Integration Questions

Question Number 145081 by qaz last updated on 02/Jul/21

$$\int_{\mathrm{0}} ^{\infty} \left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{2}\centerdot\mathrm{4}}−\frac{\mathrm{x}^{\mathrm{7}} }{\mathrm{2}\centerdot\mathrm{4}\centerdot\mathrm{6}}+...\right)\centerdot\left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{4}^{\mathrm{2}} \centerdot\mathrm{6}^{\mathrm{2}} }+...\right)\mathrm{dx}=\sqrt{\mathrm{e}} \\$$

Answered by mindispower last updated on 02/Jul/21

$$\int_{\mathrm{0}} ^{\infty} \underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}\right)!!}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{\mathrm{2}{k}} }{\left(\left(\mathrm{2}{k}\right)!!\right)^{\mathrm{2}} }{dx} \\$$$$\left(\mathrm{2}{n}\right)!!=\mathrm{2}^{{n}} .{n}! \\$$$$=\int_{\mathrm{0}} ^{\infty} \underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}^{{n}} .{n}!}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{\mathrm{2}{k}} }{\left(\mathrm{2}^{{k}} {k}!\right)^{\mathrm{2}} }{dx} \\$$$$=\int_{\mathrm{0}} ^{\infty} {xe}^{\frac{−{x}^{\mathrm{2}} }{\mathrm{2}}} .\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} }{dx} \\$$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}{n}+\mathrm{1}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx},{x}=\sqrt{\mathrm{2}{u}} \\$$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \mathrm{2}^{{n}} {u}^{{n}} {e}^{−{u}} .{du} \\$$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{2}^{{n}} \left({n}!\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} {u}^{{n}+\mathrm{1}−\mathrm{1}} {e}^{−{u}} {du} \\$$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\Gamma\left({n}+\mathrm{1}\right)}{\mathrm{2}^{{n}} \left({n}!\right)^{\mathrm{2}} }\hat {=}\underset{{n}\geqslant\mathrm{0}} {\sum}.\frac{{n}!}{\mathrm{2}^{{n}} .\left({n}!\right)^{\mathrm{2}} }=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} }{{n}!}={e}^{\frac{\mathrm{1}}{\mathrm{2}}} =\sqrt{{e}} \\$$$$\\$$$$\\$$