# 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(t)=int_a^t f(x)dx.

If it happens that A(t) approaches a definite number as t->b^-, then we say that the area of the region S is A and we write A=lim_(t->b^-)int_a^t f(x)dx. 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(x)dx=lim_(t->b^-)int_a^t f(x)dx 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(x)dx=lim_(t->a^+)int_t^b f(x)dx if this limit exists (as a finite number).
3. The improper integral int_a^b f(x)dx 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(x)dx and int_c^bf(x)dx are convergent, then int_a^b f(x)dx=int_a^c f(x)dx+int_c^b f(x)dx. Only when BOTH integrals are convergent, convergent is int_a^b f(x)dx. Otherwise it is divergent.

Example 1. Find int_3^7 1/sqrt(x-3)dx.

Integral is improper because f(x)=1/sqrt(x-3) has infinite discontinuity at x=3.

Thus, int_3^7 1/sqrt(x-3) dx=lim_(t->3^+) int_t^7 1/sqrt(x-3)dx=lim_(t->3^+)(2sqrt(x-3)|_t^7)=2lim_(t->3^+)(sqrt(7-3)-sqrt(t-3))=

=2(sqrt(7-3)-sqrt(t-3))=4.

Example 2. Find int_0^(pi/2)tan(x)dx if possible.

Integral is divergent because lim_(t->(pi/2)^-)tan(x)=oo.

Therefore, int_0^(pi/2)tan(x)dx=lim_(t->(pi/2)^-)int_0^t tan(x)dx=lim_(t->(pi/2)^-)(-ln|cos(x)||_0^t)=

=lim_(t->(pi/2)^-)(-ln|cos(t)|+ln|cos(0)|)=lim_(t->(pi/2)^-)(-ln|cos(t)|)=oo.

Thus, integral is divergent.

Example 3. Evaluate int_0^3 1/(x-1)dx.

This integral is improper because x=1 is a vertical asymptote. Thus, int_0^3 1/(x-1)dx=int_0^1 1/(x-1)dx+int_1^3 1/(x-1)dx.

int_0^1 1/(x-1)dx=lim_(t->1^-)int_0^t 1/(x-1)dx=lim_(t->1^-)(ln|x-1||_0^t)=lim_(t->1^-)(ln|t-1|-ln|0-1|)=

=lim_(t->1^-)ln|t-1|=-oo.

Therefore, int_0^1 1/(x-1)dx is divergent. This implies that int_0^3 1/(x-1)dx 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 1/(x-1)dx=ln|x-1||_0^3=ln|3-1|-ln|0-1|=ln(2).

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^(oo) 1/x^3dx.

Note that f(x)=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^(oo) 1/x^3dx=int_0^1 1/x^3 dx+int_1^(oo) 1/x^3 dx.

Handle first integral: int_0^1 1/x^3dx=lim_(t->0^+)int_t^1 1/x^3 dx=lim_(t->0^+)(-1/2 1/x^2|_t^1)=-1/2 lim_(t->0^+)(1/1^2-1/t^2)=

=1/2 lim_(t->0^+)(1/t^2-1)=oo.

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