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Section Improper Integrals
There are two types of improper integrals; the only type weβll be looking at in differential equations are the type where the upper limit of the integral is infinity. When we see improper integrals, we re-write the integral as a limit as follows.
\begin{equation*}
\int_0^{\infty} f(x)dx = \lim_{b \to \infty} \int_0^b f(x) dx
\end{equation*}
We donβt have to use \(b\) for the placeholder in the limit, but we shouldnβt use any of the variables that is already in the integrand; in the above, we should not use \(x\) or \(f\text{.}\) Once weβve written the improper integral as a limit, we integrate and use the fundamental theorem of calculus, keeping the limit as we go, and we evaluate the limit at the very end. Hereβs an example.
π Example 326 . Compute \(\int_1^{\infty} \frac{1}{x^2}dx\) .
\begin{align*}
\int_1^{\infty} \frac{1}{x^2}dx
\amp = \lim_{b \to \infty} \int_1^b \frac{1}{x^2}dx \\
\amp = \lim_{b \to \infty} \int_1^b x^{-2} dx \\
\amp = \lim_{b \to \infty} \Bigg[ -x^{-1} \Bigg]_1^b \\
\amp = \lim_{b \to \infty} \left[ -\frac{1}{x} \right]_1^b \\
\amp = \lim_{b \to \infty} \left[ -\frac{1}{b}
- \left(-\frac{1}{1}\right) \right] \\
\amp = \lim_{b \to \infty} \left[ -\frac{1}{b} +1 \right] \\
\amp = \lim_{b \to \infty} \left[ -\frac{1}{b}\right]
+ \lim_{b \to \infty} 1 \\
\amp = 0 + 1 \\
\amp = 1
\end{align*}
If you want more of a refresher, check out your Calculus 2 book (available online-- just search for "APEX calculus").
Evaluate the following integrals. Use proper limit notation.
\(\ds \int_0^{\infty}e^{-3t}dt \qquad\) Solution .
\begin{align*}
\int_0^{\infty}e^{-3t}dt
\amp = \lim_{b \to \infty}\int_0^b e^{-3t}dt \\
\amp = \lim_{b \to \infty}\left[-\frac{1}{3}e^{-3t} \right]_0^b \\
\amp = \lim_{b \to \infty}\left[-\frac{1}{3}e^{-3b}
+ \frac{1}{3}e^{-3\cdot 0} \right] \\
\amp = \lim_{b \to \infty}\left[-\frac{1}{3}e^{-3b}
+ \frac{1}{3}\cdot 1 \right] \\
\amp = \lim_{b \to \infty}\left[-\frac{1}{3}e^{-3b}\right]
+ \lim_{b \to \infty}\left[\frac{1}{3} \right] \\
\amp = -\frac{1}{3} \lim_{b \to \infty}\left[e^{-3b}\right]
+ \frac{1}{3} \\
\amp = -\frac{1}{3} \cdot 0+ \frac{1}{3} \\
\amp = \frac{1}{3}
\end{align*}
Answer .
\(\ds \int_0^{\infty}e^{-st}dt\text{,}\) where
\(s\) is a constant and
\(s>0 \qquad\) Solution .
\begin{align*}
\int_0^{\infty}e^{-st}dt
\amp = \lim_{b \to \infty}\int_0^b e^{-st}dt \\
\amp = \lim_{b \to \infty}\left[-\frac{1}{s}e^{-st} \right]_0^b \\
\amp = \lim_{b \to \infty}\left[-\frac{1}{s}e^{-sb}
+ \frac{1}{s}e^{-s\cdot 0} \right] \\
\amp = \lim_{b \to \infty}\left[-\frac{1}{s}e^{-sb}
+ \frac{1}{s}\cdot 1 \right] \\
\amp = \lim_{b \to \infty}\left[-\frac{1}{s}e^{-sb}\right]
+ \lim_{b \to \infty}\left[\frac{1}{s} \right] \\
\amp = -\frac{1}{s} \lim_{b \to \infty}\left[e^{-sb}\right]
+ \frac{1}{s} \\
\amp = -\frac{1}{s} \cdot 0+ \frac{1}{3} \\
\amp = \frac{1}{s}
\end{align*}
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