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Question Number 135395 by 0731619177 last updated on 12/Mar/21
Answered by Dwaipayan Shikari last updated on 12/Mar/21
I(b)=∫_0 ^1 sin(log((1/x)))((x^b −x^a )/(log(x)))dx I′(b)=∫_0 ^1 x^b sin(log((1/x)))dx=∫_1 ^∞ t^(2−b) sin(log(t)) log(t)=u =∫_0 ^∞ e^(−u(b−2)+u) sin(u)du=(1/(2i))∫_0 ^∞ e^(−u(b−3−i)) −e^(−u(b−3+i)) du =(1/(2i))((1/(b−3−i))−(1/(b−3+i)))=(1/(b^2 −6b−10)) I(b)=∫(1/(b^2 −6b+10))db=tan^(−1) (b−3)+C I(a)=tan^(−1) (a−3)+C=0⇒C=−tan^(−1) (a−3) I(b)=tan^(−1) (b−3)−tan^(−1) (a−3)=tan^(−1) ((b−a)/(ab−3(b+a)+10))
$${I}\left({b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left({log}\left(\frac{\mathrm{1}}{{x}}\right)\right)\frac{{x}^{{b}} −{x}^{{a}} }{{log}\left({x}\right)}{dx} \\ $$$${I}'\left({b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{b}} {sin}\left({log}\left(\frac{\mathrm{1}}{{x}}\right)\right){dx}=\int_{\mathrm{1}} ^{\infty} {t}^{\mathrm{2}−{b}} {sin}\left({log}\left({t}\right)\right)\:\:\:\:{log}\left({t}\right)={u} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−{u}\left({b}−\mathrm{2}\right)+{u}} {sin}\left({u}\right){du}=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} {e}^{−{u}\left({b}−\mathrm{3}−{i}\right)} −{e}^{−{u}\left({b}−\mathrm{3}+{i}\right)} {du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\frac{\mathrm{1}}{{b}−\mathrm{3}−{i}}−\frac{\mathrm{1}}{{b}−\mathrm{3}+{i}}\right)=\frac{\mathrm{1}}{{b}^{\mathrm{2}} −\mathrm{6}{b}−\mathrm{10}} \\ $$$${I}\left({b}\right)=\int\frac{\mathrm{1}}{{b}^{\mathrm{2}} −\mathrm{6}{b}+\mathrm{10}}{db}={tan}^{−\mathrm{1}} \left({b}−\mathrm{3}\right)+{C} \\ $$$${I}\left({a}\right)={tan}^{−\mathrm{1}} \left({a}−\mathrm{3}\right)+{C}=\mathrm{0}\Rightarrow{C}=−{tan}^{−\mathrm{1}} \left({a}−\mathrm{3}\right) \\ $$$${I}\left({b}\right)={tan}^{−\mathrm{1}} \left({b}−\mathrm{3}\right)−{tan}^{−\mathrm{1}} \left({a}−\mathrm{3}\right)={tan}^{−\mathrm{1}} \frac{{b}−{a}}{{ab}−\mathrm{3}\left({b}+{a}\right)+\mathrm{10}} \\ $$
Answered by Olaf last updated on 12/Mar/21
1) Let C = ∫_0 ^(π/2) ln(cosx)dx and S = ∫_0 ^(π/2) ln(sinx)dx Let u = (π/2)−x S = ∫_(π/2) ^0 ln(sin((π/2)−u))(−du) = C C+S = ∫_0 ^(π/2) (ln(cosx)+ln(sinx))dx 2S = ∫_0 ^(π/2) ln(sinxcosx)dx 2S = ∫_0 ^(π/2) ln((1/2)sin2x)dx Let u = 2x 2S = ∫_0 ^π ln((1/2)sinu)((du/2)) 2S = −(π/2)ln2+(1/2)[∫_0 ^(π/2) ln(sinu)du+∫_(π/2) ^π ln(sinu)du] Let v = π−u ∫_(π/2) ^π ln(sinu)du = ∫_(π/2) ^0 ln(sin(π−v))(−dv) = S 2S = −(π/2)ln2+(1/2)(S+S) S = −(π/2)ln2
$$\left.\mathrm{1}\right) \\ $$$$\mathrm{Let}\:\mathrm{C}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cos}{x}\right){dx}\:\mathrm{and}\:\mathrm{S}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}{x}\right){dx} \\ $$$$\mathrm{Let}\:{u}\:=\:\frac{\pi}{\mathrm{2}}−{x} \\ $$$$\mathrm{S}\:=\:\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \mathrm{ln}\left(\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−{u}\right)\right)\left(−{du}\right)\:=\:\mathrm{C} \\ $$$$\mathrm{C}+\mathrm{S}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{ln}\left(\mathrm{cos}{x}\right)+\mathrm{ln}\left(\mathrm{sin}{x}\right)\right){dx} \\ $$$$\mathrm{2S}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}{x}\mathrm{cos}{x}\right){dx} \\ $$$$\mathrm{2S}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}{x}\right){dx} \\ $$$$\mathrm{Let}\:{u}\:=\:\mathrm{2}{x} \\ $$$$\mathrm{2S}\:=\:\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}{u}\right)\left(\frac{{du}}{\mathrm{2}}\right) \\ $$$$\mathrm{2S}\:=\:−\frac{\pi}{\mathrm{2}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{2}}\left[\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}{u}\right){du}+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \mathrm{ln}\left(\mathrm{sin}{u}\right){du}\right] \\ $$$$\mathrm{Let}\:{v}\:=\:\pi−{u} \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \mathrm{ln}\left(\mathrm{sin}{u}\right){du}\:=\:\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \mathrm{ln}\left(\mathrm{sin}\left(\pi−{v}\right)\right)\left(−{dv}\right)\:=\:\mathrm{S} \\ $$$$\mathrm{2S}\:=\:−\frac{\pi}{\mathrm{2}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{S}+\mathrm{S}\right) \\ $$$$\mathrm{S}\:=\:−\frac{\pi}{\mathrm{2}}\mathrm{ln2} \\ $$
Answered by Dwaipayan Shikari last updated on 12/Mar/21
∫_0 ^(π/2) log(sinx)dx =∫_0 ^1 ((log(t))/( (√(1−t^2 ))))dt=Ψ′(0) Where Ψ(α)=∫_0 ^1 (t^a /( (√(1−t^2 ))))dt=(1/2)∫_0 ^1 (u^((a/2)−(1/2)) /( (√(1−u))))du =((Γ(((a+1)/2))Γ((1/2)))/(2Γ((a/2)+1))) Ψ′(a)=((√π)/4).((Γ((a/2)+1)Γ′(((a+1)/2))−Γ′((a/2)+1)Γ(((a+1)/2)))/(Γ^2 ((a/2)+1))) Ψ′(0)=((√π)/4)((√π)(ψ((1/2))−ψ(1))=(π/4)(−2log(2))=−(π/2)log(2)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sinx}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left({t}\right)}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}=\Psi'\left(\mathrm{0}\right) \\ $$$${Where}\:\Psi\left(\alpha\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{a}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\frac{{a}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} }{\:\sqrt{\mathrm{1}−{u}}}{du} \\ $$$$=\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$$\Psi'\left({a}\right)=\frac{\sqrt{\pi}}{\mathrm{4}}.\frac{\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\Gamma'\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\Gamma'\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$$\Psi'\left(\mathrm{0}\right)=\frac{\sqrt{\pi}}{\mathrm{4}}\left(\sqrt{\pi}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}\right)\right)=\frac{\pi}{\mathrm{4}}\left(−\mathrm{2}{log}\left(\mathrm{2}\right)\right)=−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right)\right. \\ $$
Answered by mathmax by abdo last updated on 12/Mar/21
1) let I=∫_0 ^(π/2) ln(sinx)dx ⇒I =∫_0 ^(π/2) ln(((e^x −e^(−x) )/(2i)))dx =−(π/2)ln(2i)+∫_0 ^(π/2) ln(e^x −e^(−x) )dx =−(π/2)ln(2)−(π/2)(((iπ)/2))+∫_0 ^(π/2) ln(e^x −e^(−x) )dx =−(π/2)ln(2)−i(π^2 /4)+∫_0 ^(π/2) xdx +∫_0 ^(π/2) ln(1−e^(−2x) )dx but ln^′ (1−u)=((−1)/(1−u))=−Σ_(n=0) ^∞ u^n ⇒ln(1−u)=−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) =−Σ_(n=1) ^∞ (u^n /n) ⇒ln(1−e^(−2x) ) =−Σ_(n=1) ^∞ (e^(−2nx) /n) ⇒ ∫_0 ^(π/2) ln(1−e^(−2x) )dx =−Σ_(n=1) ^∞ (1/n)∫_0 ^(π/2) e^(−2nx) dx =−Σ_(n=1) ^∞ (1/(n(−2n))) [e^(−2nx) ]_0 ^(π/2) =(1/2)Σ_(n=1) ^∞ (1/n^2 )((−1)^n −1) =−Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒I=−(π/2)ln(2)−((iπ^2 )/4) +(π^2 /8)−Σ_(n=0) ^∞ (1/((2n+1)^2 )) (π^2 /8)−Σ_(n=0) ^∞ (1/((2n+1)^2 ))=0 and I is real ⇒I =−(π/2)ln2
$$\left.\mathrm{1}\right)\:\mathrm{let}\:\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sinx}\right)\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }{\mathrm{2i}}\right)\mathrm{dx} \\ $$$$=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2i}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} \right)\mathrm{dx} \\ $$$$=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{i}\pi}{\mathrm{2}}\right)+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} \right)\mathrm{dx} \\ $$$$=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{i}\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{xdx}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}−\mathrm{e}^{−\mathrm{2x}} \right)\mathrm{dx}\:\:\mathrm{but} \\ $$$$\mathrm{ln}^{'} \left(\mathrm{1}−\mathrm{u}\right)=\frac{−\mathrm{1}}{\mathrm{1}−\mathrm{u}}=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{u}^{\mathrm{n}} \:\Rightarrow\mathrm{ln}\left(\mathrm{1}−\mathrm{u}\right)=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{u}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}} \\ $$$$=−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{u}^{\mathrm{n}} }{\mathrm{n}}\:\Rightarrow\mathrm{ln}\left(\mathrm{1}−\mathrm{e}^{−\mathrm{2x}} \right)\:=−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{2nx}} }{\mathrm{n}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}−\mathrm{e}^{−\mathrm{2x}} \right)\mathrm{dx}\:=−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{2nx}} \:\mathrm{dx} \\ $$$$=−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}\left(−\mathrm{2n}\right)}\:\left[\mathrm{e}^{−\mathrm{2nx}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\left(\left(−\mathrm{1}\right)^{\mathrm{n}} −\mathrm{1}\right) \\ $$$$=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\mathrm{I}=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)−\frac{\mathrm{i}\pi^{\mathrm{2}} }{\mathrm{4}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0}\:\:\mathrm{and}\:\mathrm{I}\:\mathrm{is}\:\mathrm{real}\:\Rightarrow\mathrm{I}\:=−\frac{\pi}{\mathrm{2}}\mathrm{ln2} \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 12/Mar/21
3) let ϕ(a)=−∫_0 ^1 sin(lnx)((x^b −x^a )/(lnx))dx ⇒ ϕ(a)=∫_0 ^1 sin(lnx)((x^a −x^b )/(lnx))dx changement lnx=−t give ϕ(a) =−∫_0 ^∞ sin(−t)((e^(−at) −e^(−bt) )/(−t))(−e^(−t) )dt =∫_0 ^∞ ((sint)/t)(e^(−(a+1)t) −e^(−(b+1)t) )dt ⇒ ϕ^′ (a)=−∫_0 ^∞ sint e^(−(a+1)t) dt =−Im(∫_0 ^∞ e^(it−(a+1)t) dt) we have
$$\left.\mathrm{3}\right)\:\mathrm{let}\:\varphi\left(\mathrm{a}\right)=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{sin}\left(\mathrm{lnx}\right)\frac{\mathrm{x}^{\mathrm{b}} −\mathrm{x}^{\mathrm{a}} }{\mathrm{lnx}}\mathrm{dx}\:\Rightarrow \\ $$$$\varphi\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\left(\mathrm{lnx}\right)\frac{\mathrm{x}^{\mathrm{a}} −\mathrm{x}^{\mathrm{b}} }{\mathrm{lnx}}\mathrm{dx}\:\mathrm{changement}\:\:\mathrm{lnx}=−\mathrm{t}\:\mathrm{give} \\ $$$$\varphi\left(\mathrm{a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\mathrm{sin}\left(−\mathrm{t}\right)\frac{\mathrm{e}^{−\mathrm{at}} −\mathrm{e}^{−\mathrm{bt}} }{−\mathrm{t}}\left(−\mathrm{e}^{−\mathrm{t}} \right)\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sint}}{\mathrm{t}}\left(\mathrm{e}^{−\left(\mathrm{a}+\mathrm{1}\right)\mathrm{t}} −\mathrm{e}^{−\left(\mathrm{b}+\mathrm{1}\right)\mathrm{t}} \right)\mathrm{dt}\:\Rightarrow \\ $$$$\varphi^{'} \left(\mathrm{a}\right)=−\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{sint}\:\mathrm{e}^{−\left(\mathrm{a}+\mathrm{1}\right)\mathrm{t}} \:\mathrm{dt}\:\:=−\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{\mathrm{it}−\left(\mathrm{a}+\mathrm{1}\right)\mathrm{t}} \mathrm{dt}\right)\:\mathrm{we}\:\mathrm{have} \\ $$
Commented by mathmax by abdo last updated on 12/Mar/21
we have ∫_0 ^∞ e^((−(a+1)+i)t) dt =[(1/(−(a+1)+i))e^((−(a+1)+i)t) ]_0 ^∞ =−(1/(−(a+1)+i)) =(1/(a+1−i)) =((a+1+i)/((a+1)^2 +1)) ⇒ ϕ^′ (a)=−(1/((a+1)^2 +1)) ⇒ϕ(a)=−∫ (da/((a+1)^2 +1)) +K ∫ (da/((a+1)^2 +1))=_(a+1=z) ∫ (dz/(z^2 +1)) =arctanz =arctan(a+1) ⇒ ϕ(a)=K−arctan(a+1) ϕ(−1)=K =∫_0 ^∞ ((sint)/t)(1−e^(−(b+1)t) )dt rest to find the value of K ...be continued...
