Improper Integrals

Type 1 Improper Integral: $\infty$ in the bounds

When the Improper Integral has $\infty$ in one of the bounds

Let $f$ be a function and let $a$ be a real number.

(Type 1) We define ${\int }_a^{\infty }f(x)\mathrm{d}x\to {\lim }_{t\to \infty }{\int }_a^{t}f(x)$

When the limit exists: “The improper integral is convergent”

The value of a convergent improper integral is the value of the limit.

When the limit D.N.E: “The improper integral is divergent”

Rewriting Improper Integrals with a $\infty$ in a single bound

Get rid of the infinity in the bounds and add a limit with bound $t$ (or any other unused variable). ${\int }_0^{\infty }f(x)\to {\lim }_{t\to \infty }{\int }_0^{t}f(x)$

When the improper integral is from $-\infty$ to $\infty$

(Type 1) We define ${\int }_{-\infty }^{\infty }f(x)\mathrm{d}x$

Split up the integral to define it. let $a$ be any real number.

If ${\int }_a^{\infty }f(x)\mathrm{d}x$ is convergent AND If ${\int }_{-\infty }^{a}f(x)\mathrm{d}x$ is convergent

Then ${\int }_{-\infty }^{\infty }f(x)\mathrm{d}x$ is convergent: ${\int }_{-\infty }^{\infty }f(x)\mathrm{d}x={\int }_{-\infty }^{a}f(x)\mathrm{d}x+{\int }_a^{\infty }f(x)\mathrm{d}x$

If either of the split up integrals are divergent, the whole term is divergent as well.

Type 2 Improper Integral: discontinuous function

When the function is not continuous at: $a$: ${\int }_a^{b}f(x)\mathrm{d}x\to {\lim }_{t\to a^{+}}{\int }_t^{b}f(x)\mathrm{d}x$ $b$: ${\int }_a^{b}f(x)\mathrm{d}x\to {\lim }_{t\to b^{-}}{\int }_a^{t}f(x)\mathrm{d}x$