0-pi-2-e-2x-tanx-dx-

Question Number 147060 by ArielVyny last updated on 17/Jul/21

\int_{0}^{\frac{π}{2}} e^{2 x} \sqrt{t a n x} d x

Answered by mathmax by abdo last updated on 17/Jul/21

Ψ = \int_{0}^{\frac{π}{2}} e^{2 x} \sqrt{tanx} dx changement \sqrt{tanx} = z give x = \arctan (z^{2})

Ψ = \int_{0}^{\infty} e^{2 \arctan (z^{2})} z \times \frac{2 z}{1 + z^{4}} dz = 2 \int_{0}^{\infty} \frac{z^{2} e^{2 \arctan (z^{2})}}{1 + z^{4}} dz

φ (z) = \frac{z^{2} e^{2 \arctan (z^{2})}}{z^{4} + 1} \Rightarrow φ (z) = \frac{z^{2} e^{2 \arctan (z^{2})}}{(z^{2} - i) (z^{2} + i)}

= \frac{z^{2} e^{2 \arctan (z^{2})}}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})} we can use rdsidus

but this way is not sure \dots!

\int_{R} φ (z) dz = 2 i π {Res (φ, e^{\frac{i π}{4}}) + Res (φ, - e^{- \frac{i π}{4}})}

Res (φ, e^{\frac{i π}{4}}) = \frac{{ie}^{2 \arctan (i)}}{{2 e}^{\frac{i π}{4}} (2 i)} = \frac{1}{4} e^{- \frac{i π}{4}} e^{2 \arctan (i)}

Res (φ, - e^{- \frac{i π}{4}}) = \frac{(- i) e^{2 \arctan (- i)}}{- {2 e}^{- \frac{i π}{4}} (- 2 i)} = - \frac{1}{4} e^{\frac{i π}{4}} e^{2 \arctan (- i)}

\Rightarrow \int_{- \infty}^{+ \infty} φ (z) dz = \frac{i π}{2} {e^{- \frac{i π}{4}} . e^{2 \arctan (i)} + e^{\frac{i π}{4}} e^{2 \arctan (- i)}}

\dots be continued \dots .