Determine whether the following improper integrals converge or diverge:

Solution 
(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.