let-0-lt-a-lt-1-calculate-0-ln-t-t-a-1-1-t-dt-and-0-ln-2-t-t-a-1-1-t-dt-

Question Number 73238 by mathmax by abdo last updated on 08/Nov/19

let 0<a<1 calculate ∫_0 ^∞ ((ln(t)t^(a−1) )/(1+t))dt and ∫_0 ^∞ ((ln^2 (t)t^(a−1) )/(1+t))dt

$${let}\:\mathrm{0}<{a}<\mathrm{1}\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({t}\right){t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}^{\mathrm{2}} \left({t}\right){t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt} \\ $$

Commented by mathmax by abdo last updated on 10/Nov/19

we have ∫_0 ^∞ (t^(a−1) /(1+t))dt =(π/(sin(πa)))=f(a) if 0<a<1 (this result is proved in the plstform) ⇒ f^′ (a) =∫_0 ^∞ (∂/∂a)((t^(a−1) /(1+t)))dt =∫_0 ^∞ (∂/∂a)((e^((a−1)lnt) /(1+t)))dt =∫_0 ^∞ ((lnt ×t^(a−1) )/(1+t)) but f^′ (a)=π×((−π cos(πa))/(sin^2 (πa))) =−π^2 ((cos(πa))/(sin^2 (πa))) also f^((2)) (a) =∫_0 ^∞ ((ln^2 (t)t^(a−1) )/(1+t))dt and f^((2)) (a)= −π^2 ((−πsin(πa)sin^2 (πa)−2sin(πa)π cos(πa))/(sin^4 (πa))) =π^3 ×((sin^2 (πa)+2cos(πa))/(sin^3 (πa))) ⇒∫_0 ^∞ ((ln(t)t^(a−1) )/(1+t))dt=−((π^2 cos(πa))/(sin^2 (πa))) and ∫_0 ^∞ ((ln^2 t t^(a−1) )/(1+t))dt =π^3 ×((sin^2 (πa)+2cos(πa))/(sin^3 (πa))) .

$${we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}={f}\left({a}\right)\:{if}\:\mathrm{0}<{a}<\mathrm{1}\:\:\:\left({this}\:{result}\:{is}\:{proved}\:{in}\right. \\ $$$$\left.{the}\:{plstform}\right)\:\Rightarrow\:{f}^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\partial}{\partial{a}}\left(\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\right){dt}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\partial}{\partial{a}}\left(\frac{{e}^{\left({a}−\mathrm{1}\right){lnt}} }{\mathrm{1}+{t}}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{lnt}\:×{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:\:{but}\:\:{f}^{'} \left({a}\right)=\pi×\frac{−\pi\:{cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)}\:=−\pi^{\mathrm{2}} \:\frac{{cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)} \\ $$$${also}\:{f}^{\left(\mathrm{2}\right)} \left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}^{\mathrm{2}} \left({t}\right){t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:{and}\:{f}^{\left(\mathrm{2}\right)} \left({a}\right)= \\ $$$$−\pi^{\mathrm{2}} \frac{−\pi{sin}\left(\pi{a}\right){sin}^{\mathrm{2}} \left(\pi{a}\right)−\mathrm{2}{sin}\left(\pi{a}\right)\pi\:{cos}\left(\pi{a}\right)}{{sin}^{\mathrm{4}} \left(\pi{a}\right)} \\ $$$$=\pi^{\mathrm{3}} ×\frac{{sin}^{\mathrm{2}} \left(\pi{a}\right)+\mathrm{2}{cos}\left(\pi{a}\right)}{{sin}^{\mathrm{3}} \left(\pi{a}\right)}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({t}\right){t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=−\frac{\pi^{\mathrm{2}} {cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)} \\ $$$${and}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}^{\mathrm{2}} {t}\:{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\pi^{\mathrm{3}} ×\frac{{sin}^{\mathrm{2}} \left(\pi{a}\right)+\mathrm{2}{cos}\left(\pi{a}\right)}{{sin}^{\mathrm{3}} \left(\pi{a}\right)}\:. \\ $$$$ \\ $$

Answered by mind is power last updated on 09/Nov/19