advanced-calculus-0-pi-2-x-tan-x-1-2-dx-

Question Number 136572 by mnjuly1970 last updated on 23/Mar/21

....advanced calculus.... 𝛗=∫_0 ^( (π/2)) x.(tan(x))^(1/2) dx=??

$$\:\:\:\:\:\:\:\:\:\:\:\:\:….{advanced}\:\:\:\:{calculus}…. \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {x}.\left({tan}\left({x}\right)\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx}=?? \\ $$$$ \\ $$

Commented by Olaf last updated on 24/Mar/21

Do you know if it can be calculated sir ? I can calculate ∫_0 ^(π/2) (√(tanx)) dx but not ∫_0 ^(π/2) x(√(tanx)) dx

$$\mathrm{Do}\:\mathrm{you}\:\mathrm{know}\:\mathrm{if}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{calculated} \\ $$$$\mathrm{sir}\:?\:\mathrm{I}\:\mathrm{can}\:\mathrm{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{tan}{x}}\:{dx} \\ $$$$\mathrm{but}\:\mathrm{not}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\sqrt{\mathrm{tan}{x}}\:{dx} \\ $$

Answered by Dwaipayan Shikari last updated on 24/Mar/21

∫_0 ^(π/2) x(√(tanx)) dx ∫_0 ^(π/2) (√(tanx)) dx=((Γ((3/4))Γ((1/4)))/(2Γ(1)))=(π/( (√2))) =(π/2)((π/( (√2))))−∫_0 ^(π/2) ∫(√(tanx)) dx ∫(√(tanx)) dx=2∫(t^2 /(1+t^4 ))dt=(1/( (√i)))tan^(−1) ((t/( (√i))))−(1/( (√(−i))))tan^(−1) ((t/( (√(−i))))) =(π^2 /(2(√2)))−∫_0 ^(π/2) (1/( (√i)))tan^(−1) ((t/( (√i))))−(1/( (√(−i))))tan^(−1) ((t/( (√(−i)))))dt =(π^2 /(2(√2)))−(π/2)(e^(−(π/4)i) tan^(−1) (π/(2(√i)))−e^((π/4)i) tan^(−1) (π/(2(√(−i)))))+(1/(2i))log(((4i−π^2 )/(4i+π^2 )))

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\sqrt{{tanx}}\:{dx}\:\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{tanx}}\:{dx}=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{2}\Gamma\left(\mathrm{1}\right)}=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int\sqrt{{tanx}}\:{dx} \\ $$$$\int\sqrt{{tanx}}\:{dx}=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}=\frac{\mathrm{1}}{\:\sqrt{{i}}}{tan}^{−\mathrm{1}} \left(\frac{{t}}{\:\sqrt{{i}}}\right)−\frac{\mathrm{1}}{\:\sqrt{−{i}}}{tan}^{−\mathrm{1}} \left(\frac{{t}}{\:\sqrt{−{i}}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\:\sqrt{{i}}}{tan}^{−\mathrm{1}} \left(\frac{{t}}{\:\sqrt{{i}}}\right)−\frac{\mathrm{1}}{\:\sqrt{−{i}}}{tan}^{−\mathrm{1}} \left(\frac{{t}}{\:\sqrt{−{i}}}\right){dt} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{2}}\left({e}^{−\frac{\pi}{\mathrm{4}}{i}} {tan}^{−\mathrm{1}} \frac{\pi}{\mathrm{2}\sqrt{{i}}}−{e}^{\frac{\pi}{\mathrm{4}}{i}} {tan}^{−\mathrm{1}} \frac{\pi}{\mathrm{2}\sqrt{−{i}}}\right)+\frac{\mathrm{1}}{\mathrm{2}{i}}{log}\left(\frac{\mathrm{4}{i}−\pi^{\mathrm{2}} }{\mathrm{4}{i}+\pi^{\mathrm{2}} }\right) \\ $$

Commented by mnjuly1970 last updated on 24/Mar/21

$${thanks}\:{alot}\:{mr}\:{payan}…{grateful}.. \\ $$

Answered by Ñï= last updated on 24/Mar/21

φ=∫_0 ^(π/2) x(√(tan x))dx=2∫_0 ^∞ (t^2 /(1+t^4 ))tan^(−1) t^2 dt φ(s)=2∫_0 ^∞ (t^2 /(1+t^4 ))tan^(−1) (st^2 )dt φ(s)′=2∫_0 ^∞ (t^4 /((1+t^4 )(1+s^2 t^4 )))dt=2∫_0 ^∞ ((1/(1+s^2 t^4 ))−(1/(1+t^4 )))(1/(1−s^2 ))dt =(2/(1−s^2 ))∫_0 ^∞ {(s^(−1/2) /4)∙(u^(−3/4) /(1+u))−(1/4)∙(u^(−3/4) /(1+u))}du =(1/(2(1−s^2 )))(s^(−1/2) −1)Γ((1/4))Γ((3/4)) =(π/( (√2)(1−s^2 )))(s^(−1/2) −1) φ(1)=(π/( (√2)))∫_0 ^1 ((s^(−1/2) −1)/(1−s^2 ))ds=^(s^(1/2) →s) (π/( (√2)))∫_0 ^1 ((1/(1+s))+((1−s)/(1+s^2 )))ds=(π/( (√2)))((1/2)ln2+(π/4))

$$\phi=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {x}\sqrt{\mathrm{tan}\:{x}}{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\mathrm{tan}^{−\mathrm{1}} {t}^{\mathrm{2}} {dt} \\ $$$$\phi\left({s}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\mathrm{tan}\:^{−\mathrm{1}} \left({st}^{\mathrm{2}} \right){dt} \\ $$$$\phi\left({s}\right)'=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{4}} }{\left(\mathrm{1}+{t}^{\mathrm{4}} \right)\left(\mathrm{1}+{s}^{\mathrm{2}} {t}^{\mathrm{4}} \right)}{dt}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{1}+{s}^{\mathrm{2}} {t}^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }\right)\frac{\mathrm{1}}{\mathrm{1}−{s}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}−{s}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \left\{\frac{{s}^{−\mathrm{1}/\mathrm{2}} }{\mathrm{4}}\centerdot\frac{{u}^{−\mathrm{3}/\mathrm{4}} }{\mathrm{1}+{u}}−\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{{u}^{−\mathrm{3}/\mathrm{4}} }{\mathrm{1}+{u}}\right\}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−{s}^{\mathrm{2}} \right)}\left({s}^{−\mathrm{1}/\mathrm{2}} −\mathrm{1}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}−{s}^{\mathrm{2}} \right)}\left({s}^{−\mathrm{1}/\mathrm{2}} −\mathrm{1}\right) \\ $$$$\phi\left(\mathrm{1}\right)=\frac{\pi}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{s}^{−\mathrm{1}/\mathrm{2}} −\mathrm{1}}{\mathrm{1}−{s}^{\mathrm{2}} }{ds}\overset{{s}^{\mathrm{1}/\mathrm{2}} \rightarrow{s}} {=}\frac{\pi}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{s}}+\frac{\mathrm{1}−{s}}{\mathrm{1}+{s}^{\mathrm{2}} }\right){ds}=\frac{\pi}{\:\sqrt{\mathrm{2}}}\left(\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}+\frac{\pi}{\mathrm{4}}\right) \\ $$

Commented by mnjuly1970 last updated on 24/Mar/21

$$\:\:\:{thanks}\:{alot}…. \\ $$$${very}\:{nice}\:{sir}\:??? \\ $$

Commented by Ñï= last updated on 24/Mar/21

$${the}\:{other}\:{people}\:{did}\:{it}.{i}\:{upload}\:{picture},{but}\:\:{not}\:{display}. \\ $$

Commented by mnjuly1970 last updated on 24/Mar/21

$$\:\:\:\:{mercey}\:…{thank}\:{you}\:{so}\:{much}.. \\ $$

Answered by maths mind last updated on 24/Mar/21

f(a,b)=∫_0 ^π ((cos(ax))/(sin^b (x)))dx=∫_0 ^π 2^(b−1) i^b ((e^(iax) +e^(−iax) )/((1−e^(−2ix) )^b )).e^(−ibx) dx =2^(b−1) i^b ∫_0 ^π ((e^(i(a−b)x) +e^(−i(a+b)x) )/((1−e^(−2ix) )^b ))dx let g(s)=∫_0 ^π (e^(isx) /((1−e^(−2ix) )^b ))dx= (1/((1−e^(−2ix) )^b ))=1+Σ_(n≥1) ((Π_(k=0) ^(n−1) (b+k))/(n!))e^(−2inx) g(s)=∫_0 ^π e^(isx) dx+Σ_(n≥1) (((b)_n )/(n!))∫_0 ^π e^(x(−2in+is)) dx =((e^(isπ) −1)/(is))+Σ_(n≥1) (((b)_n )/(n!)).((e^(iπs) −1)/(−2in+is)) f(a,b)=2^(b−1) i^b (g(i(a−b))+g(−i(a+b))) =2^(b−1) .i^(b−1) (((e^(i(a−b)𝛑) −1)/(a−b))+((e^(−i(a+b)) −1)/(−a−b))) +Σ_(n≥1) (((b)_n )/(n!)).((1−e^(iπ(a−b)) )/(2n+b−a))+Σ_(n≥1) (((b)_n )/(n!)).((1−e^(iπ(−a−b)) )/(2n+a+b)) =2^(b−1) i^(b−1) ((1/(b−a))+Σ_(n≥1) (((b)_n )/(n!)).(1/(2(n+(((b−a))/2)))))(1−e^(iπ(a−b)) )+ 2^(b−1) i^(b−1) ((1/(a+b))+Σ_(n≥1) (((b)_n )/(n!)).(1/(2(n+(((a+b))/2)))))(−e^(−iπ(a+b)) +1) =2^(b−1) i^(b−1) ((1/(b−a))+Σ_(n≥1) (((b)_n (((b−a)/2))_n )/((a−b)(((−a+b+2)/2)))))(−e^(iπ(a−b)) +1) +2^(b−1) i^(b−1) ((1/(a+b))+Σ_(n≥1) (((b)_n (((a+b)/2))_n )/((a+b)(((a+b+2)/2))_n )))(−e^(−iπ(a+b)) +1) =2^(b−1) i^(b−1) ._2 F_1 (b,((b−a)/2);((b−a+2)/2);1).((1−e^(iπ(a−b)) )/(b−a)) +2^(b−1) i^(b−1) ._2 F_1 (b,((a+b)/2);((a+b+2)/2);1).((1−e^(−iπ(a+b)) )/(a+b)) _2 F_1 (a,b;c,1)=((Γ(c)Γ(c−a−b))/(Γ(c−a)Γ(c−b))) wr get 2^(b−1) i^(b−1) (((Γ(((b−a+2)/2))Γ(1−b))/(Γ(1−(a/2)−(b/2)))))((1−e^(iπ(a−b)) )/(b−a)) +2^(b−1) i^(b−1) (((Γ(((a+b+2)/2))Γ(1−b))/(Γ(1+((a−b)/2))))).((1−e^(−i𝛑(a+b)) )/(a+b)) =2^(b−1) i^(b−1) .((Γ(1−b)Γ(((a+b)/2)))/(Γ(1+((a−b)/2)))).((1−e^(−i𝛑(a+b)) )/2)+((Γ(1−b)Γ(((b−a)/2)))/(Γ(1−((a+b)/2)))).((1−e^(iπ(a−b)) )/2) =2^(b−1) .i^(b−1) (((Γ(1−b))/(Γ(1+((a−b)/2))Γ(1−((a+b)/2)))).iπe^(−i(π/2)(a+b)) ) +2^(b−1) i^(b−1) (((Γ(1−b))/(Γ(1+((a−b)/2))Γ(1−((a+b)/2)))).iπe^(i(π/2)(((a−b)/2))) ) =2^(b−1) .((Γ(1−b).2πcos(((πa)/2)))/(Γ(1+((a−b)/2))Γ(1−((a+b)/2))))=π((2^b cos(((πa)/2))Γ(1−b))/(Γ(1+((a−b)/2))Γ(1−((a+b)/2)))) φ=∫_0 ^(π/2) x(√(tg(x)))dx=(√2)∫_0 ^π ((xsin(x))/( (√(sin(2x)))))dx =(1/(2(√2)))∫_0 ^(π/2) ((ysin((y/2)))/( (√(sin(y))))) f(a,b)=∫_0 ^π ((cos(ax))/(sin^b (x)))dx,∂^a f(a,b)=∫_0 ^π ((−xsin(ax))/(sin^b (x)))dx (a,b)=((1/2);(1/2)) we get φ=−(1/(2(√2)))(∂f/∂a)((1/2);(1/2))=(π/(4sin((π/4))))(Ψ((1/2))−Ψ((1/4))) =(π/(2(√2)))(−2ln(2)−𝛄+(π/2)+3ln(2)+γ)=(π^2 /(4(√2)))+((πln(2))/(2(√2)))=∫_0 ^(π/2) x(√(tan(x)))dx

$${f}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\pi} \frac{{cos}\left({ax}\right)}{{sin}^{{b}} \left({x}\right)}{dx}=\int_{\mathrm{0}} ^{\pi} \mathrm{2}^{\boldsymbol{{b}}−\mathrm{1}} \boldsymbol{{i}}^{\boldsymbol{{b}}} \frac{\boldsymbol{{e}}^{\boldsymbol{{iax}}} +\boldsymbol{{e}}^{−\boldsymbol{{iax}}} }{\left(\mathrm{1}−\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{ix}}} \right)^{\boldsymbol{{b}}} }.\boldsymbol{{e}}^{−\boldsymbol{{ibx}}} \boldsymbol{{dx}} \\ $$$$=\mathrm{2}^{\boldsymbol{{b}}−\mathrm{1}} \boldsymbol{{i}}^{\boldsymbol{{b}}} \int_{\mathrm{0}} ^{\pi} \frac{{e}^{{i}\left({a}−{b}\right){x}} +{e}^{−{i}\left({a}+{b}\right){x}} }{\left(\mathrm{1}−{e}^{−\mathrm{2}{ix}} \right)^{{b}} }{dx} \\ $$$${let}\:{g}\left({s}\right)=\int_{\mathrm{0}} ^{\pi} \frac{{e}^{{isx}} }{\left(\mathrm{1}−{e}^{−\mathrm{2}{ix}} \right)^{{b}} }{dx}= \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{e}^{−\mathrm{2}{ix}} \right)^{{b}} }=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({b}+{k}\right)}{{n}!}{e}^{−\mathrm{2}{inx}} \\ $$$${g}\left({s}\right)=\int_{\mathrm{0}} ^{\pi} {e}^{{isx}} {dx}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({b}\right)_{{n}} }{{n}!}\int_{\mathrm{0}} ^{\pi} {e}^{{x}\left(−\mathrm{2}{in}+{is}\right)} {dx} \\ $$$$=\frac{{e}^{{is}\pi} −\mathrm{1}}{{is}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({b}\right)_{{n}} }{{n}!}.\frac{{e}^{{i}\pi{s}} −\mathrm{1}}{−\mathrm{2}{in}+{is}} \\ $$$${f}\left({a},{b}\right)=\mathrm{2}^{{b}−\mathrm{1}} {i}^{{b}} \left({g}\left({i}\left({a}−{b}\right)\right)+{g}\left(−{i}\left({a}+{b}\right)\right)\right) \\ $$$$=\mathrm{2}^{{b}−\mathrm{1}} .{i}^{{b}−\mathrm{1}} \left(\frac{\boldsymbol{{e}}^{\boldsymbol{{i}}\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)\boldsymbol{\pi}} −\mathrm{1}}{{a}−\boldsymbol{{b}}}+\frac{{e}^{−{i}\left({a}+{b}\right)} −\mathrm{1}}{−{a}−{b}}\right) \\ $$$$+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({b}\right)_{{n}} }{{n}!}.\frac{\mathrm{1}−{e}^{{i}\pi\left({a}−\boldsymbol{{b}}\right)} }{\mathrm{2}\boldsymbol{{n}}+\boldsymbol{{b}}−{a}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({b}\right)_{{n}} }{{n}!}.\frac{\mathrm{1}−{e}^{{i}\pi\left(−{a}−{b}\right)} }{\mathrm{2}{n}+{a}+{b}} \\ $$$$=\mathrm{2}^{{b}−\mathrm{1}} {i}^{{b}−\mathrm{1}} \left(\frac{\mathrm{1}}{{b}−{a}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({b}\right)_{{n}} }{{n}!}.\frac{\mathrm{1}}{\mathrm{2}\left({n}+\frac{\left({b}−{a}\right)}{\mathrm{2}}\right)}\right)\left(\mathrm{1}−{e}^{{i}\pi\left({a}−{b}\right)} \right)+ \\ $$$$\mathrm{2}^{{b}−\mathrm{1}} {i}^{{b}−\mathrm{1}} \left(\frac{\mathrm{1}}{{a}+{b}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({b}\right)_{{n}} }{{n}!}.\frac{\mathrm{1}}{\mathrm{2}\left({n}+\frac{\left({a}+{b}\right)}{\mathrm{2}}\right)}\right)\left(−{e}^{−{i}\pi\left({a}+{b}\right)} +\mathrm{1}\right) \\ $$$$=\mathrm{2}^{{b}−\mathrm{1}} {i}^{{b}−\mathrm{1}} \left(\frac{\mathrm{1}}{{b}−{a}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({b}\right)_{{n}} \left(\frac{{b}−{a}}{\mathrm{2}}\right)_{{n}} }{\left({a}−{b}\right)\left(\frac{−{a}+{b}+\mathrm{2}}{\mathrm{2}}\right)}\right)\left(−{e}^{{i}\pi\left({a}−{b}\right)} +\mathrm{1}\right) \\ $$$$+\mathrm{2}^{{b}−\mathrm{1}} {i}^{{b}−\mathrm{1}} \left(\frac{\mathrm{1}}{{a}+{b}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({b}\right)_{{n}} \left(\frac{{a}+{b}}{\mathrm{2}}\right)_{{n}} }{\left({a}+{b}\right)\left(\frac{{a}+{b}+\mathrm{2}}{\mathrm{2}}\right)_{{n}} }\right)\left(−{e}^{−{i}\pi\left({a}+{b}\right)} +\mathrm{1}\right) \\ $$$$=\mathrm{2}^{{b}−\mathrm{1}} {i}^{{b}−\mathrm{1}} ._{\mathrm{2}} {F}_{\mathrm{1}} \left({b},\frac{{b}−{a}}{\mathrm{2}};\frac{{b}−{a}+\mathrm{2}}{\mathrm{2}};\mathrm{1}\right).\frac{\mathrm{1}−{e}^{{i}\pi\left({a}−{b}\right)} }{{b}−{a}} \\ $$$$+\mathrm{2}^{{b}−\mathrm{1}} {i}^{{b}−\mathrm{1}} ._{\mathrm{2}} {F}_{\mathrm{1}} \left({b},\frac{{a}+{b}}{\mathrm{2}};\frac{{a}+{b}+\mathrm{2}}{\mathrm{2}};\mathrm{1}\right).\frac{\mathrm{1}−{e}^{−{i}\pi\left({a}+{b}\right)} }{{a}+{b}} \\ $$$$\:_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b};{c},\mathrm{1}\right)=\frac{\Gamma\left({c}\right)\Gamma\left({c}−{a}−{b}\right)}{\Gamma\left({c}−{a}\right)\Gamma\left({c}−{b}\right)} \\ $$$${wr}\:{get}\:\mathrm{2}^{{b}−\mathrm{1}} {i}^{{b}−\mathrm{1}} \left(\frac{\Gamma\left(\frac{{b}−{a}+\mathrm{2}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−{b}\right)}{\Gamma\left(\mathrm{1}−\frac{{a}}{\mathrm{2}}−\frac{{b}}{\mathrm{2}}\right)}\right)\frac{\mathrm{1}−{e}^{{i}\pi\left({a}−{b}\right)} }{{b}−{a}} \\ $$$$+\mathrm{2}^{{b}−\mathrm{1}} {i}^{{b}−\mathrm{1}} \left(\frac{\Gamma\left(\frac{{a}+{b}+\mathrm{2}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−{b}\right)}{\Gamma\left(\mathrm{1}+\frac{{a}−{b}}{\mathrm{2}}\right)}\right).\frac{\mathrm{1}−\boldsymbol{{e}}^{−\boldsymbol{{i}\pi}\left({a}+{b}\right)} }{{a}+{b}} \\ $$$$=\mathrm{2}^{{b}−\mathrm{1}} {i}^{{b}−\mathrm{1}} .\frac{\Gamma\left(\mathrm{1}−{b}\right)\Gamma\left(\frac{{a}+{b}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{1}+\frac{{a}−{b}}{\mathrm{2}}\right)}.\frac{\mathrm{1}−{e}^{−\boldsymbol{{i}\pi}\left({a}+{b}\right)} }{\mathrm{2}}+\frac{\Gamma\left(\mathrm{1}−{b}\right)\Gamma\left(\frac{{b}−{a}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{1}−\frac{{a}+{b}}{\mathrm{2}}\right)}.