Determine whether the following improper integrals converge or diverge:
(a) \(\int_{1}^{\infty }\dfrac{1}{x}\,dx{:} \lim\limits_{b\,\rightarrow \,\infty }\int_{1}^{b}\dfrac{1}{x} \,dx=\lim\limits_{b\,\rightarrow \,\infty }\big[\ln \,\vert x \vert \big] _{1}^{b}=\lim\limits_{b\,\rightarrow \,\infty }[ \ln b-\ln 1] = \infty\).
The limit is infinite, so \(\int_{1}^{\infty }\dfrac{1}{x}\,dx\) diverges.
(b) \(\int_{-\infty }^{0} e^{x}\,dx: \lim\limits_{a\, \rightarrow \,-\infty }\int_{a}^{0}e^{x}\,dx=\lim\limits_{a\,\rightarrow \,-\infty }\big[e^{x}\big]_{a}^{0}=\lim\limits_{a\,\rightarrow \,-\infty }(1-e^{a}) = 1-0=1.\) Since the limit exists, \(\int_{-\infty }^{0}e^{x}\,dx\) converges and equals \(1\).
(c) \[ \begin{eqnarray*} \int_{\pi/2}^{\infty }\sin x \,dx: \lim\limits_{b\,\rightarrow \,\infty }\int_{\pi/2}^{b}\sin x\, dx &=& \lim\limits_{b\,\rightarrow \,\infty }\,\big[ -\cos x \big]^b_{\pi/2}\\ &=& \lim\limits_{b\,\rightarrow \,\infty } [ - \cos b +0 ] \\ &=& - \lim\limits_{b\,\rightarrow \,\infty }\cos b \end{eqnarray*} \]
This limit does not exist, since as \(b\rightarrow \infty\), the value of \(\cos b\) oscillates between \(-1\) and \(1\). So, \( \int_{\pi/2}^{\infty }\sin x \,dx\) diverges.