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Question Number 139455 by mnjuly1970 last updated on 27/Apr/21

# calculus# evaluate: 𝛗:=Ξ£_(k=1) ^∞ (((βˆ’1)^(kβˆ’1) Ξ“ ((k/2)))/(k Ξ“(((k+1)/2)))) =?

$$\:\:\:\:\:\:\:#\:{calculus}# \\ $$$$\:\:{evaluate}: \\ $$$$\:\:\boldsymbol{\phi}:=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{k}βˆ’\mathrm{1}} \:\Gamma\:\left(\frac{{k}}{\mathrm{2}}\right)}{{k}\:\Gamma\left(\frac{{k}+\mathrm{1}}{\mathrm{2}}\right)}\:=? \\ $$

Answered by Dwaipayan Shikari last updated on 27/Apr/21

Ξ£_(k=1) ^∞ (((βˆ’1)^(k+1) Ξ“((k/2)))/(kΞ“(((k+1)/2))))=(1/( (βˆšΟ€)))Ξ£_(k=1) ^∞ (βˆ’1)^(k+1) (1/k)∫_0 ^1 x^((k/2)βˆ’1) (1βˆ’x)^(βˆ’(1/2)) dx =(1/( (βˆšΟ€)))∫_0 ^1 ((log(1+(√x)))/(x(√(1βˆ’x))))dx=(2/( (βˆšΟ€)))∫_0 ^1 ((log(1+u))/(u(√(1βˆ’u^2 ))))du=(2/( (βˆšΟ€)))∫_0 ^(Ο€/2) ((log(1+sinΞΈ))/(sinΞΈ))dΞΈ =(2/( (βˆšΟ€)))Ο‚(1) Ο‚(g)=∫_0 ^(Ο€/2) ((log(1+gsinΞΈ))/(sinΞΈ))dΞΈβ‡’Ο‚β€²(g)=∫_0 ^(Ο€/2) (1/(1+gsinΞΈ))dΞΈ =2∫_0 ^1 (1/(1+((2gt)/(1+t^2 )))).(1/(1+t^2 ))dt=2∫_0 ^1 (1/((t+g)^2 +((√(1βˆ’g^2 )))^2 ))dt=(2/( (√(1βˆ’g^2 ))))(tan^(βˆ’1) (√((1+g)/(1βˆ’g)))βˆ’tan^(βˆ’1) (g/( (√(1βˆ’g^2 ))))) Ο‚(g)=2∫(1/( (√(1βˆ’g^2 ))))tan^(βˆ’1) (√((1+g)/(1βˆ’g))) dgβˆ’(2/( (√(1βˆ’g^2 ))))tan^(βˆ’1) (g/( (√(1βˆ’g^2 )))) =2∫(1/( (√(1βˆ’g^2 ))))tan^(βˆ’1) (((1+gβˆ’g)/( (√(1βˆ’g^2 ))))/(1+((g+g^2 )/(1βˆ’g^2 ))))dg=2∫(1/( (√(1βˆ’g^2 ))))tan^(βˆ’1) ((√(1βˆ’g^2 ))/(1+g))dg g=cosΞΆ β‡’2∫((βˆ’sinΞΆ)/(sinΞΆ))tan^(βˆ’1) (tan(ΞΆ/2))dΞΆ=βˆ’(ΞΆ^2 /2)+C=βˆ’(((cos^(βˆ’1) (g))^2 )/2)+C Ο‚(0)=βˆ’(Ο€^2 /8)+C=0β‡’C=(Ο€^2 /8) Ο‚(1)=βˆ’(0/2)+(Ο€^2 /8) So (2/( (βˆšΟ€)))∫_0 ^(Ο€/2) ((log(1+sinΞΈ))/(sinΞΈ))dΞΈ=(2/( (βˆšΟ€))).(Ο€^2 /8)=(Ο€^(3/2) /4)

