Convergence-of-I-0-e-t-e-t-e-2t-sint-dt-

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

C o n v e r g e n c e o f I = \int_{0}^{\infty} \frac{e^{t}}{e^{- t} + e^{2 t} ∣ s i n t ∣} d t

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

I = \int_{0}^{\infty} \frac{e^{- t}}{e^{- 3 t} + ∣ s i n t ∣} d t = \int_{0}^{+ \infty} \frac{e^{- t}}{e^{- 3 t} + ∣ s i n t ∣} d t w e h a v e

∣ s i n t ∣⩽ 1 \Rightarrow e^{- 3 t} + ∣ s i n t ∣⩽ 1 + e^{- 3 t} \Rightarrow \frac{1}{e^{- 3 t} + ∣ s i n t ∣} ⩾ \frac{1}{1 + e^{- 3 t}} \Rightarrow

\int_{0}^{\infty} \frac{e^{- t}}{e^{- 3 t} + ∣ s i n t ∣} ⩾ \int_{0}^{\infty} \frac{e^{- t}}{1 + e^{- 3 t}} d t a n d t h i s i n t e g r a l e d i v e r g e s

b e c a u s e a t + \infty \frac{e^{- t}}{1 + e^{- 3 t}} \sim e^{2 t}