# Type I (Infinite Intervals)

## Related calculator: Definite and Improper Integral Calculator

When we defined definite integral ${\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$ we assumed that interval [a,b] is finite and that function f doesn't have infinite discontinuities. Let's extend integral to the case where interval [a,b] is infinite.

Consider the infinite region S that lies below the curve ${y}={f{{\left({x}\right)}}}$, above the x-axis and to the right of the line x=a.

You might think that, since S is infinite in extent, its area must be infinite, but let’s take a closer look. The area of the part of S that lies to the

left of the line x=t is ${A}{\left({t}\right)}={\int_{{a}}^{{t}}}{f{{\left({x}\right)}}}{d}{x}={F}{\left({t}\right)}-{F}{\left({a}\right)}$ where ${F}$ is antiderivative of f.

For example, if ${f{{\left({x}\right)}}}=\frac{{1}}{{{x}}^{{2}}}$ and ${a}={1}$ then ${A}{\left({t}\right)}={\int_{{1}}^{{t}}}\frac{{1}}{{{x}}^{{2}}}{d}{x}={1}-\frac{{1}}{{t}}$.

Observe that in this case $\lim_{{{t}\to\infty}}{A}{\left({t}\right)}=\lim_{{{t}\to\infty}}{\left({1}-\frac{{1}}{{t}}\right)}={1}$.

So, we define the integral of f (not necessarily a positive function) over an infinite interval as the limit of integrals over finite intervals.

Definition of an Improper Integral of Type 1

1. If ${\int_{{a}}^{{t}}}{f{{\left({x}\right)}}}{d}{x}$ exists for every number ${t}\ge{a}$, then ${\int_{{a}}^{{\infty}}}{f{{\left({x}\right)}}}{d}{x}=\lim_{{{t}\to\infty}}{\int_{{a}}^{{t}}}{f{{\left({x}\right)}}}{d}{x}$ provided this limit exists (as a finite number).
2. If ${\int_{{t}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$ exists for every number ${t}\le{b}$, then ${\int_{{-\infty}}^{{{b}}}}{f{{\left({x}\right)}}}{d}{x}=\lim_{{{t}\to\infty}}{\int_{{t}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$ provided this limit exists (as a finite number).
3. The improper integrals ${\int_{{a}}^{{\infty}}}{f{{\left({x}\right)}}}{d}{x}$ and ${\int_{{-\infty}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$ are called convergent are called convergent if the corresponding limit exists and divergent if the limit doesn't exist.
4. If both ${\int_{{a}}^{{\infty}}}{f{{\left({x}\right)}}}{d}{x}$ and ${\int_{{-\infty}}^{{a}}}{f{{\left({x}\right)}}}{d}{x}$ are convergent then ${\int_{{-\infty}}^{{\infty}}}{f{{\left({x}\right)}}}{d}{x}={\int_{{-\infty}}^{{a}}}{f{{\left({x}\right)}}}{d}{x}+{\int_{{a}}^{{\infty}}}{f{{\left({x}\right)}}}{d}{x}$. Note, that any real number can be used. Once again only when BOTH integrals are convergent, convergent is also an integral ${\int_{{-\infty}}^{{\infty}}}{f{{\left({x}\right)}}}{d}{x}$. Otherwise, it is divergent.

Any of the improper integrals in Definition can be interpreted as an area provided that f is a positive function.

Example 1. For which p is integral ${\int_{{1}}^{{\infty}}}\frac{{1}}{{{x}}^{{p}}}{d}{x}$ convergent?

Let's handle two cases: p=1 and ${p}\ne{1}$.

If p=1 then ${\int_{{1}}^{{\infty}}}\frac{{1}}{{x}}{d}{x}=\lim_{{{t}\to\infty}}{\int_{{1}}^{{t}}}\frac{{1}}{{x}}{d}{x}=\lim_{{{t}\to\infty}}{\left({\ln}{\left|{x}\right|}{{\mid}_{{1}}^{{t}}}\right)}=\lim_{{{t}\to\infty}}{\left({\ln{{\left({t}\right)}}}-{\ln{{\left({1}\right)}}}\right)}=\infty$.

