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 193231 by C2coder last updated on 07/Jun/23

Answered by leodera last updated on 08/Jun/23

F(a) = ∫_(−∞) ^∞ ((e^(−x^2 ) sin^2 (ax^2 ))/x^2 )dx F′(a) = ∫_(−∞) ^∞ e^(−x^2 ) sin (2ax^2 )dx F′(a) = 2∫_0 ^∞ e^(−x^2 ) sin (2ax^2 )dx let u = 2ax^2 ⇒ (1/(2(√(2a))))u^(−(1/2)) du = dx F′(a) = (1/( (√(2a))))∫_(−∞) ^∞ u^(−(1/2)) e^(−(1/(2a))u) sin(u)du F′(a) = Im (1/( (√(2a))))∫_0 ^∞ u^(−(1/2)) e^(−((1/2)−i)u) du F′(a) = Im(1/( (√(2a)))){((Γ((1/2)))/(((1/2)−i)^(1/2) ))} F′(a) = Im(√(π/(2a))){(1/( (((√5)/2))^(1/2) e^(−i((tan^(−1) (2))/2)) ))} F′(a) =− (√(π/(a(√5))))sin(((tan^(−1) (2))/2)) F(a) = 2(√((aπ)/( (√5))))sin (((tan^(−1) 2)/2)) + C F(0) = 0 so C = 0 F(1) = 2(√(π/( (√5))))sin (((tan^(−1) (2))/2))

$${F}\left({a}\right)\:=\:\int_{−\infty} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } \mathrm{sin}\:^{\mathrm{2}} \left({ax}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx} \\ $$$${F}'\left({a}\right)\:=\:\int_{−\infty} ^{\infty} {e}^{−{x}^{\mathrm{2}} } \mathrm{sin}\:\left(\mathrm{2}{ax}^{\mathrm{2}} \right){dx} \\ $$$${F}'\left({a}\right)\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } \mathrm{sin}\:\left(\mathrm{2}{ax}^{\mathrm{2}} \right){dx} \\ $$$$ \\ $$$${let}\:{u}\:=\:\mathrm{2}{ax}^{\mathrm{2}} \:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}{a}}}{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} {du}\:=\:{dx} \\ $$$${F}'\left({a}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{a}}}\int_{−\infty} ^{\infty} {u}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−\frac{\mathrm{1}}{\mathrm{2}{a}}{u}} \mathrm{sin}\left({u}\right){du} \\ $$$${F}'\left({a}\right)\:=\:\:{Im}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{a}}}\int_{\mathrm{0}} ^{\infty} {u}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−\left(\frac{\mathrm{1}}{\mathrm{2}}−{i}\right){u}} {du} \\ $$$${F}'\left({a}\right)\:=\:{Im}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{a}}}\left\{\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\frac{\mathrm{1}}{\mathrm{2}}−{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\right\} \\ $$$${F}'\left({a}\right)\:=\:{Im}\sqrt{\frac{\pi}{\mathrm{2}{a}}}\left\{\frac{\mathrm{1}}{\:\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{i}\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)}{\mathrm{2}}} }\right\} \\ $$$${F}'\left({a}\right)\:=−\:\sqrt{\frac{\pi}{{a}\sqrt{\mathrm{5}}}}\mathrm{sin}\left(\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)}{\mathrm{2}}\right) \\ $$$${F}\left({a}\right)\:=\:\mathrm{2}\sqrt{\frac{{a}\pi}{\:\sqrt{\mathrm{5}}}}\mathrm{sin}\:\left(\frac{\mathrm{tan}^{−\mathrm{1}} \mathrm{2}}{\mathrm{2}}\right)\:+\:{C} \\ $$$${F}\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$${so}\:{C}\:=\:\mathrm{0} \\ $$$${F}\left(\mathrm{1}\right)\:=\:\mathrm{2}\sqrt{\frac{\pi}{\:\sqrt{\mathrm{5}}}}\mathrm{sin}\:\left(\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)}{\mathrm{2}}\right) \\ $$$$ \\ $$

Commented by C2coder last updated on 08/Jun/23

thanks alot man

$${thanks}\:{alot}\:{man} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com