Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

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

ConvergenceofI=∫0∞ete−t+e2t∣sint∣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=∫0∞e−te−3t+∣sint∣dt=∫0+∞e−te−3t+∣sint∣dtwehave∣sint∣⩽1⇒e−3t+∣sint∣⩽1+e−3t⇒1e−3t+∣sint∣⩾11+e−3t⇒∫0∞e−te−3t+∣sint∣⩾∫0∞e−t1+e−3tdtandthisintegraledivergesbecauseat+∞e−t1+e−3t∼e2t

Terms of Service

Privacy Policy

Contact: info@tinkutara.com