Evaluating Improper Integrals Examples 1

We will now look at some examples of evaluating improper integrals. Recall that:

(1)
\begin{align} \int_a^{\infty} f(x) \: dx = \lim_{b \to \infty} \int_a^b f(x) \: dx \end{align}
(2)
\begin{align} \int_{-\infty}^b f(x) \: dx = \lim_{a \to -\infty} \int_a^f f(x) \: dx \end{align}

Also recall that if we a function f that is continuous on the interval [a, b) and is discontinuous at b, then:

(3)
\begin{align} \int_a^b f(x) \: dx = \lim_{t \to b^-} \int_a^t f(x) \: dx \end{align}

Similarly, if we have a function that is discontinous at a and is continuous on the interval (a, b], then:

(4)
\begin{align} \int_a^b f(x) \: dx = \lim_{t \to a^+} \int_t^b f(x) \: dx \end{align}

## Example 1

Evaluate the integral $\int_0^3 \frac{1}{x - 1} \: dx$.

We first notice that this function is NOT continuous on the entire interval [0, 3]. Instead, we should notice that there is an asymptote precisely when x = 1, and that 0 ≤ 1 ≤ 3. Hence it follows that:

(5)
\begin{align} \quad \int_0^3 \frac{1}{x - 1} \: dx = \lim_{t \to 1^-} \int_0^1 \frac{1}{x - 1} \: dx + \lim_{t \to 1^+} \int_1^3 \frac{1}{x - 1} \: dx \\ \end{align}

Now let's evaluate the first limit:

(6)
\begin{align} \lim_{t \to 1^-} \int_0^1 \frac{1}{x - 1} \: dx = \lim_{t \to 1^-} [ \ln | t - 1 | - \ln | - 1| ] \\ \lim_{t \to 1^-} \int_0^1 \frac{1}{x - 1} \: dx = \lim_{t \to 1^-} \ln | 1 - t | \\ \lim_{t \to 1^-} \int_0^1 \frac{1}{x - 1} \: dx = - \infty \end{align}

Since one of our limits does not exist, then our original integral must be divergent. The graph below illustrates that area that we attempted to evaluate:

## Example 2

Evaluate the integral $\int_0^1 \ln x \: dx$.

Once again we notice that there exists a vertical asymptote at x = 0. Hence it follows that:

(7)
\begin{align} \int_0^1 \ln x \: dx = \lim_{t \to 0^+} \int_{t}^1 \ln x \: dx \\ \int_0^1 \ln x \: dx = \lim_{t \to 0^+} [ t \ln t - 1 + t ] \end{align}

We must now apply L'Hospital's Rule to evaluate the first term of this limit as follows:

(8)
\begin{align} \lim_{t \to 0^+} t \ln t = \lim_{t \to 0^+} \frac{\ln t}{1/t} = \lim_{t \to 0^+} \frac{1/t}{-1/t^2} = \lim{-t} = 0 \end{align}

Hence it follows that:

(9)
\begin{align} \int_0^1 \ln x \: dx = \lim_{t \to 0^+} [ t \ln t - 1 + t ] = -1 \end{align}

So $\int_0^1 \ln x \: dx = -1$, as illustrated below: