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Type I (Infinite Intervals)

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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.