Definition of Improper Integrals
Improper Integrals are of two types. We can define Improper Integral of type 1 as an improper integral if the interval of integration is not finite (infinite). Improper Integral of type 2 can be defined as an improper integral if the function to integrate is not continuous in the interval of integration. For example integral (zero to infinity) e^-x is an improper integral of type 1 as the upper limit of integration is infinite and integral (zero to one) dx/x is an improper integral of type 2 because 1/x is not continuous at zero.
Evaluating Improper Integrals
Let us consider some Improper Integral examples to understand the evaluation of Improper Integral.
Evaluate Integral (1 to infinity) dx/x^2. This is a type 1 improper integral, we evaluate it by finding lim(t to infinity) Integral (1 to t)dx/x^2. Here, we need to evaluate the integral part first, integral (1 to t) dx/x^2 which gives [1-1/t] and then we apply the limit. lim(t->infinity)[1-1/t]=1 and hence finally we get,
Integral(1 to infinity) dx/x^2=1
Evaluate the Integral (-infinity to 0)2 ^(5x)dx
Solution: Integral (-infinity to 0)2 ^(5x)dx = lim(n->infinity) integral (-n to 0)2 ^(5x) dx
First we shall evaluate the integral part, which gives
Lim(n->infinity)[1/5(2^(5x)/ln 2)][ -n to 0]
= 2^5(0)/5ln 2 – lim(n->infinity) [2^5(-n)/5 ln 2]
Finally we get, 1/5 ln 2 as the answer
Improper Integral Problems
While Solving Improper integrals, there are essentially three cases which are considered,
1. If Integral(a to t)f(x)dx exists for every t > a, then Integral(a to infinity) f(x) dx = lim(t->infinity)Integral (a to t) f(x) dx, provided the limit exists and is finite
2. If Integral(t to b)f(x) dx exists for every t>b, then, Integral (-infinity to b)f(x)dx = lim(t->inifity)Integral(t to b) f(x)dx, provided the limit exists and is finite
3. If Integral(-infinity to c)f(x) dx and Integral (c to infinity)f(x) dx are both convergent, then,
Integral(-infinity to infinity)f(x)dx = Integral(-infinity to c)f(x)dx + Integral(c to infinity) f(x)dx (where c is any number) In this case we need to note that both of the integrals are to be convergent for this integral to be convergent. If either of the two integrals is divergent then so is this integral
Evaluate Integral (0 to pi) sec^2 (x) dx. This is a type 2 improper integral, sec^2(x) is not continuous at pi/2. Integral(0 to pi)sec^2(x) dx = integral(0 to pi/2)sec^2(x)dx + integral(pi/2 to pi)sec^2(x) dx. We evaluate first integral(0 to pi/2) sec^2(x) dx which gives, lim (t ->pi/2 -)integral(0 to t) sec^2(x)dx
Integral(0 to t)sec^2(x)dx= tant – tan 0= tan t. Therefore, we get, integral (0 to pi/2) sec^2(x)dx = lim(t->pi/2 -)[tan t] = infinity. It follows that integral(0 to pi/2) sec^2(x) dx diverges and hence integral (0 to pi) sec^2(x)dx also diverges.
