Question Number 122186 by physicstutes last updated on 14/Nov/20
$$\mathrm{Obtain}\:\mathrm{a}\:\mathrm{reduction}\:\mathrm{formulae}\:\mathrm{for}\: \\ $$$$\:\:\:{I}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left(\mathrm{ln}\:{x}\right)^{{n}} {dx}\: \\ $$$$\mathrm{find}\:{I}_{\mathrm{2}} =\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} \:{dx} \\ $$
Answered by Olaf last updated on 14/Nov/20
![I_n = ∫_0 ^1 ln^n xdx I_0 = 1 and for n ≥ 1 : I_n = [xln^n x]_0 ^1 −∫_0 ^1 x(n(1/x)ln^(n−1) x)dx I_n = −n∫_0 ^1 ln^(n−1) xdx = −nI_(n−1) ⇒ I_n = (−1)^n n!I_0 = (−1)^n n! If n = 2, I_2 = (−1)^2 2! = 2](https://www.tinkutara.com/question/Q122188.png)
$$\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}^{{n}} {xdx} \\ $$$$\mathrm{I}_{\mathrm{0}} \:=\:\mathrm{1}\:\mathrm{and}\:\mathrm{for}\:{n}\:\geqslant\:\mathrm{1}\:: \\ $$$$\mathrm{I}_{{n}} \:=\:\left[{x}\mathrm{ln}^{{n}} {x}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left({n}\frac{\mathrm{1}}{{x}}\mathrm{ln}^{{n}−\mathrm{1}} {x}\right){dx} \\ $$$$\mathrm{I}_{{n}} \:=\:−{n}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}^{{n}−\mathrm{1}} {xdx}\:=\:−{n}\mathrm{I}_{{n}−\mathrm{1}} \\ $$$$\Rightarrow\:\mathrm{I}_{{n}} \:=\:\left(−\mathrm{1}\right)^{{n}} {n}!\mathrm{I}_{\mathrm{0}} \:=\:\left(−\mathrm{1}\right)^{{n}} {n}! \\ $$$$\mathrm{If}\:{n}\:=\:\mathrm{2},\:\mathrm{I}_{\mathrm{2}} \:=\:\left(−\mathrm{1}\right)^{\mathrm{2}} \mathrm{2}!\:=\:\mathrm{2} \\ $$
Commented by physicstutes last updated on 14/Nov/20
$$\mathrm{This}\:\mathrm{is}\:\mathrm{incredible}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{Olaf}. \\ $$
Answered by Dwaipayan Shikari last updated on 14/Nov/20
$$\int_{\infty} ^{\mathrm{0}} {t}^{{n}} {xdt}\:\:\:\:\:\:{logx}={t}\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{{dt}}{{dx}} \\ $$$$=\int_{\infty} ^{\mathrm{0}} {t}^{{n}} {e}^{{t}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {u}^{{n}} {e}^{−{t}} {dt}\:\:\:\:{t}=−{u} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \Gamma\left({n}+\mathrm{1}\right)=\left(−\mathrm{1}\right)^{{n}} {n}!\: \\ $$$${as}\:{per}\:{question} \\ $$$${It}\:{is}\:\left(−\mathrm{1}\right)^{\mathrm{2}} \mathrm{2}!=\mathrm{2} \\ $$
Commented by physicstutes last updated on 14/Nov/20
$$\mathrm{thats}\:\mathrm{right}! \\ $$