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Question Number 76782 by mathmax by abdo last updated on 30/Dec/19

$$calculate{\int }_0^{\mathrm{\infty }}{e}^{-x}\frac{sin\left(x^2\right)}{x^2}dx$$

Answered by mind is power last updated on 04/Apr/20

$$A={\int }_0^{+\mathrm{\infty }}{e}^{-x}\frac{sin\left(x^2\right)}{x^2}dx$$$$sin\left(x^2\right)=\sum _{n⩾0}\frac{{(-1)}^n.x^{4n+2}}{(2n+1)!}$$$$A={\int }_0^{+\mathrm{\infty }}\sum _{n⩾0}\frac{{(-1)}^n{x}^{4n}{e}^{-x}dx}{(2n+1)!}$$$$A=\sum _{n⩾0}\frac{{(-1)}^n}{(2n+1)!}{\int }_0^{+\mathrm{\infty }}{x}^{4n}{e}^{-x}dx$$$$=\sum _{n⩾0}\frac{{(-1)}^n\mathrm{\Gamma }(4n+1)}{(2n+1)!}=\sum _{n⩾0}\frac{{(-1)}^n.\left(4n\right)!}{(2n+1)!}$$$$=\sum _{n⩾0}.\frac{{(-1)}^n.\underset{k=1}{\prod ^n}\left(4k\right).\underset{k=0}{\prod ^{n-1}}(4k+1).\underset{k=0}{\prod ^{n-1}}(4k+2).\underset{k=0}{\prod ^{n-1}}(4k+3)}{.\underset{k=0}{\prod ^{n-1}}(2k+1).\underset{k=0}{\prod ^{n-1}}(2k+2)}$$$$=\sum _{n⩾0}\frac{{(-1)}^n.4^{4n}.n!.\underset{k=0}{\prod ^{n-1}}(k+\frac{1}{4}).\underset{k=0}{\prod ^{n-1}}(\frac{1}{2}+k).\underset{k=0}{\prod ^{n-1}}(k+\frac{3}{4})}{2^{2n}\underset{k=0}{\prod ^{n-1}}(\frac{1}{2}+k).n!}$$$$=\sum _{n⩾0}\frac{.\underset{k=0}{\prod ^{n-1}}(1+k).\underset{k=0}{\prod ^{n-1}}(\frac{1}{4}+k).\underset{k=0}{\prod ^{n-1}}(\frac{3}{4}+k)}{\underset{k=0}{\prod ^{n-1}}(\frac{1}{2}+k).}.\frac{{(-4^3)}^n}{n!}$$$$={}_3{F}_1(1,\frac{1}{4},\frac{3}{4};\frac{1}{2};-4^3)$$$$$$$$$$

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