Question Number 75770 by Ajao yinka last updated on 16/Dec/19
Commented by ~blr237~ last updated on 17/Dec/19
![A=∫_0 ^∞ (dx/(e^x (√(sinh2x)))) =(√2)∫_0 ^∞ ((e^(−x) dx)/(√(e^(2x) −e^(−2x) ))) =(√2) ∫_0 ^∞ ((e^(−2x) dx)/(√(1−(e^(−2x) )^2 ))) = ((−1)/(√2)) [ arsin(e^(−2x) )]_0 ^∞ =(π/(2(√2)))](Q75786.png)
$$\mathrm{A}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}} \sqrt{\mathrm{sinh2x}}}\:=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{−\mathrm{x}} \mathrm{dx}}{\sqrt{\mathrm{e}^{\mathrm{2x}} −\mathrm{e}^{−\mathrm{2x}} }}\: \\ $$$$=\sqrt{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{−\mathrm{2x}} \mathrm{dx}}{\sqrt{\mathrm{1}−\left(\mathrm{e}^{−\mathrm{2x}} \right)^{\mathrm{2}} }}\:=\:\frac{−\mathrm{1}}{\sqrt{\mathrm{2}}}\:\left[\:\mathrm{arsin}\left(\mathrm{e}^{−\mathrm{2x}} \right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\: \\ $$$$ \\ $$
Commented by ~blr237~ last updated on 17/Dec/19
![let consider for b≠0 and a≠−1 ,f(a,b)=∫_0 ^∞ (x^a /(b+x))dx f(a,b)=∫_0 ^∞ ((b^a ((x/b))^a )/(1+((x/b))))d((x/b))=b^a ∫_0 ^∞ (t^a /(1+t))dt using the Mellin function ∫_0 ^∞ (t^a /(1+t))dt=(π/(sin(π(a+1)))) so f(a,b)=((πb^a )/(sin(π(a+1)))) Now study (∂f/∂a)(a,b)=∫_0 ^∞ ((x^a lnx)/(b+x))dx (∂^n f/(∂b^(n−1) ∂a))(a,b)=∫_0 ^∞ (−1)^n (n−1) ((x^a lnx)/((b+x)^n ))dx So ∫_0 ^∞ ((lnx)/((1+x)^n ))dx=(((−1)^n )/(n−1)) (∂^n f/(∂b^(n−1) ∂a))(0,1) ∫_0 ^∞ (((1−3x))/((1+x)^5 ))(logx)^2 dx=∫_0 ^∞ [(4/((1+x)^5 ))(logx)^2 −(3/((1+x)^4 ))(logx)^2 ]dx = −(∂^6 f/(∂b^4 ∂a^2 ))(0,1)−(∂^5 f/(∂b^3 ∂a^2 ))(0,1)](Q75788.png)
$$\mathrm{let}\:\mathrm{consider}\:\mathrm{for}\:\mathrm{b}\neq\mathrm{0}\:\mathrm{and}\:\mathrm{a}\neq−\mathrm{1}\:,\mathrm{f}\left(\mathrm{a},\mathrm{b}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{a}} }{\mathrm{b}+\mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{f}\left(\mathrm{a},\mathrm{b}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{b}^{\mathrm{a}} \left(\frac{\mathrm{x}}{\mathrm{b}}\right)^{\mathrm{a}} }{\mathrm{1}+\left(\frac{\mathrm{x}}{\mathrm{b}}\right)}\mathrm{d}\left(\frac{\mathrm{x}}{\mathrm{b}}\right)=\mathrm{b}^{\mathrm{a}} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{a}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{Mellin}\:\mathrm{function}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}=\frac{\pi}{\mathrm{sin}\left(\pi\left(\mathrm{a}+\mathrm{1}\right)\right)}\: \\ $$$$\mathrm{so}\:\:\mathrm{f}\left(\mathrm{a},\mathrm{b}\right)=\frac{\pi\mathrm{b}^{\mathrm{a}} }{\mathrm{sin}\left(\pi\left(\mathrm{a}+\mathrm{1}\right)\right)} \\ $$$$\mathrm{Now}\:\mathrm{study}\:\:\frac{\partial\mathrm{f}}{\partial\mathrm{a}}\left(\mathrm{a},\mathrm{b}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{a}} \mathrm{lnx}}{\mathrm{b}+\mathrm{x}}\mathrm{dx} \\ $$$$\frac{\partial^{\mathrm{n}} \mathrm{f}}{\partial\mathrm{b}^{\mathrm{n}−\mathrm{1}} \partial\mathrm{a}}\left(\mathrm{a},\mathrm{b}\right)=\int_{\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{n}−\mathrm{1}\right)\:\frac{\mathrm{x}^{\mathrm{a}} \mathrm{lnx}}{\left(\mathrm{b}+\mathrm{x}\right)^{\mathrm{n}} }\mathrm{dx} \\ $$$$\mathrm{So}\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{lnx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} }\mathrm{dx}=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}−\mathrm{1}}\:\frac{\partial^{\mathrm{n}} \mathrm{f}}{\partial\mathrm{b}^{\mathrm{n}−\mathrm{1}} \partial\mathrm{a}}\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{1}−\mathrm{3x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{5}} }\left(\mathrm{logx}\right)^{\mathrm{2}} \mathrm{dx}=\int_{\mathrm{0}} ^{\infty} \left[\frac{\mathrm{4}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{5}} }\left(\mathrm{logx}\right)^{\mathrm{2}} −\frac{\mathrm{3}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{4}} }\left(\mathrm{logx}\right)^{\mathrm{2}} \right]\mathrm{dx} \\ $$$$\:\:\:\:=\:−\frac{\partial^{\mathrm{6}} \mathrm{f}}{\partial\mathrm{b}^{\mathrm{4}} \partial\mathrm{a}^{\mathrm{2}} }\left(\mathrm{0},\mathrm{1}\right)−\frac{\partial^{\mathrm{5}} \mathrm{f}}{\partial\mathrm{b}^{\mathrm{3}} \partial\mathrm{a}^{\mathrm{2}} }\left(\mathrm{0},\mathrm{1}\right) \\ $$
Commented by ~blr237~ last updated on 17/Dec/19
![let consider for b≠0 and a≠−1 ,f(a,b)=∫_0 ^∞ (x^a /(b+x))dx f(a,b)=∫_0 ^∞ ((b^a ((x/b))^a )/(1+((x/b))))d((x/b))=b^a ∫_0 ^∞ (t^a /(1+t))dt using the Mellin function ∫_0 ^∞ (t^a /(1+t))dt=(π/(sin(π(a+1)))) so f(a,b)=((πb^a )/(sin(π(a+1)))) Now study (∂f/∂a)(a,b)=∫_0 ^∞ ((x^a lnx)/(b+x))dx (∂^n f/(∂b^(n−1) ∂a))(a,b)=∫_0 ^∞ (−1)^n (n−1) ((x^a lnx)/((b+x)^n ))dx So ∫_0 ^∞ ((lnx)/((1+x)^n ))dx=(((−1)^n )/(n−1)) (∂^n f/(∂b^(n−1) ∂a))(0,1) ∫_0 ^∞ (((1−3x))/((1+x)^5 ))(logx)^2 dx=∫_0 ^∞ [(4/((1+x)^5 ))(logx)^2 −(3/((1+x)^4 ))(logx)^2 ]dx = −(∂^6 f/(∂b^4 ∂a^2 ))(0,1)−(∂^5 f/(∂b^3 ∂a^2 ))(0,1)](Q75789.png)
$$\mathrm{let}\:\mathrm{consider}\:\mathrm{for}\:\mathrm{b}\neq\mathrm{0}\:\mathrm{and}\:\mathrm{a}\neq−\mathrm{1}\:,\mathrm{f}\left(\mathrm{a},\mathrm{b}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{a}} }{\mathrm{b}+\mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{f}\left(\mathrm{a},\mathrm{b}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{b}^{\mathrm{a}} \left(\frac{\mathrm{x}}{\mathrm{b}}\right)^{\mathrm{a}} }{\mathrm{1}+\left(\frac{\mathrm{x}}{\mathrm{b}}\right)}\mathrm{d}\left(\frac{\mathrm{x}}{\mathrm{b}}\right)=\mathrm{b}^{\mathrm{a}} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{a}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{Mellin}\:\mathrm{function}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}=\frac{\pi}{\mathrm{sin}\left(\pi\left(\mathrm{a}+\mathrm{1}\right)\right)}\: \\ $$$$\mathrm{so}\:\:\mathrm{f}\left(\mathrm{a},\mathrm{b}\right)=\frac{\pi\mathrm{b}^{\mathrm{a}} }{\mathrm{sin}\left(\pi\left(\mathrm{a}+\mathrm{1}\right)\right)} \\ $$$$\mathrm{Now}\:\mathrm{study}\:\:\frac{\partial\mathrm{f}}{\partial\mathrm{a}}\left(\mathrm{a},\mathrm{b}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{a}} \mathrm{lnx}}{\mathrm{b}+\mathrm{x}}\mathrm{dx} \\ $$$$\frac{\partial^{\mathrm{n}} \mathrm{f}}{\partial\mathrm{b}^{\mathrm{n}−\mathrm{1}} \partial\mathrm{a}}\left(\mathrm{a},\mathrm{b}\right)=\int_{\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{n}−\mathrm{1}\right)\:\frac{\mathrm{x}^{\mathrm{a}} \mathrm{lnx}}{\left(\mathrm{b}+\mathrm{x}\right)^{\mathrm{n}} }\mathrm{dx} \\ $$$$\mathrm{So}\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{lnx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} }\mathrm{dx}=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}−\mathrm{1}}\:\frac{\partial^{\mathrm{n}} \mathrm{f}}{\partial\mathrm{b}^{\mathrm{n}−\mathrm{1}} \partial\mathrm{a}}\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{1}−\mathrm{3x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{5}} }\left(\mathrm{logx}\right)^{\mathrm{2}} \mathrm{dx}=\int_{\mathrm{0}} ^{\infty} \left[\frac{\mathrm{4}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{5}} }\left(\mathrm{logx}\right)^{\mathrm{2}} −\frac{\mathrm{3}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{4}} }\left(\mathrm{logx}\right)^{\mathrm{2}} \right]\mathrm{dx} \\ $$$$\:\:\:\:=\:−\frac{\partial^{\mathrm{6}} \mathrm{f}}{\partial\mathrm{b}^{\mathrm{4}} \partial\mathrm{a}^{\mathrm{2}} }\left(\mathrm{0},\mathrm{1}\right)−\frac{\partial^{\mathrm{5}} \mathrm{f}}{\partial\mathrm{b}^{\mathrm{3}} \partial\mathrm{a}^{\mathrm{2}} }\left(\mathrm{0},\mathrm{1}\right) \\ $$
Commented by Ajao yinka last updated on 17/Dec/19
Superb
Commented by Ajao yinka last updated on 17/Dec/19
So what's final answer?
Commented by ~blr237~ last updated on 17/Dec/19
$$\mathrm{you}\:\mathrm{find}\:\mathrm{out}\:\mathrm{the}\:\mathrm{exprezsion}\:\mathrm{of}\:\:\mathrm{that}\:\mathrm{derivate}\left(\mathrm{it}\:\mathrm{will}\:\mathrm{be}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{start}\:\:\mathrm{derivation}\:\mathrm{on}\:\mathrm{b}\right) \\ $$$$\mathrm{after}\:\mathrm{you}\:\mathrm{just}\:\mathrm{replace}\:\left(\mathrm{a},\mathrm{b}\right)\:\mathrm{by}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$
Answered by ~blr237~ last updated on 17/Dec/19
![B=∫_((−π)/4) ^(π/4) log(sinx+cosx)dx observe that sinx+cosx=(√2) cos(x−(π/4)) B=∫_(−(π/2)) ^0 log((√2) cosu)du with u=x−(π/4) B=−(π/4)log2 +∫_0 ^(π/2) ln(cosu)du secondly observe that B=∫_(−(π/4)) ^(π/4) log(cost−sint)dt with t=−x so 2B=∫_((−π)/4) ^(π/4) [log(cosx+sinx)+log(cosx−sinx)]dx =∫ _((−π)/4)^(π/4) log(cos2x)dx =2∫_0 ^(π/4) log(cos2x)dx =∫_0 ^(π/2) log(cosu)du with u=2x Finally we have B=−(π/4)log2 +2B so B=(π/4)log2](Q75787.png)
