# 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**

- 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).
- 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).
- 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. - 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.