Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

calculate ∫_0 ^1 xe^(−x^2 ) arctan((2/x))dx

$$\mathrm{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{arctan}\left(\frac{\mathrm{2}}{\mathrm{x}}\right)\mathrm{dx} \\ $$

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

=[−(1/2)e^(−x^2 ) tan^(−1) ((2/x))+(1/2)∫_0 ^1 (−((2e^(−x^2 ) )/(x^2 +4))) ∫_0 ^1 (e^(−x^2 ) /(x^2 +4))dx =∫_0 ^1 (1/4).(Σ_(k≥0) (−(x/2))^k e^(−x^2 ) )dx =Σ_(k≥0) (((−1)^k )/2^(k+2) )∫_0 ^1 x^k e^(−x^2 ) dx ∫_0 ^1 x^s e^(−x^2 ) dx=∫_0 ^1 Σ_(k≥0) ((x^s (−x^2 )^k )/(k!)) =Σ_(k≥0) (((−1)^k )/(k!(s+2k+1)))=Σ_(k≥0) (1/(2(k+((s+1)/2)))).(((−1)^k )/(k!)) =(1/(s+1))Σ_(k≥0) (((s+1)/2)/((k+((s+1)/2)))).(((−1)^k )/(k!)) =(1/(s+1))(1+Σ_(k≥1) ((Π_(j=0) ^(k−1) (j+((s+1)/2)))/(Π_(j=0) ^(k−1) (k+((s+3)/2)))).(((−1)^k )/(k!)))=(1/(s+1)) _1 F_1 (((s+1)/2);((s+3)/2),−1) we get Σ_(k≥0) (((−1)^k )/2^(k+2) ).(1/((k+1))). _1 F_1 (((k+1)/2);((k+3)/2);−1)

$$=\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{x}^{\mathrm{2}} } \mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{x}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\frac{\mathrm{2}{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +\mathrm{4}}\right)\right. \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +\mathrm{4}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}}.\left(\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\frac{{x}}{\mathrm{2}}\right)^{{k}} {e}^{−{x}^{\mathrm{2}} } \right){dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}^{{k}+\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{s}} {e}^{−{x}^{\mathrm{2}} } {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{{s}} \left(−{x}^{\mathrm{2}} \right)^{{k}} }{{k}!} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({s}+\mathrm{2}{k}+\mathrm{1}\right)}=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{2}\left({k}+\frac{{s}+\mathrm{1}}{\mathrm{2}}\right)}.\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!} \\ $$$$=\frac{\mathrm{1}}{{s}+\mathrm{1}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\frac{{s}+\mathrm{1}}{\mathrm{2}}}{\left({k}+\frac{{s}+\mathrm{1}}{\mathrm{2}}\right)}.\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!} \\ $$$$=\frac{\mathrm{1}}{{s}+\mathrm{1}}\left(\mathrm{1}+\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\underset{{j}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left({j}+\frac{{s}+\mathrm{1}}{\mathrm{2}}\right)}{\underset{{j}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left({k}+\frac{{s}+\mathrm{3}}{\mathrm{2}}\right)}.\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!}\right)=\frac{\mathrm{1}}{{s}+\mathrm{1}}\:\:_{\mathrm{1}} {F}_{\mathrm{1}} \left(\frac{{s}+\mathrm{1}}{\mathrm{2}};\frac{{s}+\mathrm{3}}{\mathrm{2}},−\mathrm{1}\right) \\ $$$${we}\:\:{get}\:\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}^{{k}+\mathrm{2}} }.\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)}.\:_{\mathrm{1}} {F}_{\mathrm{1}} \left(\frac{{k}+\mathrm{1}}{\mathrm{2}};\frac{{k}+\mathrm{3}}{\mathrm{2}};−\mathrm{1}\right) \\ $$$$ \\ $$

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

thank you sir mind.

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

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

withe pleasur

$${withe}\:{pleasur} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com