$$\mathrm{we}\:\mathrm{have}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{\left(−\left(\mathrm{a}+\mathrm{1}\right)+\mathrm{i}\right)\mathrm{t}} \:\mathrm{dt}\:=\left[\frac{\mathrm{1}}{−\left(\mathrm{a}+\mathrm{1}\right)+\mathrm{i}}\mathrm{e}^{\left(−\left(\mathrm{a}+\mathrm{1}\right)+\mathrm{i}\right)\mathrm{t}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=−\frac{\mathrm{1}}{−\left(\mathrm{a}+\mathrm{1}\right)+\mathrm{i}}\:=\frac{\mathrm{1}}{\mathrm{a}+\mathrm{1}−\mathrm{i}}\:=\frac{\mathrm{a}+\mathrm{1}+\mathrm{i}}{\left(\mathrm{a}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\varphi^{'} \left(\mathrm{a}\right)=−\frac{\mathrm{1}}{\left(\mathrm{a}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\varphi\left(\mathrm{a}\right)=−\int\:\frac{\mathrm{da}}{\left(\mathrm{a}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}\:+\mathrm{K} \\ $$$$\int\:\:\frac{\mathrm{da}}{\left(\mathrm{a}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}=_{\mathrm{a}+\mathrm{1}=\mathrm{z}} \:\:\:\int\:\:\frac{\mathrm{dz}}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\:=\mathrm{arctanz}\:=\mathrm{arctan}\left(\mathrm{a}+\mathrm{1}\right)\:\Rightarrow \\ $$$$\varphi\left(\mathrm{a}\right)=\mathrm{K}−\mathrm{arctan}\left(\mathrm{a}+\mathrm{1}\right) \\ $$$$\varphi\left(−\mathrm{1}\right)=\mathrm{K}\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sint}}{\mathrm{t}}\left(\mathrm{1}−\mathrm{e}^{−\left(\mathrm{b}+\mathrm{1}\right)\mathrm{t}} \right)\mathrm{dt}\:\:\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{K} \\ $$$$…\mathrm{be}\:\mathrm{continued}… \\ $$
Answered by Ar Brandon last updated on 20/Mar/21
I=∫_0 ^(π/2) ln(sinx)dx I(α)=∫_0 ^(π/2) sin^α xdx ⇒I′(α)=∫_0 ^(π/2) sin^α xln(sinx)dx ⇒∫_0 ^(π/2) ln(sinx)dx=I′(0) I(α)=∫_0 ^(π/2) sin^α xdx=((Γ(((α+1)/2))Γ((1/2)))/(2Γ((α/2)+1))) I′(α)=((√π)/2)∙((Γ((α/2)+1)Γ′(((α+1)/2))−Γ(((α+1)/2))Γ′((α/2)+1))/(Γ^2 ((α/2)+1))) =((√π)/2)∙(1/2)((Γ((α/2)+1)Γ(((α+1)/2))[ψ(((α+1)/2))−ψ((α/2)+1)])/(Γ^2 ((α/2)+1))) I′(0)=((√π)/4)Γ((1/2))[ψ((1/2))−ψ(1)]=(π/4)[(−γ−2ln2)−(−γ)]=−((πln2)/2) ★E......n★
$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sinx}\right)\mathrm{dx} \\ $$$$\mathrm{I}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\alpha} \mathrm{xdx}\:\Rightarrow\mathrm{I}'\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\alpha} \mathrm{xln}\left(\mathrm{sinx}\right)\mathrm{dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sinx}\right)\mathrm{dx}=\mathrm{I}'\left(\mathrm{0}\right) \\ $$$$\mathrm{I}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\alpha} \mathrm{xdx}=\frac{\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$$\mathrm{I}'\left(\alpha\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\centerdot\frac{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)\Gamma'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)−\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma'\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\pi}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\frac{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\left[\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)\right]}{\Gamma^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$$\mathrm{I}'\left(\mathrm{0}\right)=\frac{\sqrt{\pi}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left[\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}\right)\right]=\frac{\pi}{\mathrm{4}}\left[\left(−\gamma−\mathrm{2ln2}\right)−\left(−\gamma\right)\right]=−\frac{\pi\mathrm{ln2}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\bigstar\mathrm{E}……\mathrm{n}\bigstar \\ $$

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