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Question Number 99824 by mathmax by abdo last updated on 23/Jun/20

$$\mathrm{calculate}{\int }_0^1{\mathrm{xe}}^{-{\mathrm{x}}^2}\arctan \left(\frac{2}{\mathrm{x}}\right)\mathrm{dx}$$

Answered by maths mind last updated on 23/Jun/20

$$=[-\frac{1}{2}{e}^{-x^2}{\tan }^{-1}\left(\frac{2}{x}\right)+\frac{1}{2}{\int }_0^1(-\frac{2e^{-x^2}}{x^2+4})$$$${\int }_0^1\frac{e^{-x^2}}{x^2+4}dx$$$$={\int }_0^1\frac{1}{4}.\left(\sum _{k⩾0}{(-\frac{x}{2})}^k{e}^{-x^2}\right)dx$$$$=\sum _{k⩾0}\frac{{(-1)}^k}{2^{k+2}}{\int }_0^1{x}^k{e}^{-x^2}dx$$$${\int }_0^1{x}^s{e}^{-x^2}dx={\int }_0^1\sum _{k⩾0}\frac{x^s{(-x^2)}^k}{k!}$$$$=\sum _{k⩾0}\frac{{(-1)}^k}{k!(s+2k+1)}=\sum _{k⩾0}\frac{1}{2(k+\frac{s+1}{2})}.\frac{{(-1)}^k}{k!}$$$$=\frac{1}{s+1}\sum _{k⩾0}\frac{\frac{s+1}{2}}{(k+\frac{s+1}{2})}.\frac{{(-1)}^k}{k!}$$$$=\frac{1}{s+1}(1+\sum _{k⩾1}\frac{\underset{j=0}{\prod ^{k-1}}(j+\frac{s+1}{2})}{\underset{j=0}{\prod ^{k-1}}(k+\frac{s+3}{2})}.\frac{{(-1)}^k}{k!})=\frac{1}{s+1}{}_1{F}_1(\frac{s+1}{2};\frac{s+3}{2},-1)$$$$weget\sum _{k⩾0}\frac{{(-1)}^k}{2^{k+2}}.\frac{1}{(k+1)}.{}_1{F}_1(\frac{k+1}{2};\frac{k+3}{2};-1)$$$$$$

Commented by mathmax by abdo last updated on 23/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}\mathrm{mind}.$$

Commented by maths mind last updated on 23/Jun/20

$$withepleasur$$

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