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Question Number 99576 by mathmax by abdo last updated on 21/Jun/20
1)let f(a) =∫_0 ^∞ (( t^(a−1) ln(t))/(1+t)) dt with 0<a<1 prove that f(a)is convergent and determine it value 2)calculate∫_0 ^∞ ((lnt)/((1+t)(√t)))dt 3)calculate∫_0 ^∞ ((lnt)/((^3 (√t^2 ))(1+t)))dt
$$\left.\mathrm{1}\right)\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\:\mathrm{t}^{\mathrm{a}−\mathrm{1}} \mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}}\:\mathrm{dt}\:\:\:\mathrm{with}\:\mathrm{0}<\mathrm{a}<\mathrm{1}\:\:\:\mathrm{prove}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{a}\right)\mathrm{is}\:\mathrm{convergent}\:\mathrm{and}\:\mathrm{determine} \\ $$$$\mathrm{it}\:\mathrm{value} \\ $$$$\left.\mathrm{2}\right)\mathrm{calculate}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{lnt}}{\left(\mathrm{1}+\mathrm{t}\right)\sqrt{\mathrm{t}}}\mathrm{dt} \\ $$$$\left.\mathrm{3}\right)\mathrm{calculate}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{lnt}}{\left(^{\mathrm{3}} \sqrt{\mathrm{t}^{\mathrm{2}} }\right)\left(\mathrm{1}+\mathrm{t}\right)}\mathrm{dt} \\ $$
Answered by maths mind last updated on 22/Jun/20
x=ln(t) =∫_(−∞) ^(+∞) ((e^(ax) t)/(e^x +1))dx =−∫_0 ^(+∞) ((e^(−ay) y)/(1+e^(−y) ))dy+∫_0 ^(+∞) ((e^(ay) y)/(1+e^y ))dy =∫_0 ^(+∞) ((y(e^((a−1)y) −e^(−ay) ))/(1+e^(−y) ))dy let f(β)=∫_0 ^(+∞) ((ye^(−βy) )/(1+e^(−y) )) =∫_0 ^∞ ye^(−βy) (Σ_(k≥0) (−e^(−y) )^k dy =∫_0 ^(+∞) Σ_(k≥0) y(−1)^k e^((−β−k)y) dy =Σ_(k≥0) (−1)^k ∫_0 ^(+∞) ye^(−(k+β)y) =Σ_(k≥0) (−1)^k ∫_0 ^(+∞) ((te^(−t) )/((k+β)^2 ))=Σ_(k≥0) (((−1)^k )/((k+β)^2 )) =Σ_(k≥0) (1/(4(k+(β/2))^2 ))−Σ_(k≥0) (1/(4(k+((1+β)/2))^2 ))=(1/4)(ζ(2,(β/2))−ζ(2,((1+β)/2))) ζ(s,q)=Σ_(n∈N) (1/((n+q)^s )) Hurwitz zeta Function n+q≠0 f(a)=(1/4)(ζ(2,((1−a)/2))−ζ(2,((2−a)/2))−ζ(2,(a/2))+ζ(2,((1+a)/2))) f(a)=(1/4)(ζ(2,((1−a)/2))+ζ(2,((1+a)/2))) =(1/4)Σ_(n≥0) (1/((n+((1−a)/2))^2 ))+(1/((n+((1+a)/2))^2 =G))−(1/4)Σ(1/((n+((2−a)/2))^2 ))+(1/((n+(a/2))^2 =F)) G=(1/4).Σ_(n≥1) (1/((n−((1+a)/2))^2 ))+(1/((n+((1+a)/2))^2 ))+(1/4)(1/((((1+a)/2))^2 )). πcot(πx)=(1/x)+Σ_(n≥1) (1/((x+n)))+(1/((x−n))) ⇒π^2 (−1−cot^2 (πx))=−(1/x^2 )+Σ_(n≥1) ((−1)/((x+n)^2 ))−(1/((x−n)^2 )) ⇒G=−(1/4)(−π^2 −π^2 cot^2 (π(((a+1)/2))) sam we find F ⇒F=(1/4)(−π^2 −π^2 cot^2 (((πa)/2)))⇒f(a)=−(π^2 /4)(tg^2 (((πa)/2))−cot^2 (((πa)/2))) tg^2 (x)−cot^2 (x)=((sin^4 (x)−cos^4 (x))/(cos^2 (x)sin^2 (x)))=−4((cos(2x))/(sin^2 (2x))) we get=((π^2 cos(πa))/(sin^2 (πa)))=f(a) 2) ∫((ln(t))/( (√t)(t+1)))dt= f((1/2)) ∫((ln(t))/(t^(2/3) (1+t)))dt=f((1/3))
$${x}={ln}\left({t}\right) \\ $$$$=\int_{−\infty} ^{+\infty} \frac{{e}^{{ax}} {t}}{{e}^{{x}} +\mathrm{1}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−{ay}} {y}}{\mathrm{1}+{e}^{−{y}} }{dy}+\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{{ay}} {y}}{\mathrm{1}+{e}^{{y}} }{dy} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{y}\left({e}^{\left({a}−\mathrm{1}\right){y}} −{e}^{−{ay}} \right)}{\mathrm{1}+{e}^{−{y}} }{dy} \\ $$$${let}\:{f}\left(\beta\right)=\int_{\mathrm{0}} ^{+\infty} \frac{{ye}^{−\beta{y}} }{\mathrm{1}+{e}^{−{y}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} {ye}^{−\beta{y}} \left(\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−{e}^{−{y}} \right)^{{k}} {dy}\right. \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \underset{{k}\geqslant\mathrm{0}} {\sum}{y}\left(−\mathrm{1}\right)^{{k}} {e}^{\left(−\beta−{k}\right){y}} \:{dy} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{+\infty} {ye}^{−\left({k}+\beta\right){y}} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{+\infty} \frac{{te}^{−{t}} }{\left({k}+\beta\right)^{\mathrm{2}} }=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\beta\right)^{\mathrm{2}} } \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{4}\left({k}+\frac{\beta}{\mathrm{2}}\right)^{\mathrm{2}} }−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{4}\left({k}+\frac{\mathrm{1}+\beta}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\left(\zeta\left(\mathrm{2},\frac{\beta}{\mathrm{2}}\right)−\zeta\left(\mathrm{2},\frac{\mathrm{1}+\beta}{\mathrm{2}}\right)\right) \\ $$$$\zeta\left({s},{q}\right)=\underset{{n}\in\mathbb{N}} {\sum}\frac{\mathrm{1}}{\left({n}+{q}\right)^{{s}} }\:\:\:{Hurwitz}\:{zeta}\:{Function}\:\:{n}+{q}\neq\mathrm{0} \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\zeta\left(\mathrm{2},\frac{\mathrm{1}−{a}}{\mathrm{2}}\right)−\zeta\left(\mathrm{2},\frac{\mathrm{2}−{a}}{\mathrm{2}}\right)−\zeta\left(\mathrm{2},\frac{{a}}{\mathrm{2}}\right)+\zeta\left(\mathrm{2},\frac{\mathrm{1}+{a}}{\mathrm{2}}\right)\right) \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\zeta\left(\mathrm{2},\frac{\mathrm{1}−{a}}{\mathrm{2}}\right)+\zeta\left(\mathrm{2},\frac{\mathrm{1}+{a}}{\mathrm{2}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}−{a}}{\mathrm{2}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}+{a}}{\mathrm{2}}\right)^{\mathrm{2}} ={G}}−\frac{\mathrm{1}}{\mathrm{4}}\Sigma\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{2}−{a}}{\mathrm{2}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({n}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} ={F}} \\ $$$${G}=\frac{\mathrm{1}}{\mathrm{4}}.\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({n}−\frac{\mathrm{1}+{a}}{\mathrm{2}}\right)^{\mathrm{2}} \:}+\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}+{a}}{\mathrm{2}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\frac{\mathrm{1}}{\left(\frac{\mathrm{1}+{a}}{\mathrm{2}}\right)^{\mathrm{2}} }. \\ $$$$\pi{cot}\left(\pi{x}\right)=\frac{\mathrm{1}}{{x}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({x}+{n}\right)}+\frac{\mathrm{1}}{\left({x}−{n}\right)} \\ $$$$\Rightarrow\pi^{\mathrm{2}} \left(−\mathrm{1}−{cot}^{\mathrm{2}} \left(\pi{x}\right)\right)=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{−\mathrm{1}}{\left({x}+{n}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({x}−{n}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{G}=−\frac{\mathrm{1}}{\mathrm{4}}\left(−\pi^{\mathrm{2}} −\pi^{\mathrm{2}} {cot}^{\mathrm{2}} \left(\pi\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\right)\right. \\ $$$${sam}\:{we}\:{find}\:{F} \\ $$$$\Rightarrow{F}=\frac{\mathrm{1}}{\mathrm{4}}\left(−\pi^{\mathrm{2}} −\pi^{\mathrm{2}} {cot}^{\mathrm{2}} \left(\frac{\pi{a}}{\mathrm{2}}\right)\right)\Rightarrow{f}\left({a}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\left({tg}^{\mathrm{2}} \left(\frac{\pi{a}}{\mathrm{2}}\right)−{cot}^{\mathrm{2}} \left(\frac{\pi{a}}{\mathrm{2}}\right)\right) \\ $$$${tg}^{\mathrm{2}} \left({x}\right)−{cot}^{\mathrm{2}} \left({x}\right)=\frac{{sin}^{\mathrm{4}} \left({x}\right)−{cos}^{\mathrm{4}} \left({x}\right)}{{cos}^{\mathrm{2}} \left({x}\right){sin}^{\mathrm{2}} \left({x}\right)}=−\mathrm{4}\frac{{cos}\left(\mathrm{2}{x}\right)}{{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)} \\ $$$${we}\:{get}=\frac{\pi^{\mathrm{2}} {cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)}={f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:\int\frac{{ln}\left({t}\right)}{\:\sqrt{{t}}\left({t}+\mathrm{1}\right)}{dt}=\:\:{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\int\frac{{ln}\left({t}\right)}{{t}^{\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{1}+{t}\right)}{dt}={f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$ \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 22/Jun/20
