Does $\displaystyle \int_0^{\infty} \frac{e^x \sin{x}}{x} \, dx \tag*{}$ converge?
Since $\lim_{x \to 0} \frac{\sin{x}}{x} = 1$, the integrand has a removable discontinuity at $x = 0$. Hence, we only need to investigate convergence of the integral at $\infty$. Noting that $\sin{x}$ attains positive and negative values, we break up the integral accordingly. $\begin{align*} \displaystyle \int_0^{\infty} \frac{e^x \sin{x}}{x} \, dx &= \sum_{n=0}^{\infty} \int_{n\pi}^{(n+1)\pi} \frac{e^x \sin{x}}{x} \, dx\\ &= \displaystyle \sum_{n=0}^{\infty} \int_0^{\pi} \frac{e^{t + n \pi} \sin(t + n \pi)}{t + n \pi} \, dt, \text{ via } x = t + n \pi\\ &= \displaystyle \sum_{n=0}^{\infty} (-1)^n e^{n \pi} \int_0^{\pi} \frac{e^t \sin{t}}{t + n \pi} \, dt. \end{align*} \tag*{}$ The upshot of the calculation above is that we have rewritten the improper integral in question as an alternating series (in which the integrals are each positive and finite)! Now, we apply the Alternating Series Test to show that the series (and thus the improper integral) diverges. In particular, we show that $\displaystyle \lim_{n \to \infty} e^{n \pi} \int_0^{\pi} \frac{e^t \sin{t}}{t + n \pi} \, dt = \infty. \tag*{}$ To see this, note that since $t \geq 0$, $\displaystyle e^{n \pi} \int_0^{\pi} \frac{e^t \sin{t}}{t + n \pi} \, dt \geq e^{n \pi} \int_0^{\pi} \frac{e^t \sin{t}}{0 + n \pi} \, dt = \frac{e^{n \pi}}{n \pi} J, \tag*{}$ where $J = \int_0^{\pi} e^t \sin{t} \, dt$ is a constant. Since by L’Hopital’s Rule we have $\displaystyle \lim_{n \to \infty} \frac{e^{n \pi}}{n \pi} J = \infty, \tag*{}$ the desired claim about the limit follows.
Hence, we conclude that $\displaystyle \int_0^{\infty} \frac{e^x \sin{x}}{x} \, dx \tag*{}$ is divergent.
I get the opposite and Wolfram alpha too. Excuse my poor English. Many thanks.