$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{k}+\mathrm{1}} \Gamma\left(\frac{{k}}{\mathrm{2}}\right)}{{k}\Gamma\left(\frac{{k}+\mathrm{1}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\:\sqrt{\pi}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(βˆ’\mathrm{1}\right)^{{k}+\mathrm{1}} \frac{\mathrm{1}}{{k}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\frac{{k}}{\mathrm{2}}βˆ’\mathrm{1}} \left(\mathrm{1}βˆ’{x}\right)^{βˆ’\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}+\sqrt{{x}}\right)}{{x}\sqrt{\mathrm{1}βˆ’{x}}}{dx}=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}+{u}\right)}{{u}\sqrt{\mathrm{1}βˆ’{u}^{\mathrm{2}} }}{du}=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{log}\left(\mathrm{1}+{sin}\theta\right)}{{sin}\theta}{d}\theta \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\pi}}\varsigma\left(\mathrm{1}\right) \\ $$$$\varsigma\left({g}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{log}\left(\mathrm{1}+{gsin}\theta\right)}{{sin}\theta}{d}\theta\Rightarrow\varsigma'\left({g}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+{gsin}\theta}{d}\theta \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}{gt}}{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left({t}+{g}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{1}βˆ’{g}^{\mathrm{2}} }\right)^{\mathrm{2}} }{dt}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}βˆ’{g}^{\mathrm{2}} }}\left({tan}^{βˆ’\mathrm{1}} \sqrt{\frac{\mathrm{1}+{g}}{\mathrm{1}βˆ’{g}}}βˆ’{tan}^{βˆ’\mathrm{1}} \frac{{g}}{\:\sqrt{\mathrm{1}βˆ’{g}^{\mathrm{2}} }}\right) \\ $$$$\varsigma\left({g}\right)=\mathrm{2}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}βˆ’{g}^{\mathrm{2}} }}{tan}^{βˆ’\mathrm{1}} \sqrt{\frac{\mathrm{1}+{g}}{\mathrm{1}βˆ’{g}}}\:{dg}βˆ’\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}βˆ’{g}^{\mathrm{2}} }}{tan}^{βˆ’\mathrm{1}} \frac{{g}}{\:\sqrt{\mathrm{1}βˆ’{g}^{\mathrm{2}} }} \\ $$$$=\mathrm{2}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}βˆ’{g}^{\mathrm{2}} }}{tan}^{βˆ’\mathrm{1}} \frac{\frac{\mathrm{1}+{g}βˆ’{g}}{\:\sqrt{\mathrm{1}βˆ’{g}^{\mathrm{2}} }}}{\mathrm{1}+\frac{{g}+{g}^{\mathrm{2}} }{\mathrm{1}βˆ’{g}^{\mathrm{2}} }}{dg}=\mathrm{2}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}βˆ’{g}^{\mathrm{2}} }}{tan}^{βˆ’\mathrm{1}} \frac{\sqrt{\mathrm{1}βˆ’{g}^{\mathrm{2}} }}{\mathrm{1}+{g}}{dg} \\ $$$${g}={cos}\zeta\:\:\Rightarrow\mathrm{2}\int\frac{βˆ’{sin}\zeta}{{sin}\zeta}{tan}^{βˆ’\mathrm{1}} \left({tan}\frac{\zeta}{\mathrm{2}}\right){d}\zeta=βˆ’\frac{\zeta^{\mathrm{2}} }{\mathrm{2}}+{C}=βˆ’\frac{\left({cos}^{βˆ’\mathrm{1}} \left({g}\right)\right)^{\mathrm{2}} }{\mathrm{2}}+{C} \\ $$$$\varsigma\left(\mathrm{0}\right)=βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{8}}+{C}=\mathrm{0}\Rightarrow{C}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\varsigma\left(\mathrm{1}\right)=βˆ’\frac{\mathrm{0}}{\mathrm{2}}+\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$${So}\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{log}\left(\mathrm{1}+{sin}\theta\right)}{{sin}\theta}{d}\theta=\frac{\mathrm{2}}{\:\sqrt{\pi}}.\frac{\pi^{\mathrm{2}} }{\mathrm{8}}=\frac{\pi^{\mathrm{3}/\mathrm{2}} }{\mathrm{4}} \\ $$

Commented by mnjuly1970 last updated on 27/Apr/21

thanks alot mr payan...

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

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