Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 120593 by snipers237 last updated on 01/Nov/20

   lim_(z→0)   ((Γ(z)+Γ(−z))/2) =^?  −γ

$$ \\ $$$$\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\Gamma\left({z}\right)+\Gamma\left(−{z}\right)}{\mathrm{2}}\:\overset{?} {=}\:−\gamma \\ $$

Answered by mnjuly1970 last updated on 02/Nov/20

solution:      note:=  Γ(z+1)=zΓ(z)        note:=ψ(z)=(d/dz)(ln(Γ(z))       lim_(z→0) (((Γ(z)+Γ(−z))/2))        =lim_(z→0) ((1/z)(((zΓ(z)+zΓ(−z))/2)))  =lim_(z→0)  (1/2)(((Γ(z+1)−Γ(1−z))/z))     =^(hopital rule) lim_(z→0) (1/2)(((Γ′(z+1)+Γ′(1−z))/1))      =(1/2)(Γ′(1)+Γ^′ (1))=Γ′(1)       =Γ(1)ψ(1)=−γ : euler constant.               ...♣m.n.july.1970♣...

$${solution}: \\ $$$$\:\:\:\:{note}:=\:\:\Gamma\left({z}+\mathrm{1}\right)={z}\Gamma\left({z}\right)\: \\ $$$$\:\:\:\:\:{note}:=\psi\left({z}\right)=\frac{{d}}{{dz}}\left({ln}\left(\Gamma\left({z}\right)\right)\right. \\ $$$$\:\:\:\:\:{lim}_{{z}\rightarrow\mathrm{0}} \left(\frac{\Gamma\left({z}\right)+\Gamma\left(−{z}\right)}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:={lim}_{{z}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}}{{z}}\left(\frac{{z}\Gamma\left({z}\right)+{z}\Gamma\left(−{z}\right)}{\mathrm{2}}\right)\right) \\ $$$$={lim}_{{z}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\Gamma\left({z}+\mathrm{1}\right)−\Gamma\left(\mathrm{1}−{z}\right)}{{z}}\right)\:\: \\ $$$$\:\overset{{hopital}\:{rule}} {=}{lim}_{{z}\rightarrow\mathrm{0}} \frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\Gamma'\left({z}+\mathrm{1}\right)+\Gamma'\left(\mathrm{1}−{z}\right)}{\mathrm{1}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\Gamma'\left(\mathrm{1}\right)+\Gamma^{'} \left(\mathrm{1}\right)\right)=\Gamma'\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:=\Gamma\left(\mathrm{1}\right)\psi\left(\mathrm{1}\right)=−\gamma\::\:{euler}\:{constant}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:...\clubsuit{m}.{n}.{july}.\mathrm{1970}\clubsuit... \\ $$$$ \\ $$

Commented by I want to learn more last updated on 02/Nov/20

Help sir.

$$\mathrm{Help}\:\mathrm{sir}. \\ $$

Commented by I want to learn more last updated on 02/Nov/20

Commented by Dwaipayan Shikari last updated on 03/Nov/20

∫(5x^4 +4x^3 −x^2 )^(−(1/2)) dx  =∫(1/( (√(5x^4 +4x^3 −x^2 ))))dx  =∫(1/(x(√(5x^2 +4x−1))))dx=∫(1/(x(√((5x−1)(x+1)))))dx  =∫((1/x^2 )/( (√((5−(1/x))(1+(1/x))))))dx    t=1+(1/x)  =−∫(dt/( (√(t(6−t)))))  =−∫(dt/( (√(9−(t−3)^2 ))))               t−3=3sin𝛉⇒1=3cos𝛉(d𝛉/dt)  =−∫((3cos𝛉d𝛉)/( (√(9−9sin^2 𝛉))))  =−∫d𝛉=−𝛉+C  =−sin^(−1) (((t−3)/3))+C  =−sin^(−1) (((1+(1/x)−3)/3))+c  =−sin^(−1) (((−2x+1)/(3x)))+c

$$\int\left(\mathrm{5}{x}^{\mathrm{4}} +\mathrm{4}\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{{x}}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \boldsymbol{{dx}} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{4}\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{{x}}^{\mathrm{2}} }}\boldsymbol{{dx}} \\ $$$$=\int\frac{\mathrm{1}}{\boldsymbol{{x}}\sqrt{\mathrm{5}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{x}}−\mathrm{1}}}\boldsymbol{{dx}}=\int\frac{\mathrm{1}}{\boldsymbol{{x}}\sqrt{\left(\mathrm{5}\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}+\mathrm{1}\right)}}\boldsymbol{{dx}} \\ $$$$=\int\frac{\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} }}{\:\sqrt{\left(\mathrm{5}−\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)}}\boldsymbol{{dx}}\:\:\:\:\boldsymbol{{t}}=\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{x}}} \\ $$$$=−\int\frac{\boldsymbol{{dt}}}{\:\sqrt{\boldsymbol{{t}}\left(\mathrm{6}−\boldsymbol{{t}}\right)}} \\ $$$$=−\int\frac{\boldsymbol{{dt}}}{\:\sqrt{\mathrm{9}−\left(\boldsymbol{{t}}−\mathrm{3}\right)^{\mathrm{2}} }}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{t}}−\mathrm{3}=\mathrm{3}\boldsymbol{{sin}\theta}\Rightarrow\mathrm{1}=\mathrm{3}\boldsymbol{{cos}\theta}\frac{\boldsymbol{{d}\theta}}{\boldsymbol{{dt}}} \\ $$$$=−\int\frac{\mathrm{3}\boldsymbol{{cos}\theta}{d}\boldsymbol{\theta}}{\:\sqrt{\mathrm{9}−\mathrm{9}\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\theta}}} \\ $$$$=−\int\boldsymbol{{d}\theta}=−\boldsymbol{\theta}+{C} \\ $$$$=−\boldsymbol{{sin}}^{−\mathrm{1}} \left(\frac{\boldsymbol{{t}}−\mathrm{3}}{\mathrm{3}}\right)+\boldsymbol{{C}} \\ $$$$=−\boldsymbol{{sin}}^{−\mathrm{1}} \left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{x}}}−\mathrm{3}}{\mathrm{3}}\right)+\boldsymbol{{c}} \\ $$$$=−\boldsymbol{{sin}}^{−\mathrm{1}} \left(\frac{−\mathrm{2}\boldsymbol{{x}}+\mathrm{1}}{\mathrm{3}\boldsymbol{{x}}}\right)+\boldsymbol{{c}} \\ $$

Commented by I want to learn more last updated on 03/Nov/20

Wow, thanks sir, i appreciate.

$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$

Commented by Dwaipayan Shikari last updated on 03/Nov/20

With pleasure bro. I am not a sir!

$${With}\:{pleasure}\:{bro}.\:{I}\:{am}\:{not}\:{a}\:{sir}! \\ $$

Commented by I want to learn more last updated on 09/Nov/20

Ohh, ma

$$\mathrm{Ohh},\:\mathrm{ma} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com