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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 bymohammad17 last updated on 27/Dec/20

$${sir}\:{is}\:{the}\:{value}\:{infinte}\:{or}\:{finite}\:? \\$$

Commented byDwaipayan 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

$$\underset{{k}\geqslant\mathrm{0}} {\sum}\int^{\pi^{\mathrm{2}} } {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} .\frac{\left(−{x}\right)^{{k}} }{{k}!}{dx} \\$$ $$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!}\int_{\mathrm{0}} ^{\pi^{\mathrm{2}} } {x}^{{k}−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\$$ $$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!}\left[\frac{{x}^{{k}+\frac{\mathrm{1}}{\mathrm{2}}} }{{k}+\frac{\mathrm{1}}{\mathrm{2}}}\right]_{\mathrm{0}} ^{\pi} =\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} \pi^{\mathrm{2}{k}+\mathrm{1}} }{\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right){k}!} \\$$ $$=\mathrm{2}\pi+\mathrm{2}\pi\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\underset{{j}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{j}\right)}{\underset{{j}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{3}}{\mathrm{2}}+{j}\right)}.\frac{\left(−\pi^{\mathrm{2}} \right)^{{k}} }{{k}!} \\$$ $$=\mathrm{2}\pi\left(\mathrm{1}+\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{k}} }.\frac{\left(−\pi^{\mathrm{2}} \right)^{{k}} }{{k}!}\right) \\$$ $$=\mathrm{2}\pi_{\mathrm{1}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};−\pi^{\mathrm{2}} \right) \\$$ $$\\$$