Question Number 53950 by maxmathsup by imad last updated on 27/Jan/19
$$\:{calculate}\:\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\Gamma\left({x}\right)\Gamma\left(\mathrm{1}−{x}\right){dx}\:\:\:{with}\:\Gamma\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} {dt}\:\:\:\:{with}\:{x}>\mathrm{0}\:. \\ $$
Commented by maxmathsup by imad last updated on 30/Jan/19
![we have for 0<x<1 Γ(x).Γ(1−x)=(π/(sin(πx)))( complments formula) ⇒ ∫_(1/3) ^(1/2) Γ(x).Γ(1−x)dx =π∫_(1/3) ^(1/2) (dx/(sin(πx))) =_(πx =t) π ∫_(π/3) ^(π/2) (dt/(πsin(t))) =∫_(π/3) ^(π/2) (dt/(sint)) =_(tan((t/2))=u) ∫_(1/( (√3))) ^1 (1/((2u)/(1+u^2 ))) ((2du)/(1+u^2 )) =∫_(1/( (√3))) ^1 (du/u) =[ln∣u∣]_(1/( (√3))) ^1 =−ln((1/( (√3)))) =ln((√3)) .](https://www.tinkutara.com/question/Q54205.png)
$${we}\:{have}\:{for}\:\:\mathrm{0}<{x}<\mathrm{1}\:\:\:\Gamma\left({x}\right).\Gamma\left(\mathrm{1}−{x}\right)=\frac{\pi}{{sin}\left(\pi{x}\right)}\left(\:\:{complments}\:{formula}\right)\:\Rightarrow \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\Gamma\left({x}\right).\Gamma\left(\mathrm{1}−{x}\right){dx}\:=\pi\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{{dx}}{{sin}\left(\pi{x}\right)}\:=_{\pi{x}\:={t}} \:\:\:\pi\:\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{\pi{sin}\left({t}\right)} \\ $$$$=\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{{sint}}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\mathrm{1}} \:\:\frac{{du}}{{u}}\:=\left[{ln}\mid{u}\mid\right]_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\mathrm{1}} =−{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$={ln}\left(\sqrt{\mathrm{3}}\right)\:. \\ $$