x)dx:=\lim_{\epsilon\searrow 0}\int_a^{b-\epsilon}f(x)dx.$$

Analogously, if $f:(a,b]\to\mathbb R$

• is Riemann-integrable for all closed intervals $[a+\epsilon,b]$ for all $\epsilon > 0$ and
• the limit $\lim_{\epsilon\searrow 0}\int_{a+\epsilon}^{b}f(x)dx$ exists,

then we set

$$\int_a^b f(x)dx:=\lim_{\epsilon\searrow 0}\int_{a+\epsilon}^bf(x)dx.$$

### Case 3: Mixed case, both endpoints of the improper integral are critical

Let $f:(a,b)\to\mathbb R$ for $a\in\mathbb R\cup \{-\infty\}$ and $b\in\mathbb R\cup \{+\infty\}$ ($a,b$ being elements of the extended real numbers) is a function and let $c\in(a,b)$ be given1. If

• $f$ is Riemann-integrable for all closed intervals $[\alpha,\beta]\subset (a,b)$ and
• the limits $\lim_{\alpha\searrow a}\int_\alpha^{c}f(x)dx$ and $\lim_{\beta\nearrow b}\int_c^{\beta}f(x)dx$ both exist,

then we set

$$\int_a^b f(x)dx:=\lim_{\alpha\searrow a}\int_\alpha^{c}f(x)dx+\lim_{\beta\nearrow b}\int_c^{\beta}f(x)dx.$$

1 Note that this definition is independent of the specific choice of $c\in(a,b).$ This is because by hypothesis, $f$ is Riemann-integrable for all closed intervals $[\alpha,\beta]\subset (a,b)$ and the integrals can be summed to the same value on any adjacent intervals.

| | | | | created: 2020-01-24 09:06:20 | modified: 2020-01-25 08:29:42 | by: bookofproofs | references: [581]

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