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Question Number 79913 by Henri Boucatchou last updated on 29/Jan/20

Convergence of I=∫_0 ^( ∞) (e^t /(e^(−t) +e^(2t) ∣sint∣))dt

$$\:{Convergence}\:\:{of}\:\:{I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{{t}} }{{e}^{−{t}} +{e}^{\mathrm{2}{t}} \mid{sint}\mid}{dt} \\ $$

Commented by mathmax by abdo last updated on 29/Jan/20

I =∫_0 ^∞ (e^(−t) /(e^(−3t) +∣sint∣))dt =∫_0 ^(+∞) (e^(−t) /(e^(−3t) +∣sint∣))dt we have ∣sint∣≤1 ⇒e^(−3t) +∣sint∣≤1+e^(−3t) ⇒(1/(e^(−3t) +∣sint∣))≥(1/(1+e^(−3t) )) ⇒ ∫_0 ^∞ (e^(−t) /(e^(−3t) +∣sint∣)) ≥∫_0 ^∞ (e^(−t) /(1+e^(−3t) )) dt and this integrale diverges because at +∞ (e^(−t) /(1+e^(−3t) )) ∼e^(2t)

$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} }{{e}^{−\mathrm{3}{t}} \:+\mid{sint}\mid}{dt}\:=\int_{\mathrm{0}} ^{+\infty} \:\frac{{e}^{−{t}} }{{e}^{−\mathrm{3}{t}} \:+\mid{sint}\mid}{dt}\:\:\:{we}\:{have} \\ $$$$\mid{sint}\mid\leqslant\mathrm{1}\:\Rightarrow{e}^{−\mathrm{3}{t}} \:+\mid{sint}\mid\leqslant\mathrm{1}+{e}^{−\mathrm{3}{t}} \:\Rightarrow\frac{\mathrm{1}}{{e}^{−\mathrm{3}{t}} +\mid{sint}\mid}\geqslant\frac{\mathrm{1}}{\mathrm{1}+{e}^{−\mathrm{3}{t}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} }{{e}^{−\mathrm{3}{t}} \:+\mid{sint}\mid}\:\geqslant\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} }{\mathrm{1}+{e}^{−\mathrm{3}{t}} }\:{dt}\:\:{and}\:{this}\:{integrale}\:{diverges}\: \\ $$$${because}\:{at}\:+\infty\:\:\frac{{e}^{−{t}} }{\mathrm{1}+{e}^{−\mathrm{3}{t}} }\:\sim{e}^{\mathrm{2}{t}} \\ $$

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