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Question Number 110118 by mathdave last updated on 27/Aug/20
prove that ∫_0 ^1 Γ(1−(x/2))Γ(1+(x/2))dx=(4/π)G where G(catalan constant)
$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \Gamma\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right){dx}=\frac{\mathrm{4}}{\pi}{G} \\ $$$${where}\:{G}\left({catalan}\:{constant}\right) \\ $$
Commented by Sarah85 last updated on 27/Aug/20
please show how to get ∫_0 ^π sin (a sin (x)) dx=πH_0 (a) where H_0 (a) is the Struve−H−Function
$$ \\ $$$$\mathrm{please}\:\mathrm{show}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get} \\ $$$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{sin}\:\left({a}\:\mathrm{sin}\:\left({x}\right)\right)\:{dx}=\pi\mathrm{H}_{\mathrm{0}} \:\left({a}\right) \\ $$$$\mathrm{where}\:\mathrm{H}_{\mathrm{0}} \:\left({a}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{Struve}−\mathrm{H}−\mathrm{Function} \\ $$
Commented by maths mind last updated on 30/Aug/20
H_t (a)=Σ_(m≥0) (((−1)^m )/(Γ(m+(3/2))Γ(m+t+(3/2))))((z/2))^(2m+t+1) sin(asin(x))=Σ_(k≥0) (−1)^k (((asin(x))^((2k+1)) )/((2k+1)!)) ∫_0 ^π sin^m (x)dx=∫_0 ^(π/2) {sin^m (x)+cos^m (x)}dx =2∫_0 ^(π/2) sin^m (x)dx=β(((m+1)/2),(1/2))=((Γ(((m+1)/2))Γ((1/2)))/(Γ((m/2)+1))) =Σ_(k≥0) (((−1)^k Γ(k+1)Γ((1/2)))/(Γ(2k+2).Γ(k+(3/2))))a^(2k+1) Γ(2k+2)=Π_(j=1) ^k (2j).Π_(j=0) ^k (2j+1) =2^k Γ(k+1).2^(k+1) Γ((k/2)+(3/2))=2^(2k+1) Γ(k+1)Γ(((k+3)/2)) =Σ_(k≥0) (((−1)^k Γ(k+1)Γ((1/2)))/(2^(2k+1) Γ(k+1)Γ(((k+3)/2)).Γ(((k+3)/2))))a^(2k+1) =Γ((1/2))Σ_(k≥0) (((−1)^k )/(Γ((k/2)+0+(3/2))Γ((k/2)+(3/2)))).((a/2))^(2k+1+0) _(=H_0 (a)) =(√π).H_0 (a)
$$ \\ $$$${H}_{{t}} \left({a}\right)=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{m}} }{\Gamma\left({m}+\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left({m}+{t}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{m}+{t}+\mathrm{1}} \\ $$$${sin}\left({asin}\left({x}\right)\right)=\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} \frac{\left({asin}\left({x}\right)\right)^{\left(\mathrm{2}{k}+\mathrm{1}\right)} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$$\int_{\mathrm{0}} ^{\pi} {sin}^{{m}} \left({x}\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left\{{sin}^{{m}} \left({x}\right)+{cos}^{{m}} \left({x}\right)\right\}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{m}} \left({x}\right){dx}=\beta\left(\frac{{m}+\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\Gamma\left(\frac{{m}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{m}}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} \Gamma\left({k}+\mathrm{1}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}{k}+\mathrm{2}\right).\Gamma\left({k}+\frac{\mathrm{3}}{\mathrm{2}}\right)}{a}^{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\Gamma\left(\mathrm{2}{k}+\mathrm{2}\right)=\underset{{j}=\mathrm{1}} {\overset{{k}} {\prod}}\left(\mathrm{2}{j}\right).\underset{{j}=\mathrm{0}} {\overset{{k}} {\prod}}\left(\mathrm{2}{j}+\mathrm{1}\right) \\ $$$$=\mathrm{2}^{{k}} \Gamma\left({k}+\mathrm{1}\right).\mathrm{2}^{{k}+\mathrm{1}} \Gamma\left(\frac{{k}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{2}^{\mathrm{2}{k}+\mathrm{1}} \Gamma\left({k}+\mathrm{1}\right)\Gamma\left(\frac{{k}+\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} \Gamma\left({k}+\mathrm{1}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}^{\mathrm{2}{k}+\mathrm{1}} \Gamma\left({k}+\mathrm{1}\right)\Gamma\left(\frac{{k}+\mathrm{3}}{\mathrm{2}}\right).\Gamma\left(\frac{{k}+\mathrm{3}}{\mathrm{2}}\right)}{a}^{\mathrm{2}{k}+\mathrm{1}} \\ $$$$=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\Gamma\left(\frac{{k}}{\mathrm{2}}+\mathrm{0}+\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{{k}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\right)}.\left(\frac{{a}}{\mathrm{2}}\underset{={H}_{\mathrm{0}} \left({a}\right)} {\right)}^{\mathrm{2}{k}+\mathrm{1}+\mathrm{0}} \\ $$$$=\sqrt{\pi}.{H}_{\mathrm{0}} \left({a}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Sarah85 last updated on 30/Aug/20
thank you very much!