$$\mathrm{B}=\int_{\frac{−\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{log}\left(\mathrm{sinx}+\mathrm{cosx}\right)\mathrm{dx} \\ $$$$\mathrm{observe}\:\mathrm{that}\:\mathrm{sinx}+\mathrm{cosx}=\sqrt{\mathrm{2}}\:\mathrm{cos}\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{B}=\int_{−\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \mathrm{log}\left(\sqrt{\mathrm{2}}\:\mathrm{cosu}\right)\mathrm{du}\:\:\:\mathrm{with}\:\:\mathrm{u}=\mathrm{x}−\frac{\pi}{\mathrm{4}}\: \\ $$$$\mathrm{B}=−\frac{\pi}{\mathrm{4}}\mathrm{log2}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cosu}\right)\mathrm{du} \\ $$$$\mathrm{secondly}\:\mathrm{observe}\:\mathrm{that}\: \\ $$$$\mathrm{B}=\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{log}\left(\mathrm{cost}−\mathrm{sint}\right)\mathrm{dt}\:\:\:\:\mathrm{with}\:\:\mathrm{t}=−\mathrm{x} \\ $$$$\mathrm{so}\:\mathrm{2B}=\int_{\frac{−\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \left[\mathrm{log}\left(\mathrm{cosx}+\mathrm{sinx}\right)+\mathrm{log}\left(\mathrm{cosx}−\mathrm{sinx}\right)\right]\mathrm{dx} \\ $$$$\:\:\:=\int\:_{\frac{−\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{log}\left(\mathrm{cos2x}\right)\mathrm{dx}\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{log}\left(\mathrm{cos2x}\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{cosu}\right)\mathrm{du}\:\:\:\mathrm{with}\:\:\mathrm{u}=\mathrm{2x} \\ $$$$\mathrm{Finally}\:\:\mathrm{we}\:\mathrm{have}\:\: \\ $$$$\mathrm{B}=−\frac{\pi}{\mathrm{4}}\mathrm{log2}\:+\mathrm{2B}\: \\ $$$$\mathrm{so}\:\:\mathrm{B}=\frac{\pi}{\mathrm{4}}\mathrm{log2} \\ $$$$ \\ $$
Commented by Ajao yinka last updated on 17/Dec/19
Nice
Answered by mind is power last updated on 17/Dec/19
$$\int_{\frac{−\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{x}\right)+\mathrm{cos}\left(\mathrm{x}\right)\right)\mathrm{dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{log}\left(\sqrt{\mathrm{2}}\mathrm{sin}\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\sqrt{\mathrm{2}}\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$=\frac{\pi}{\mathrm{2}}\mathrm{log}\left(\sqrt{\mathrm{2}}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{cos}\left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$\int_{\mathrm{0}} ^{\pi} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{2u}\right)\right)\mathrm{du}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\mathrm{log}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}=−\frac{\pi}{\mathrm{2}}\mathrm{log}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\int_{−\frac{\pi}{\mathrm{4}}} ^{+\frac{\pi}{\mathrm{4}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)+\mathrm{cos}\left(\mathrm{u}\right)\right)\mathrm{du}=\frac{\pi}{\mathrm{2}}\mathrm{log}\left(\sqrt{\mathrm{2}}\right)−\frac{\pi}{\mathrm{2}}\mathrm{log}\left(\mathrm{2}\right) \\ $$$$\frac{\pi}{\mathrm{2}}\mathrm{log}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right) \\ $$
Answered by mind is power last updated on 17/Dec/19
![∫_0 ^(+∞) ((log(x))/((1+x)^n )) n≥2 if n =1 ,0diverge x=(1/y)⇒ =∫_0 ^(+∞) −((log(y))/((1+y)^n )).y^(n−2) dy if n=2 ⇒∫_0 ^(+∞) ((log(y))/((1+y)^2 ))dy=0 n≥3, by part U_n =∫_0 ^(+∞) ((log(x))/((1+x)^n ))dx ((log(x))/((1+x)^n ))=[((xlog(x)−x)/((1+x)^n ))]+n∫((xlog(x)−x)/((1+x)^(n+1) ))dx =n∫_0 ^(+∞) (((x+1)log(x)−log(x))/((1+x)^(n+1) ))dx−n∫_0 ^(+∞) {(1/((1+x)^n ))−(1/((1+x)^(n+1) ))}dx ==n∫_0 ^(+∞) ((log(x))/((1+x)^n ))−n∫_0 ^(+∞) ((log(x))/((1+x)^(n+1) ))dx−n[−(1/(n−1)).(1/((1+x)^(n−1) ))+(1/n).(1/((1+x)^n ))]_0 ^(+∞) ⇒U_n =nU_n −nU_(n+1) −n{−(1/(n−1))+(1/n)} ⇒(1−n)U_n =−nU_(n+1) +(1/((n−1))) ⇔U_(n+1) =((n−1)/n)U_n +(1/(n(n−1))) U_2 =0](Q75808.png)
