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Question Number 76782 by mathmax by abdo last updated on 30/Dec/19
calculate ∫_0 ^∞ e^(−x) ((sin(x^2 ))/x^2 )dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}} \:\:\frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx} \\ $$
Answered by mind is power last updated on 04/Apr/20
A=∫_0 ^(+∞) e^(−x) ((sin(x^2 ))/x^2 )dx sin(x^2 )=Σ_(n≥0) (((−1)^n .x^(4n+2) )/((2n+1)!)) A=∫_0 ^(+∞) Σ_(n≥0) (((−1)^n x^(4n) e^(−x) dx)/((2n+1)!)) A=Σ_(n≥0) (((−1)^n )/((2n+1)!))∫_0 ^(+∞) x^(4n) e^(−x) dx =Σ_(n≥0) (((−1)^n Γ(4n+1))/((2n+1)!))=Σ_(n≥0) (((−1)^n .(4n)!)/((2n+1)!)) =Σ_(n≥0) .(((−1)^n .Π_(k=1) ^n (4k).Π_(k=0) ^(n−1) (4k+1).Π_(k=0) ^(n−1) (4k+2).Π_(k=0) ^(n−1) (4k+3))/(.Π_(k=0) ^(n−1) (2k+1).Π_(k=0) ^(n−1) (2k+2))) =Σ_(n≥0) (((−1)^n .4^(4n) .n!.Π_(k=0) ^(n−1) (k+(1/4)).Π_(k=0) ^(n−1) ((1/2)+k).Π_(k=0) ^(n−1) (k+(3/4)))/(2^(2n) Π_(k=0) ^(n−1) ((1/2)+k).n!)) =Σ_(n≥0) ((.Π_(k=0) ^(n−1) (1+k).Π_(k=0) ^(n−1) ((1/4)+k).Π_(k=0) ^(n−1) ((3/4)+k))/(Π_(k=0) ^(n−1) ((1/2)+k).)).(((−4^3 )^n )/(n!)) = _3 F_1 (1,(1/4),(3/4);(1/2);−4^3 )
$${A}=\int_{\mathrm{0}} ^{+\infty} {e}^{−{x}} \frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx} \\ $$$${sin}\left({x}^{\mathrm{2}} \right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}+\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$${A}=\int_{\mathrm{0}} ^{+\infty} \underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{4}{n}} {e}^{−{x}} {dx}}{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$${A}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\int_{\mathrm{0}} ^{+\infty} {x}^{\mathrm{4}{n}} {e}^{−{x}} {dx} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \Gamma\left(\mathrm{4}{n}+\mathrm{1}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} .\left(\mathrm{4}{n}\right)!}{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}.\frac{\left(−\mathrm{1}\right)^{{n}} .\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{4}{k}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{4}{k}+\mathrm{1}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{4}{k}+\mathrm{2}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{4}{k}+\mathrm{3}\right)}{.\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{2}{k}+\mathrm{1}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{2}{k}+\mathrm{2}\right)} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} .\mathrm{4}^{\mathrm{4}{n}} .{n}!.\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{4}}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{k}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{3}}{\mathrm{4}}\right)}{\mathrm{2}^{\mathrm{2}{n}} \underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{k}\right).{n}!} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{.\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{1}+{k}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{4}}+{k}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{3}}{\mathrm{4}}+{k}\right)}{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{k}\right).}.\frac{\left(−\mathrm{4}^{\mathrm{3}} \right)^{{n}} }{{n}!} \\ $$$$=\:\:\:_{\mathrm{3}} {F}_{\mathrm{1}} \left(\mathrm{1},\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}};\frac{\mathrm{1}}{\mathrm{2}};−\mathrm{4}^{\mathrm{3}} \right) \\ $$$$ \\ $$$$ \\ $$

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