Menu Close

calculate-0-e-x-ln-x-dx-with-gt-0-

Tinku Tara 0 Comments
Pin




Question Number 72025 by mathmax by abdo last updated on 23/Oct/19
calculate ∫_0 ^∞ e^(−αx) ln(x)dx with α>0
calculate∫0∞e−αxln(x)dxwithα>0
Commented by mathmax by abdo last updated on 24/Oct/19
changement αx=t give ∫_0 ^∞ e^(−αx) ln(x)dx=∫_0 ^∞ e^(−t) ln((t/α))(dt/α) =(1/α)∫_0 ^∞ e^(−t) {ln(t)−lnα}dt =(1/α)∫_0 ^∞ e^(−t) ln(t)−((ln(α))/α) =(1/α)(−γ)−((ln(α))/α) =−(1/α)( γ +ln(α)) ∫_0 ^∞ e^(−t) ln(t)dt=−γ and this value is proved.
changementαx=tgive∫0∞e−αxln(x)dx=∫0∞e−tln(tα)dtα=1α∫0∞e−t{ln(t)−lnα}dt=1α∫0∞e−tln(t)−ln(α)α=1α(−γ)−ln(α)α=−1α(γ+ln(α))∫0∞e−tln(t)dt=−γandthisvalueisproved.
Answered by mind is power last updated on 23/Oct/19
u=αx ⇒∫_0 ^∞ e^(−u) ln((u/α)).(du/α) =∫_0 ^(+∞) e^(−u) ln(u)(du/α)−∫_0 ^∞ e^(−u) .((ln(α))/α)du =(1/α)∫_0 ^(+∞) e^(−u) ln(u)du+((ln(α))/α)[e^(−u) ]_0 ^(+∞) =(1/α).γ−((ln(α))/α)=(1/α)(γ−ln(α))
u=αx⇒∫0∞e−uln(uα).duα=∫0+∞e−uln(u)duα−∫0∞e−u.ln(α)αdu=1α∫0+∞e−uln(u)du+ln(α)α[e−u]0+∞=1α.γ−ln(α)α=1α(γ−ln(α))

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *