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

....advanced calculus.... 𝛗=∫_0 ^( (Ο€/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 ^(Ο€/2) (√(tanx)) dx but not ∫_0 ^(Ο€/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 ^(Ο€/2) x(√(tanx)) dx ∫_0 ^(Ο€/2) (√(tanx)) dx=((Ξ“((3/4))Ξ“((1/4)))/(2Ξ“(1)))=(Ο€/( (√2))) =(Ο€/2)((Ο€/( (√2))))βˆ’βˆ«_0 ^(Ο€/2) ∫(√(tanx)) dx ∫(√(tanx)) dx=2∫(t^2 /(1+t^4 ))dt=(1/( (√i)))tan^(βˆ’1) ((t/( (√i))))βˆ’(1/( (√(βˆ’i))))tan^(βˆ’1) ((t/( (√(βˆ’i))))) =(Ο€^2 /(2(√2)))βˆ’βˆ«_0 ^(Ο€/2) (1/( (√i)))tan^(βˆ’1) ((t/( (√i))))βˆ’(1/( (√(βˆ’i))))tan^(βˆ’1) ((t/( (√(βˆ’i)))))dt =(Ο€^2 /(2(√2)))βˆ’(Ο€/2)(e^(βˆ’(Ο€/4)i) tan^(βˆ’1) (Ο€/(2(√i)))βˆ’e^((Ο€/4)i) tan^(βˆ’1) (Ο€/(2(√(βˆ’i)))))+(1/(2i))log(((4iβˆ’Ο€^2 )/(4i+Ο€^2 )))

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\sqrt{{tanx}}\:{dx}\:\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{tanx}}\:{dx}=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{2}\Gamma\left(\mathrm{1}\right)}=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int\sqrt{{tanx}}\:{dx} \\ $$$$\int\sqrt{{tanx}}\:{dx}=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}=\frac{\mathrm{1}}{\:\sqrt{{i}}}{tan}^{βˆ’\mathrm{1}} \left(\frac{{t}}{\:\sqrt{{i}}}\right)βˆ’\frac{\mathrm{1}}{\:\sqrt{βˆ’{i}}}{tan}^{βˆ’\mathrm{1}} \left(\frac{{t}}{\:\sqrt{βˆ’{i}}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}}βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\:\sqrt{{i}}}{tan}^{βˆ’\mathrm{1}} \left(\frac{{t}}{\:\sqrt{{i}}}\right)βˆ’\frac{\mathrm{1}}{\:\sqrt{βˆ’{i}}}{tan}^{βˆ’\mathrm{1}} \left(\frac{{t}}{\:\sqrt{βˆ’{i}}}\right){dt} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}}βˆ’\frac{\pi}{\mathrm{2}}\left({e}^{βˆ’\frac{\pi}{\mathrm{4}}{i}} {tan}^{βˆ’\mathrm{1}} \frac{\pi}{\mathrm{2}\sqrt{{i}}}βˆ’{e}^{\frac{\pi}{\mathrm{4}}{i}} {tan}^{βˆ’\mathrm{1}} \frac{\pi}{\mathrm{2}\sqrt{βˆ’{i}}}\right)+\frac{\mathrm{1}}{\mathrm{2}{i}}{log}\left(\frac{\mathrm{4}{i}βˆ’\pi^{\mathrm{2}} }{\mathrm{4}{i}+\pi^{\mathrm{2}} }\right) \\ $$

Commented by mnjuly1970 last updated on 24/Mar/21

thanks alot mr payan...grateful..

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

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

Ο†=∫_0 ^(Ο€/2) x(√(tan x))dx=2∫_0 ^∞ (t^2 /(1+t^4 ))tan^(βˆ’1) t^2 dt Ο†(s)=2∫_0 ^∞ (t^2 /(1+t^4 ))tan^(βˆ’1) (st^2 )dt Ο†(s)β€²=2∫_0 ^∞ (t^4 /((1+t^4 )(1+s^2 t^4 )))dt=2∫_0 ^∞ ((1/(1+s^2 t^4 ))βˆ’(1/(1+t^4 )))(1/(1βˆ’s^2 ))dt =(2/(1βˆ’s^2 ))∫_0 ^∞ {(s^(βˆ’1/2) /4)βˆ™(u^(βˆ’3/4) /(1+u))βˆ’(1/4)βˆ™(u^(βˆ’3/4) /(1+u))}du =(1/(2(1βˆ’s^2 )))(s^(βˆ’1/2) βˆ’1)Ξ“((1/4))Ξ“((3/4)) =(Ο€/( (√2)(1βˆ’s^2 )))(s^(βˆ’1/2) βˆ’1) Ο†(1)=(Ο€/( (√2)))∫_0 ^1 ((s^(βˆ’1/2) βˆ’1)/(1βˆ’s^2 ))ds=^(s^(1/2) β†’s) (Ο€/( (√2)))∫_0 ^1 ((1/(1+s))+((1βˆ’s)/(1+s^2 )))ds=(Ο€/( (√2)))((1/2)ln2+(Ο€/4))

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

Commented by mnjuly1970 last updated on 24/Mar/21

thanks alot.... very nice sir ???

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

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

the other people did it.i upload picture,but not display.

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

Commented by mnjuly1970 last updated on 24/Mar/21

mercey ...thank you so much..