$$\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{log}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} } \\ $$$$\mathrm{n}\geqslant\mathrm{2}\:\:\mathrm{if}\:\mathrm{n}\:=\mathrm{1}\:,\mathrm{0diverge}\: \\ $$$$\mathrm{x}=\frac{\mathrm{1}}{\mathrm{y}}\Rightarrow \\ $$$$=\int_{\mathrm{0}} ^{+\infty} −\frac{\mathrm{log}\left(\mathrm{y}\right)}{\left(\mathrm{1}+\mathrm{y}\right)^{\mathrm{n}} }.\mathrm{y}^{\mathrm{n}−\mathrm{2}} \mathrm{dy} \\ $$$$\mathrm{if}\:\mathrm{n}=\mathrm{2}\:\Rightarrow\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{log}\left(\mathrm{y}\right)}{\left(\mathrm{1}+\mathrm{y}\right)^{\mathrm{2}} }\mathrm{dy}=\mathrm{0} \\ $$$$\mathrm{n}\geqslant\mathrm{3},\:\: \\ $$$$\mathrm{by}\:\mathrm{part}\:\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{log}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} }\mathrm{dx} \\ $$$$\frac{\mathrm{log}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} }=\left[\frac{\mathrm{xlog}\left(\mathrm{x}\right)−\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} }\right]+\mathrm{n}\int\frac{\mathrm{xlog}\left(\mathrm{x}\right)−\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}+\mathrm{1}} }\mathrm{dx} \\ $$$$=\mathrm{n}\int_{\mathrm{0}} ^{+\infty} \frac{\left(\mathrm{x}+\mathrm{1}\right)\mathrm{log}\left(\mathrm{x}\right)−\mathrm{log}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}+\mathrm{1}} }\mathrm{dx}−\mathrm{n}\int_{\mathrm{0}} ^{+\infty} \left\{\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}+\mathrm{1}} }\right\}\mathrm{dx} \\ $$$$==\mathrm{n}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{log}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} }−\mathrm{n}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{log}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}+\mathrm{1}} }\mathrm{dx}−\mathrm{n}\left[−\frac{\mathrm{1}}{\mathrm{n}−\mathrm{1}}.\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}−\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{n}}.\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} }\right]_{\mathrm{0}} ^{+\infty} \\ $$$$\Rightarrow\mathrm{U}_{\mathrm{n}} =\mathrm{nU}_{\mathrm{n}} −\mathrm{nU}_{\mathrm{n}+\mathrm{1}} −\mathrm{n}\left\{−\frac{\mathrm{1}}{\mathrm{n}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{n}}\right\} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{n}\right)\mathrm{U}_{\mathrm{n}} =−\mathrm{nU}_{\mathrm{n}+\mathrm{1}} +\frac{\mathrm{1}}{\left(\mathrm{n}−\mathrm{1}\right)} \\ $$$$\Leftrightarrow\mathrm{U}_{\mathrm{n}+\mathrm{1}} =\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}\mathrm{U}_{\mathrm{n}} +\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)} \\ $$$$\mathrm{U}_{\mathrm{2}} =\mathrm{0} \\ $$$$ \\ $$