Question Number 79913 by Henri Boucatchou last updated on 29/Jan/20
$$\:{Convergence}\:\:{of}\:\:{I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{{t}} }{{e}^{−{t}} +{e}^{\mathrm{2}{t}} \mid{sint}\mid}{dt} \\ $$
Commented by mathmax by abdo last updated on 29/Jan/20
$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} }{{e}^{−\mathrm{3}{t}} \:+\mid{sint}\mid}{dt}\:=\int_{\mathrm{0}} ^{+\infty} \:\frac{{e}^{−{t}} }{{e}^{−\mathrm{3}{t}} \:+\mid{sint}\mid}{dt}\:\:\:{we}\:{have} \\ $$$$\mid{sint}\mid\leqslant\mathrm{1}\:\Rightarrow{e}^{−\mathrm{3}{t}} \:+\mid{sint}\mid\leqslant\mathrm{1}+{e}^{−\mathrm{3}{t}} \:\Rightarrow\frac{\mathrm{1}}{{e}^{−\mathrm{3}{t}} +\mid{sint}\mid}\geqslant\frac{\mathrm{1}}{\mathrm{1}+{e}^{−\mathrm{3}{t}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} }{{e}^{−\mathrm{3}{t}} \:+\mid{sint}\mid}\:\geqslant\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} }{\mathrm{1}+{e}^{−\mathrm{3}{t}} }\:{dt}\:\:{and}\:{this}\:{integrale}\:{diverges}\: \\ $$$${because}\:{at}\:+\infty\:\:\frac{{e}^{−{t}} }{\mathrm{1}+{e}^{−\mathrm{3}{t}} }\:\sim{e}^{\mathrm{2}{t}} \\ $$