Thus, when p=1 integral is divergent.

If ${p}\ne{1}$ then ${\int_{{1}}^{{\infty}}}\frac{{1}}{{{x}}^{{p}}}{d}{x}=\lim_{{{t}\to\infty}}{\int_{{1}}^{{t}}}\frac{{1}}{{{x}}^{{p}}}{d}{x}=\lim_{{{t}\to\infty}}{\left(\frac{{1}}{{-{p}+{1}}}{{x}}^{{-{p}+{1}}}{{\mid}_{{1}}^{{t}}}\right)}=\frac{{1}}{{{1}-{p}}}\lim_{{{t}\to\infty}}{\left(\frac{{1}}{{{t}}^{{{p}-{1}}}}-{1}\right)}$.

If ${p}>{1}$ then ${p}-{1}>{0}$ and $\frac{{1}}{{{t}}^{{{p}-{1}}}}\to{0}$ as ${t}\to\infty$. Therefore, ${\int_{{1}}^{\infty}}\frac{{1}}{{{x}}^{{p}}}{d}{x}=\frac{{1}}{{{p}-{1}}}$ when p>1.

If ${p}<{1}$ then ${p}-{1}<{0}$ and $\frac{{1}}{{{t}}^{{{p}-{1}}}}\to\infty$ as ${t}\to\infty$. Therefore, integral is divergent.

Therefore, ${\int_{{1}}^{{\infty}}}\frac{{1}}{{{x}}^{{p}}}{d}{x}$ is convergent for p>1.

Note the interesting fact: although the curves ${y}=\frac{{1}}{{{x}}^{{2}}}$ and ${y}=\frac{{1}}{{x}}$ look very similar for x>0, the region under ${y}=\frac{{1}}{{{x}}^{{2}}}$ to the right of x=1 has finite area whereas the corresponding region under ${y}=\frac{{1}}{{x}}$ has infinite area. Note that both $\frac{{1}}{{{x}}^{{2}}}$ and $\frac{{1}}{{x}}$ approach 0 as ${x}\to\infty$ but $\frac{{1}}{{{x}}^{{2}}}$ approaches faster than $\frac{{1}}{{x}}$. The values of $\frac{{1}}{{x}}$ don't decrease fast enough for its integral to have a finite value.

Example 2. Evaluate ${\int_{{-\infty}}^{{-{1}}}}{{e}}^{{{2}{x}}}{d}{x}$.

By definition ${\int_{{-\infty}}^{{-{1}}}}{{e}}^{{{2}{x}}}{d}{x}=\lim_{{{t}\to-\infty}}{\int_{{t}}^{{-{1}}}}{{e}}^{{{2}{x}}}{d}{x}=\lim_{{{t}\to-\infty}}{\left(\frac{{1}}{{2}}{{e}}^{{{2}{x}}}{{\mid}_{{t}}^{{-{1}}}}\right)}=\lim_{{{t}\to-\infty}}{\left(\frac{{1}}{{2}}{{e}}^{{-{2}}}-\frac{{1}}{{2}}{{e}}^{{{2}{t}}}\right)}=$

$=\frac{{1}}{{2}}{{e}}^{{-{2}}}-{0}=\frac{{1}}{{{2}{{e}}^{{2}}}}$.

Example 3. Calculate if possible ${\int_{{-\infty}}^{{\infty}}}\frac{{1}}{{{{r}}^{{2}}+{4}}}{d}{r}$.

It is convenient to take ${a}={0}$ in definition.

${\int_{{-\infty}}^{{\infty}}}\frac{{1}}{{{{r}}^{{2}}+{4}}}{d}{r}={\int_{{-\infty}}^{{0}}}\frac{{1}}{{{{r}}^{{2}}+{4}}}{d}{r}+{\int_{{0}}^{{\infty}}}\frac{{1}}{{{{r}}^{{2}}+{4}}}{d}{r}$.

