Question-193231

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

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

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