# Type II (Discontinuous Integrands)

## Related calculator: Definite and Improper Integral Calculator

Suppose that f is a positive continuous function defined on a finite interval [a,b) but has a vertical asymptote at b. Let S be the unbounded region under the graph of f and above the x-axis between a and b. (For Type 1 integrals, the regions extended indefinitely in a horizontal direction. Here the region is infinite in a vertical direction.) The area of the part of S between a and t is ${A}{\left({t}\right)}={\int_{{a}}^{{t}}}{f{{\left({x}\right)}}}{d}{x}$.

If it happens that A(t) approaches a definite number as ${t}\to{{b}}^{{-{}}}$, then we say that the area of the region S is A and we write ${A}=\lim_{{{t}\to{{b}}^{{-}}}}{\int_{{a}}^{{t}}}{f{{\left({x}\right)}}}{d}{x}$. We use this equation to define an improper integral of Type 2 even when f is not a positive function, no matter what type of discontinuity f has at b.

Definition of an Improper Integral of Type 2

1. If is continuous on [a,b) and is discontinuous at b, then ${\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}=\lim_{{{t}\to{{b}}^{{-}}}}{\int_{{a}}^{{t}}}{f{{\left({x}\right)}}}{d}{x}$ if this limit exists (as a finite number).
2. If is continuous on (a,b] and is discontinuous at a, then ${\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}=\lim_{{{t}\to{{a}}^{+}}}{\int_{{t}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$ if this limit exists (as a finite number).
3. The improper integral ${\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$ is called convergent if the corresponding limit exists and divergent if the limit does not exist.
4. If f has a discontinuity at c, where a<c<b, and both ${\int_{{a}}^{{c}}}{f{{\left({x}\right)}}}{d}{x}$ and ${\int_{{c}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$ are convergent, then ${\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}={\int_{{a}}^{{c}}}{f{{\left({x}\right)}}}{d}{x}+{\int_{{c}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$. Only when BOTH integrals are convergent, convergent is ${\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{d}{x}$. Otherwise it is divergent.

Example 1. Find ${\int_{{3}}^{{7}}}\frac{{1}}{\sqrt{{{x}-{3}}}}{d}{x}$.

Integral is improper because ${f{{\left({x}\right)}}}=\frac{{1}}{\sqrt{{{x}-{3}}}}$ has infinite discontinuity at ${x}={3}$.

Thus, ${\int_{{3}}^{{7}}}\frac{{1}}{\sqrt{{{x}-{3}}}}{d}{x}=\lim_{{{t}\to{{3}}^{+}}}{\int_{{t}}^{{7}}}\frac{{1}}{\sqrt{{{x}-{3}}}}{d}{x}=\lim_{{{t}\to{{3}}^{+}}}{\left({2}\sqrt{{{x}-{3}}}{{\mid}_{{t}}^{{7}}}\right)}={2}\lim_{{{t}\to{{3}}^{+}}}{\left(\sqrt{{{7}-{3}}}-\sqrt{{{t}-{3}}}\right)}=$

$={2}{\left(\sqrt{{{7}-{3}}}-\sqrt{{{t}-{3}}}\right)}={4}$.

Example 2. Find ${\int_{{0}}^{{\frac{\pi}{{2}}}}}{\tan{{\left({x}\right)}}}{d}{x}$ if possible.

Integral is divergent because $\lim_{{{t}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}{\tan{{\left({x}\right)}}}=\infty$.

Therefore, ${\int_{{0}}^{{\frac{\pi}{{2}}}}}{\tan{{\left({x}\right)}}}{d}{x}=\lim_{{{t}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}{\int_{{0}}^{{t}}}{\tan{{\left({x}\right)}}}{d}{x}=\lim_{{{t}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}{\left(-{\ln}{\left|{\cos{{\left({x}\right)}}}\right|}{{\mid}_{{0}}^{{t}}}\right)}=$

$=\lim_{{{t}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}{\left(-{\ln}{\left|{\cos{{\left({t}\right)}}}\right|}+{\ln}{\left|{\cos{{\left({0}\right)}}}\right|}\right)}=\lim_{{{t}\to{{\left(\frac{\pi}{{2}}\right)}}^{{-}}}}{\left(-{\ln}{\left|{\cos{{\left({t}\right)}}}\right|}\right)}=\infty$.