1) let ϕ(a) =∫_0 ^∞ (t^(a−1) /(1+t))dt with 0<a<1 its eazy to prove thst ϕ is derivable and ϕ^′ (a) =∫_0 ^∞ (∂/∂a){ (e^((a−1)lnt) /(1+t))}dt =∫_0 ^∞ ((lnt t^(a−1) )/(1+t))dt =f(a) but ϕ(a) =(π/(sin(πa))) ⇒ϕ^′ (a) =π(((−πcos(πa))/(sin^2 (πa)))) =−π^2 ((cos(πa))/(sin^2 (πa))) ⇒ f(a) =−((π^2 cos(πa))/(sin^2 (πa))) 2)∫_0 ^∞ ((ln(t))/((1+t)(√t)))dt =∫_0 ^∞ ((t^(−(1/2)) ln(t))/(1+t))dt =∫_0 ^∞ ((t^((1/2)−1) ln(t))/(1+t))dt =f((1/2)) =−((π^2 cos((π/2)))/(sin^2 ((π/2)))) =0 another we do tbe changement (√t)=u ⇒∫_0 ^∞ ((lnt)/((1+t)(√t)))dt =∫_0 ^∞ ((2lnu)/((1+u^2 )u))(2u)du =4∫_0 ^∞ ((lnu)/(1+u^2 ))du =0 3) ∫_0 ^∞ ((ln(t))/((^3 (√t^2 ))(1+t)))dt =∫_0 ^∞ ((t^(−(2/3)) ln(t))/(1+t))dt =∫_0 ^∞ ((t^((1/3)−1) ln(t))/(1+t))dt =f((1/3)) =−((π^2 cos((π/3)))/(sin^2 ((π/3)))) =−(π^2 /2)×(1/((((√3)/2))^2 )) =−(π^2 /2)×(4/3) =−((2π^2 )/3)
$$\left.\mathrm{1}\right)\:\mathrm{let}\:\varphi\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\mathrm{with}\:\:\mathrm{0}<\mathrm{a}<\mathrm{1}\:\:\mathrm{its}\:\mathrm{eazy}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{thst}\:\varphi\:\mathrm{is}\:\mathrm{derivable} \\ $$$$\mathrm{and}\:\varphi^{'} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \frac{\partial}{\partial\mathrm{a}}\left\{\:\:\frac{\mathrm{e}^{\left(\mathrm{a}−\mathrm{1}\right)\mathrm{lnt}} }{\mathrm{1}+\mathrm{t}}\right\}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnt}\:\mathrm{t}^{\mathrm{a}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:=\mathrm{f}\left(\mathrm{a}\right) \\ $$$$\mathrm{but}\:\varphi\left(\mathrm{a}\right)\:=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{a}\right)}\:\Rightarrow\varphi^{'} \left(\mathrm{a}\right)\:=\pi\left(\frac{−\pi\mathrm{cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{2}} \left(\pi\mathrm{a}\right)}\right)\:=−\pi^{\mathrm{2}} \:\frac{\mathrm{cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{2}} \left(\pi\mathrm{a}\right)}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=−\frac{\pi^{\mathrm{2}} \:\mathrm{cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{2}} \left(\pi\mathrm{a}\right)} \\ $$$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{ln}\left(\mathrm{t}\right)}{\left(\mathrm{1}+\mathrm{t}\right)\sqrt{\mathrm{t}}}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \:\mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\: \\ $$$$=\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=−\frac{\pi^{\mathrm{2}} \:\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\right)}{\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)}\:=\mathrm{0} \\ $$$$\mathrm{another}\:\mathrm{we}\:\mathrm{do}\:\mathrm{tbe}\:\mathrm{changement}\:\sqrt{\mathrm{t}}=\mathrm{u}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnt}}{\left(\mathrm{1}+\mathrm{t}\right)\sqrt{\mathrm{t}}}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2lnu}}{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\mathrm{u}}\left(\mathrm{2u}\right)\mathrm{du} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnu}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\mathrm{du}\:=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{ln}\left(\mathrm{t}\right)}{\left(^{\mathrm{3}} \sqrt{\mathrm{t}^{\mathrm{2}} }\right)\left(\mathrm{1}+\mathrm{t}\right)}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{−\frac{\mathrm{2}}{\mathrm{3}}} \:\mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \:\mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\ $$$$=\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=−\frac{\pi^{\mathrm{2}} \mathrm{cos}\left(\frac{\pi}{\mathrm{3}}\right)}{\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{3}}\right)}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}×\frac{\mathrm{1}}{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}}\:=−\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}} \\ $$$$ \\ $$

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