Now ${\int_{{-\infty}}^{{0}}}\frac{{1}}{{{{r}}^{{2}}+{4}}}{d}{r}=\lim_{{{t}\to-\infty}}{\int_{{t}}^{{0}}}\frac{{1}}{{{{r}}^{{2}}+{4}}}{d}{r}=\lim_{{{t}\to-\infty}}{\left(\frac{{1}}{{2}}{{\tan}}^{{-{1}}}{\left(\frac{{r}}{{2}}\right)}{{\mid}_{{t}}^{{0}}}\right)}=\lim_{{{t}\to-\infty}}{\left(\frac{{1}}{{2}}{{\tan}}^{{-{1}}}{\left(\frac{{0}}{{2}}\right)}-\frac{{1}}{{2}}{{\tan}}^{{-{1}}}{\left(\frac{{t}}{{2}}\right)}\right)}=$

$=\lim_{{{t}\to-\infty}}{\left({0}-\frac{{1}}{{2}}{{\tan}}^{{-{1}}}{\left(\frac{{t}}{{2}}\right)}\right)}={0}-{\left(\frac{{1}}{{2}}\cdot{\left(-\frac{\pi}{{2}}\right)}\right)}=\frac{\pi}{{4}}$.

Similarly ${\int_{{0}}^{{\infty}}}\frac{{1}}{{{{r}}^{{2}}+{4}}}{d}{r}=\lim_{{{t}\to\infty}}{\int_{{0}}^{{t}}}\frac{{1}}{{{{r}}^{{2}}+{4}}}{d}{r}=\lim_{{{t}\to\infty}}{\left(\frac{{1}}{{2}}{{\tan}}^{{-{1}}}{\left(\frac{{r}}{{2}}\right)}{{\mid}_{{0}}^{{t}}}\right)}=\lim_{{{t}\to-\infty}}{\left(\frac{{1}}{{2}}{{\tan}}^{{-{1}}}{\left(\frac{{t}}{{2}}\right)}-\frac{{1}}{{2}}{{\tan}}^{{-{1}}}{\left(\frac{{0}}{{2}}\right)}\right)}=$

$=\lim_{{{t}\to\infty}}{\left(\frac{{1}}{{2}}{{\tan}}^{{-{1}}}{\left(\frac{{t}}{{2}}\right)}-{0}\right)}=\frac{{1}}{{2}}\cdot\frac{\pi}{{2}}-{0}=\frac{\pi}{{4}}$.

Since both of these integrals are convergent, the given integral is convergent and ${\int_{{-\infty}}^{{\infty}}}\frac{{1}}{{{{r}}^{{2}}+{4}}}{d}{r}=\frac{\pi}{{4}}+\frac{\pi}{{4}}=\frac{\pi}{{2}}$.

Since $\frac{{1}}{{{{x}}^{{2}}+{4}}}>{0}$, the given improper integral can be interpreted as the area of the infinite region that lies under the curve ${y}=\frac{{1}}{{{{x}}^{{2}}+{4}}}$ and above the x-axis.

Example 4. Find ${\int_{{-\infty}}^{{\infty}}}\frac{{1}}{{x}}{d}{x}$.

Here we will choose a=1.

${\int_{{-\infty}}^{{\infty}}}\frac{{1}}{{x}}{d}{x}={\int_{{-\infty}}^{{1}}}\frac{{1}}{{x}}{d}{x}+{\int_{{1}}^{{\infty}}}\frac{{1}}{{x}}{d}{x}$.

It was shown in example 1 that ${\int_{{1}}^{{\infty}}}\frac{{1}}{{x}}{d}{x}$ is divergent, so we don't need to calculate second integral. Since at least one integral is divergent then ${\int_{{\infty}}^{{-\infty}}}\frac{{1}}{{x}}{d}{x}$ is also divergent.