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Type I (Infinite Intervals)

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When we defined definite integral ∫abf(x)dx{\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(x){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(t)=∫atf(x)dx=F(t)−F(a){A}{\left({t}\right)}={\int_{{a}}^{{t}}}{f{{\left({x}\right)}}}{d}{x}={F}{\left({t}\right)}-{F}{\left({a}\right)} where F{F} is antiderivative of f.

For example, if f(x)=1x2{f{{\left({x}\right)}}}=\frac{{1}}{{{x}}^{{2}}} and a=1{a}={1} then A(t)=∫1t1x2dx=1−1t{A}{\left({t}\right)}={\int_{{1}}^{{t}}}\frac{{1}}{{{x}}^{{2}}}{d}{x}={1}-\frac{{1}}{{t}}.

Observe that in this case lim⁡t→∞A(t)=lim⁡t→∞(1−1t)=1\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 ∫atf(x)dx{\int_{{a}}^{{t}}}{f{{\left({x}\right)}}}{d}{x} exists for every number t≥a{t}\ge{a}, then ∫a∞f(x)dx=lim⁡t→∞∫atf(x)dx{\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 ∫tbf(x)dx{\int_{{t}}^{{b}}}{f{{\left({x}\right)}}}{d}{x} exists for every number t≤b{t}\le{b}, then ∫−∞bf(x)dx=lim⁡t→∞∫tbf(x)dx{\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 ∫a∞f(x)dx{\int_{{a}}^{{\infty}}}{f{{\left({x}\right)}}}{d}{x} and ∫−∞bf(x)dx{\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 ∫a∞f(x)dx{\int_{{a}}^{{\infty}}}{f{{\left({x}\right)}}}{d}{x} and ∫−∞af(x)dx{\int_{{-\infty}}^{{a}}}{f{{\left({x}\right)}}}{d}{x} are convergent then ∫−∞∞f(x)dx=∫−∞af(x)dx+∫a∞f(x)dx{\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 ∫−∞∞f(x)dx{\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 ∫1∞1xpdx{\int_{{1}}^{{\infty}}}\frac{{1}}{{{x}}^{{p}}}{d}{x} convergent?

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

If p=1 then ∫1∞1xdx=lim⁡t→∞∫1t1xdx=lim⁡t→∞(ln⁡∣x∣∣1t)=lim⁡t→∞(ln⁡(t)−ln⁡(1))=∞{\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≠1{p}\ne{1} then ∫1∞1xpdx=lim⁡t→∞∫1t1xpdx=lim⁡t→∞(1−p+1x−p+1∣1t)=11−plim⁡t→∞(1tp−1−1){\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{p}>{1} then p−1>0{p}-{1}>{0} and 1tp−1→0\frac{{1}}{{{t}}^{{{p}-{1}}}}\to{0} as t→∞{t}\to\infty. Therefore, ∫1∞1xpdx=1p−1{\int_{{1}}^{\infty}}\frac{{1}}{{{x}}^{{p}}}{d}{x}=\frac{{1}}{{{p}-{1}}} when p>1.

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

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

Note the interesting fact: although the curves y=1x2{y}=\frac{{1}}{{{x}}^{{2}}} and y=1x{y}=\frac{{1}}{{x}} look very similar for x>0, the region under y=1x2{y}=\frac{{1}}{{{x}}^{{2}}} to the right of x=1 has finite area whereas the corresponding region under y=1x{y}=\frac{{1}}{{x}} has infinite area. Note that both 1x2\frac{{1}}{{{x}}^{{2}}} and 1x\frac{{1}}{{x}} approach 0 as x→∞{x}\to\infty but 1x2\frac{{1}}{{{x}}^{{2}}} approaches faster than 1x\frac{{1}}{{x}}. The values of 1x\frac{{1}}{{x}} don't decrease fast enough for its integral to have a finite value.

Example 2. Evaluate ∫−∞−1e2xdx{\int_{{-\infty}}^{{-{1}}}}{{e}}^{{{2}{x}}}{d}{x}.

By definition ∫−∞−1e2xdx=lim⁡t→−∞∫t−1e2xdx=lim⁡t→−∞(12e2x∣t−1)=lim⁡t→−∞(12e−2−12e2t)={\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)}=

=12e−2−0=12e2=\frac{{1}}{{2}}{{e}}^{{-{2}}}-{0}=\frac{{1}}{{{2}{{e}}^{{2}}}}.

Example 3. Calculate if possible ∫−∞∞1r2+4dr{\int_{{-\infty}}^{{\infty}}}\frac{{1}}{{{{r}}^{{2}}+{4}}}{d}{r}.

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

∫−∞∞1r2+4dr=∫−∞01r2+4dr+∫0∞1r2+4dr{\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 ∫−∞01r2+4dr=lim⁡t→−∞∫t01r2+4dr=lim⁡t→−∞(12tan⁡−1(r2)∣t0)=lim⁡t→−∞(12tan⁡−1(02)−12tan⁡−1(t2))={\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→−∞(0−12tan⁡−1(t2))=0−(12⋅(−π2))=π4=\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 ∫0∞1r2+4dr=lim⁡t→∞∫0t1r2+4dr=lim⁡t→∞(12tan⁡−1(r2)∣0t)=lim⁡t→−∞(12tan⁡−1(t2)−12tan⁡−1(02))={\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→∞(12tan⁡−1(t2)−0)=12⋅π2−0=π4=\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 ∫−∞∞1r2+4dr=π4+π4=π2{\int_{{-\infty}}^{{\infty}}}\frac{{1}}{{{{r}}^{{2}}+{4}}}{d}{r}=\frac{\pi}{{4}}+\frac{\pi}{{4}}=\frac{\pi}{{2}}.

Since 1x2+4>0\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=1x2+4{y}=\frac{{1}}{{{{x}}^{{2}}+{4}}} and above the x-axis.

Example 4. Find ∫−∞∞1xdx{\int_{{-\infty}}^{{\infty}}}\frac{{1}}{{x}}{d}{x}.

Here we will choose a=1.

∫−∞∞1xdx=∫−∞11xdx+∫1∞1xdx{\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 ∫1∞1xdx{\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 ∫∞−∞1xdx{\int_{{\infty}}^{{-\infty}}}\frac{{1}}{{x}}{d}{x} is also divergent.