Thus, integral is divergent.

Example 3. Evaluate ${\int_{{0}}^{{3}}}\frac{{1}}{{{x}-{1}}}{d}{x}$.

This integral is improper because x=1 is a vertical asymptote. Thus, ${\int_{{0}}^{{3}}}\frac{{1}}{{{x}-{1}}}{d}{x}={\int_{{0}}^{{1}}}\frac{{1}}{{{x}-{1}}}{d}{x}+{\int_{{1}}^{{3}}}\frac{{1}}{{{x}-{1}}}{d}{x}$.

${\int_{{0}}^{{1}}}\frac{{1}}{{{x}-{1}}}{d}{x}=\lim_{{{t}\to{{1}}^{{-}}}}{\int_{{0}}^{{t}}}\frac{{1}}{{{x}-{1}}}{d}{x}=\lim_{{{t}\to{{1}}^{{-}}}}{\left({\ln}{\left|{x}-{1}\right|}{{\mid}_{{0}}^{{t}}}\right)}=\lim_{{{t}\to{{1}}^{{-}}}}{\left({\ln}{\left|{t}-{1}\right|}-{\ln}{\left|{0}-{1}\right|}\right)}=$

$=\lim_{{{t}\to{{1}}^{{-}}}}{\ln}{\left|{t}-{1}\right|}=-\infty$.

Therefore, ${\int_{{0}}^{{1}}}\frac{{1}}{{{x}-{1}}}{d}{x}$ is divergent. This implies that ${\int_{{0}}^{{3}}}\frac{{1}}{{{x}-{1}}}{d}{x}$ is divergent.

WARNING. If we had not noticed the asymptote in above example and had instead confused the integral with an ordinary integral, then we might have made the following erroneous calculation: ${\int_{{0}}^{{3}}}\frac{{1}}{{{x}-{1}}}{d}{x}={\ln}{\left|{x}-{1}\right|}{\left|_{{0}}^{{3}}={\ln}\right|}{3}-{1}{\left|-{\ln}\right|}{0}-{1}{\mid}={\ln{{\left({2}\right)}}}$.

This is wrong because the integral is improper and must be calculated in terms of limits.

Now, let's see how to work with integrals that belong to both types.

Example 4. Find if possible ${\int_{{0}}^{{\infty}}}\frac{{1}}{{{x}}^{{3}}}{d}{x}$.

Note that ${f{{\left({x}\right)}}}=\frac{{1}}{{{x}}^{{3}}}$ has discontinuity at x=0 and also interval is infinite. So, this integral belongs to both types. To handle it, we split it into 2 integrals. We can split it up anywhere, but pick a value that will be convenient for evaluation purposes.

${\int_{{0}}^{{\infty}}}\frac{{1}}{{{x}}^{{3}}}{d}{x}={\int_{{0}}^{{1}}}\frac{{1}}{{{x}}^{{3}}}{d}{x}+{\int_{{1}}^{{\infty}}}\frac{{1}}{{{x}}^{{3}}}{d}{x}$.

Handle first integral: ${\int_{{0}}^{{1}}}\frac{{1}}{{{x}}^{{3}}}{d}{x}=\lim_{{{t}\to{{0}}^{+}}}{\int_{{t}}^{{1}}}\frac{{1}}{{{x}}^{{3}}}{d}{x}=\lim_{{{t}\to{{0}}^{+}}}{\left(-\frac{{1}}{{2}}\frac{{1}}{{{x}}^{{2}}}{{\mid}_{{t}}^{{1}}}\right)}=-\frac{{1}}{{2}}\lim_{{{t}\to{{0}}^{+}}}{\left(\frac{{1}}{{{1}}^{{2}}}-\frac{{1}}{{{t}}^{{2}}}\right)}=$

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

So, integral is divergent and so the whole integral is also divergent.