\frac{\mathrm{1}−{e}^{{i}\pi\left({a}−{b}\right)} }{\mathrm{2}} \\ $$$$=\mathrm{2}^{{b}−\mathrm{1}} .{i}^{{b}−\mathrm{1}} \left(\frac{\Gamma\left(\mathrm{1}−{b}\right)}{\Gamma\left(\mathrm{1}+\frac{{a}−{b}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\frac{{a}+{b}}{\mathrm{2}}\right)}.{i}\pi{e}^{−{i}\frac{\pi}{\mathrm{2}}\left({a}+{b}\right)} \right) \\ $$$$+\mathrm{2}^{{b}−\mathrm{1}} {i}^{{b}−\mathrm{1}} \left(\frac{\Gamma\left(\mathrm{1}−{b}\right)}{\Gamma\left(\mathrm{1}+\frac{{a}−{b}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\frac{{a}+{b}}{\mathrm{2}}\right)}.{i}\pi{e}^{{i}\frac{\pi}{\mathrm{2}}\left(\frac{{a}−{b}}{\mathrm{2}}\right)} \right) \\ $$$$=\mathrm{2}^{{b}−\mathrm{1}} .\frac{\Gamma\left(\mathrm{1}−{b}\right).\mathrm{2}\pi{cos}\left(\frac{\pi{a}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{1}+\frac{{a}−{b}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\frac{{a}+{b}}{\mathrm{2}}\right)}=\pi\frac{\mathrm{2}^{{b}} {cos}\left(\frac{\pi{a}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−{b}\right)}{\Gamma\left(\mathrm{1}+\frac{{a}−{b}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\frac{{a}+{b}}{\mathrm{2}}\right)} \\ $$$$\phi=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\sqrt{{tg}\left({x}\right)}{dx}=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{{xsin}\left({x}\right)}{\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ysin}\left(\frac{{y}}{\mathrm{2}}\right)}{\:\sqrt{{sin}\left({y}\right)}} \\ $$$$ \\ $$$${f}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\pi} \frac{{cos}\left({ax}\right)}{{sin}^{{b}} \left({x}\right)}{dx},\partial^{{a}} {f}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\pi} \frac{−{xsin}\left({ax}\right)}{{sin}^{{b}} \left({x}\right)}{dx} \\ $$$$\left({a},{b}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}}\right)\:{we}\:{get}\:\phi=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\frac{\partial{f}}{\partial{a}}\left(\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{4}}\right)}\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right) \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\left(−\mathrm{2}{ln}\left(\mathrm{2}\right)−\boldsymbol{\gamma}+\frac{\pi}{\mathrm{2}}+\mathrm{3}{ln}\left(\mathrm{2}\right)+\gamma\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{2}}}+\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}\sqrt{\mathrm{2}}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\sqrt{{tan}\left({x}\right)}{dx} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 24/Mar/21

$${thank}\:{you}\:{so}\:{much}\:.. \\ $$

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