Question Number 127211 by mohammad17 last updated on 27/Dec/20
$$\int_{\mathrm{0}} ^{\:\pi^{\mathrm{2}} } {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{x}} {dx} \\ $$
Answered by Dwaipayan Shikari last updated on 27/Dec/20
$$\int_{\mathrm{0}} ^{\pi^{\mathrm{2}} } {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{x}} {dx}\:\:\:\:\:\:\:\:\:\:\:\:{x}={u}^{\mathrm{2}} \Rightarrow\mathrm{1}=\mathrm{2}{u}\frac{{du}}{{dx}} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\pi} {e}^{−{u}^{\mathrm{2}} } {du}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{−{u}^{\mathrm{2}} } {du}−\mathrm{2}\int_{\pi} ^{\infty} {e}^{−{u}^{\mathrm{2}} } {du} \\ $$$$=\sqrt{\pi}−\frac{\mathrm{2}{e}^{−\pi^{\mathrm{2}} } }{\mathrm{2}\pi+\frac{\mathrm{1}}{\pi+\frac{\mathrm{2}}{\mathrm{2}\pi+\frac{\mathrm{3}}{\pi+…}}}} \\ $$$${Or}\:\int_{\mathrm{0}} ^{\pi^{\mathrm{2}} } {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{x}} {dx}=\int_{\mathrm{0}} ^{\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{x}} −\int_{\pi^{\mathrm{2}} } ^{\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{x}^{\mathrm{2}} } =\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},\pi^{\mathrm{2}} \right) \\ $$$${Incomplete}\:{Gamma}\:\Gamma\left({a},{z}\right)=\int_{{z}} ^{\infty} {t}^{{a}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$
Commented by mohammad17 last updated on 27/Dec/20
$${sir}\:{is}\:{the}\:{value}\:{infinte}\:{or}\:{finite}\:? \\ $$
Commented by Dwaipayan Shikari last updated on 27/Dec/20
$${Finite},\:{It}\:{is}\:{a}\:{continued}\:{fraction} \\ $$$${Q}\mathrm{127186} \\ $$
Answered by mindispower last updated on 28/Dec/20
![Σ_(k≥0) ∫^π^2 x^(−(1/2)) .(((−x)^k )/(k!))dx =Σ_(k≥0) (((−1)^k )/(k!))∫_0 ^π^2 x^(k−(1/2)) dx =Σ_(k≥0) (((−1)^k )/(k!))[(x^(k+(1/2)) /(k+(1/2)))]_0 ^π =Σ_(k≥0) (((−1)^k π^(2k+1) )/((k+(1/2))k!)) =2π+2πΣ_(k≥1) ((Π_(j=0) ^(k−1) ((1/2)+j))/(Π_(j=0) ^(k−1) ((3/2)+j))).(((−π^2 )^k )/(k!)) =2π(1+Σ_(k≥1) ((((1/2))_k )/(((3/2))_k )).(((−π^2 )^k )/(k!))) =2π_1 F_1 ((1/2);(3/2);−π^2 )](https://www.tinkutara.com/question/Q127268.png)
$$\underset{{k}\geqslant\mathrm{0}} {\sum}\int^{\pi^{\mathrm{2}} } {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} .\frac{\left(−{x}\right)^{{k}} }{{k}!}{dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!}\int_{\mathrm{0}} ^{\pi^{\mathrm{2}} } {x}^{{k}−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!}\left[\frac{{x}^{{k}+\frac{\mathrm{1}}{\mathrm{2}}} }{{k}+\frac{\mathrm{1}}{\mathrm{2}}}\right]_{\mathrm{0}} ^{\pi} =\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} \pi^{\mathrm{2}{k}+\mathrm{1}} }{\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right){k}!} \\ $$$$=\mathrm{2}\pi+\mathrm{2}\pi\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\underset{{j}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{j}\right)}{\underset{{j}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{3}}{\mathrm{2}}+{j}\right)}.\frac{\left(−\pi^{\mathrm{2}} \right)^{{k}} }{{k}!} \\ $$$$=\mathrm{2}\pi\left(\mathrm{1}+\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{k}} }.\frac{\left(−\pi^{\mathrm{2}} \right)^{{k}} }{{k}!}\right) \\ $$$$=\mathrm{2}\pi_{\mathrm{1}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};−\pi^{\mathrm{2}} \right) \\ $$$$ \\ $$