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

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

f(a,b)=∫_0 ^Ο€ ((cos(ax))/(sin^b (x)))dx=∫_0 ^Ο€ 2^(bβˆ’1) i^b ((e^(iax) +e^(βˆ’iax) )/((1βˆ’e^(βˆ’2ix) )^b )).e^(βˆ’ibx) dx =2^(bβˆ’1) i^b ∫_0 ^Ο€ ((e^(i(aβˆ’b)x) +e^(βˆ’i(a+b)x) )/((1βˆ’e^(βˆ’2ix) )^b ))dx let g(s)=∫_0 ^Ο€ (e^(isx) /((1βˆ’e^(βˆ’2ix) )^b ))dx= (1/((1βˆ’e^(βˆ’2ix) )^b ))=1+Ξ£_(nβ‰₯1) ((Ξ _(k=0) ^(nβˆ’1) (b+k))/(n!))e^(βˆ’2inx) g(s)=∫_0 ^Ο€ e^(isx) dx+Ξ£_(nβ‰₯1) (((b)_n )/(n!))∫_0 ^Ο€ e^(x(βˆ’2in+is)) dx =((e^(isΟ€) βˆ’1)/(is))+Ξ£_(nβ‰₯1) (((b)_n )/(n!)).((e^(iΟ€s) βˆ’1)/(βˆ’2in+is)) f(a,b)=2^(bβˆ’1) i^b (g(i(aβˆ’b))+g(βˆ’i(a+b))) =2^(bβˆ’1) .i^(bβˆ’1) (((e^(i(aβˆ’b)𝛑) βˆ’1)/(aβˆ’b))+((e^(βˆ’i(a+b)) βˆ’1)/(βˆ’aβˆ’b))) +Ξ£_(nβ‰₯1) (((b)_n )/(n!)).((1βˆ’e^(iΟ€(aβˆ’b)) )/(2n+bβˆ’a))+Ξ£_(nβ‰₯1) (((b)_n )/(n!)).((1βˆ’e^(iΟ€(βˆ’aβˆ’b)) )/(2n+a+b)) =2^(bβˆ’1) i^(bβˆ’1) ((1/(bβˆ’a))+Ξ£_(nβ‰₯1) (((b)_n )/(n!)).(1/(2(n+(((bβˆ’a))/2)))))(1βˆ’e^(iΟ€(aβˆ’b)) )+ 2^(bβˆ’1) i^(bβˆ’1) ((1/(a+b))+Ξ£_(nβ‰₯1) (((b)_n )/(n!)).(1/(2(n+(((a+b))/2)))))(βˆ’e^(βˆ’iΟ€(a+b)) +1) =2^(bβˆ’1) i^(bβˆ’1) ((1/(bβˆ’a))+Ξ£_(nβ‰₯1) (((b)_n (((bβˆ’a)/2))_n )/((aβˆ’b)(((βˆ’a+b+2)/2)))))(βˆ’e^(iΟ€(aβˆ’b)) +1) +2^(bβˆ’1) i^(bβˆ’1) ((1/(a+b))+Ξ£_(nβ‰₯1) (((b)_n (((a+b)/2))_n )/((a+b)(((a+b+2)/2))_n )))(βˆ’e^(βˆ’iΟ€(a+b)) +1) =2^(bβˆ’1) i^(bβˆ’1) ._2 F_1 (b,((bβˆ’a)/2);((bβˆ’a+2)/2);1).((1βˆ’e^(iΟ€(aβˆ’b)) )/(bβˆ’a)) +2^(bβˆ’1) i^(bβˆ’1) ._2 F_1 (b,((a+b)/2);((a+b+2)/2);1).((1βˆ’e^(βˆ’iΟ€(a+b)) )/(a+b)) _2 F_1 (a,b;c,1)=((Ξ“(c)Ξ“(cβˆ’aβˆ’b))/(Ξ“(cβˆ’a)Ξ“(cβˆ’b))) wr get 2^(bβˆ’1) i^(bβˆ’1) (((Ξ“(((bβˆ’a+2)/2))Ξ“(1βˆ’b))/(Ξ“(1βˆ’(a/2)βˆ’(b/2)))))((1βˆ’e^(iΟ€(aβˆ’b)) )/(bβˆ’a)) +2^(bβˆ’1) i^(bβˆ’1) (((Ξ“(((a+b+2)/2))Ξ“(1βˆ’b))/(Ξ“(1+((aβˆ’b)/2))))).((1βˆ’e^(βˆ’i𝛑(a+b)) )/(a+b)) =2^(bβˆ’1) i^(bβˆ’1) .((Ξ“(1βˆ’b)Ξ“(((a+b)/2)))/(Ξ“(1+((aβˆ’b)/2)))).((1βˆ’e^(βˆ’i𝛑(a+b)) )/2)+((Ξ“(1βˆ’b)Ξ“(((bβˆ’a)/2)))/(Ξ“(1βˆ’((a+b)/2)))).((1βˆ’e^(iΟ€(aβˆ’b)) )/2) =2^(bβˆ’1) .i^(bβˆ’1) (((Ξ“(1βˆ’b))/(Ξ“(1+((aβˆ’b)/2))Ξ“(1βˆ’((a+b)/2)))).iΟ€e^(βˆ’i(Ο€/2)(a+b)) ) +2^(bβˆ’1) i^(bβˆ’1) (((Ξ“(1βˆ’b))/(Ξ“(1+((aβˆ’b)/2))Ξ“(1βˆ’((a+b)/2)))).iΟ€e^(i(Ο€/2)(((aβˆ’b)/2))) ) =2^(bβˆ’1) .((Ξ“(1βˆ’b).2Ο€cos(((Ο€a)/2)))/(Ξ“(1+((aβˆ’b)/2))Ξ“(1βˆ’((a+b)/2))))=Ο€((2^b cos(((Ο€a)/2))Ξ“(1βˆ’b))/(Ξ“(1+((aβˆ’b)/2))Ξ“(1βˆ’((a+b)/2)))) Ο†=∫_0 ^(Ο€/2) x(√(tg(x)))dx=(√2)∫_0 ^Ο€ ((xsin(x))/( (√(sin(2x)))))dx =(1/(2(√2)))∫_0 ^(Ο€/2) ((ysin((y/2)))/( (√(sin(y))))) f(a,b)=∫_0 ^Ο€ ((cos(ax))/(sin^b (x)))dx,βˆ‚^a f(a,b)=∫_0 ^Ο€ ((βˆ’xsin(ax))/(sin^b (x)))dx (a,b)=((1/2);(1/2)) we get Ο†=βˆ’(1/(2(√2)))(βˆ‚f/βˆ‚a)((1/2);(1/2))=(Ο€/(4sin((Ο€/4))))(Ξ¨((1/2))βˆ’Ξ¨((1/4))) =(Ο€/(2(√2)))(βˆ’2ln(2)βˆ’π›„+(Ο€/2)+3ln(2)+Ξ³)=(Ο€^2 /(4(√2)))+((Ο€ln(2))/(2(√2)))=∫_0 ^(Ο€/2) x(√(tan(x)))dx

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

Commented by mnjuly1970 last updated on 24/Mar/21

thank you so much ..

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

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