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}! \\ $$
Commented by maths mind last updated on 30/Aug/20
withe pleasur
$${withe}\:{pleasur} \\ $$
Answered by maths mind last updated on 15/Sep/20
Γ(1+(x/2))=(x/2)Γ((x/2)) ⇔∫_0 ^1 Γ(1−(x/2))Γ((x/2)).(x/2)dx =(1/2)∫_0 ^1 Γ(1−(x/2))Γ((x/2))xdx Γ(1−x)Γ(x)=(π/(sin(πx))),x∈]0,1[ we find (1/2)∫_0 ^1 π(x/(2sin(((πx)/2))))dx let y=((πx)/2)⇒dx=(2/π)dy =(1/2)∫_0 ^(π/2) .π((2y)/(πsin(y)))..((2dy)/π)=(2/π)∫_0 ^(π/2) ((ydy)/(sin(y))) ∫_0 ^(π/2) ((ydy)/(sin(y))),let tg((y/2))=u⇔∫_0 ^1 2((arctan(u))/((2u)/(1+u^2 ))).(2/(1+u^2 ))du =2∫_0 ^1 ((arctan(u))/u)du=2∫_0 ^1 Σ_(k≥0) (−1)^k (u^(2k+1) /(2k+1)).(du/u) =Σ_(k≥0) ∫_0 ^1 (−1)^k (u^(2k) /(2k+1))du=Σ_(k≥0) (((−1)^k )/((2k+1)^2 ))=G ⇒∫_0 ^(π/2) (x/(sin(x)))=2∫_0 ^1 ((tan^(−1) (x))/x)dx=2G ⇒∫_0 ^1 Γ(1+(x/2))Γ(1−(x/2))dx=(2/π)∫_0 ^(π/2) (dx/(sin(x)))=(2/π).2G=(4/π)G
$$\Gamma\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right)=\frac{{x}}{\mathrm{2}}\Gamma\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \Gamma\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)\Gamma\left(\frac{{x}}{\mathrm{2}}\right).\frac{{x}}{\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \Gamma\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)\Gamma\left(\frac{{x}}{\mathrm{2}}\right){xdx} \\ $$$$\left.\Gamma\left(\mathrm{1}−{x}\right)\Gamma\left({x}\right)=\frac{\pi}{{sin}\left(\pi{x}\right)},{x}\in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$${we}\:{find} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \pi\frac{{x}}{\mathrm{2}{sin}\left(\frac{\pi{x}}{\mathrm{2}}\right)}{dx} \\ $$$${let}\:{y}=\frac{\pi{x}}{\mathrm{2}}\Rightarrow{dx}=\frac{\mathrm{2}}{\pi}{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} .\pi\frac{\mathrm{2}{y}}{\pi{sin}\left({y}\right)}..\frac{\mathrm{2}{dy}}{\pi}=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ydy}}{{sin}\left({y}\right)} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ydy}}{{sin}\left({y}\right)},{let}\:{tg}\left(\frac{{y}}{\mathrm{2}}\right)={u}\Leftrightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}\frac{{arctan}\left({u}\right)}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}.\frac{\mathrm{2}}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arctan}\left({u}\right)}{{u}}{du}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} \frac{{u}^{\mathrm{2}{k}+\mathrm{1}} }{\mathrm{2}{k}+\mathrm{1}}.\frac{{du}}{{u}} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \frac{{u}^{\mathrm{2}{k}} }{\mathrm{2}{k}+\mathrm{1}}\mathrm{du}=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }={G} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}}{{sin}\left({x}\right)}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}}{dx}=\mathrm{2}{G} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \Gamma\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right){dx}=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{{sin}\left({x}\right)}=\frac{\mathrm{2}}{\pi}.\mathrm{2}{G}=\frac{\mathrm{4}}{\pi}{G} \\ $$$$ \\ $$$$ \\ $$

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