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Question-45705

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Question Number 45705 by Sanjarbek last updated on 15/Oct/18
Commented by maxmathsup by imad last updated on 16/Oct/18
∫ sin(x^2 )dx =(((√π)((√2)+i(√2))erf{ ((√2)+i(√2))(x/2)}+(√π)((√2)−i(√2))erf{((√(2−))i(√2))(x/2)})/8) this formulae is given by integral calculator so give me time to prof this...
$$\int\:{sin}\left({x}^{\mathrm{2}} \right){dx}\:=\frac{\sqrt{\pi}\left(\sqrt{\mathrm{2}}+{i}\sqrt{\mathrm{2}}\right){erf}\left\{\:\left(\sqrt{\mathrm{2}}+{i}\sqrt{\mathrm{2}}\right)\frac{{x}}{\mathrm{2}}\right\}+\sqrt{\pi}\left(\sqrt{\mathrm{2}}−{i}\sqrt{\mathrm{2}}\right){erf}\left\{\left(\sqrt{\mathrm{2}−}{i}\sqrt{\mathrm{2}}\right)\frac{{x}}{\mathrm{2}}\right\}}{\mathrm{8}} \\ $$$${this}\:{formulae}\:{is}\:{given}\:{by}\:{integral}\:{calculator}\:{so}\:{give}\:{me}\:{time}\:{to}\:{prof}\:{this}… \\ $$
Commented by maxmathsup by imad last updated on 16/Oct/18
erf(x)=(2/( (√π))) ∫_0 ^x e^(−t^2 ) dt and this function is used in probality and statistics....
$${erf}\left({x}\right)=\frac{\mathrm{2}}{\:\sqrt{\pi}}\:\int_{\mathrm{0}} ^{{x}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:\:\:{and}\:{this}\:{function}\:{is}\:{used}\:{in}\:{probality}\:{and}\: \\ $$$${statistics}…. \\ $$
Commented by maxmathsup by imad last updated on 16/Oct/18
let f(x)=∫_0 ^x sin(t^2 )dt and g(x)=∫_0 ^x cos(t^2 )dt⇒ g(x)−if(x)=∫_0 ^x e^(−it^2 ) dt changement (√i)t=u give ∫_0 ^x e^(−it^2 ) dt = ∫_0 ^(x(√i)) e^(−u^2 ) (du/( (√i))) =(1/e^(i(π/4)) ) ∫_0 ^(x ((1+i)/( (√2)))) e^(−u^2 ) du =e^(−((iπ)/4)) (π/2) erf((x/( (√2) ))(1+i)) =(π/2)((1/( (√2))) −(i/( (√2))))erf(((x(1+i))/( (√2)))) =(π/(2(√2)))(1−i) erf(((x(1+i))/( (√2)))) ....be continued....
$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {sin}\left({t}^{\mathrm{2}} \right){dt}\:{and}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {cos}\left({t}^{\mathrm{2}} \right){dt}\Rightarrow \\ $$$${g}\left({x}\right)−{if}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\:{e}^{−{it}^{\mathrm{2}} } {dt}\:\:\:{changement}\:\sqrt{{i}}{t}={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{{x}} \:{e}^{−{it}^{\mathrm{2}} } {dt}\:=\:\int_{\mathrm{0}} ^{{x}\sqrt{{i}}} \:\:\:{e}^{−{u}^{\mathrm{2}} } \:\frac{{du}}{\:\sqrt{{i}}}\:=\frac{\mathrm{1}}{{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:\int_{\mathrm{0}} ^{{x}\:\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}} \:{e}^{−{u}^{\mathrm{2}} } {du} \\ $$$$={e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\:\frac{\pi}{\mathrm{2}}\:{erf}\left(\frac{{x}}{\:\sqrt{\mathrm{2}}\:}\left(\mathrm{1}+{i}\right)\right)\:=\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right){erf}\left(\frac{{x}\left(\mathrm{1}+{i}\right)}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\mathrm{1}−{i}\right)\:{erf}\left(\frac{{x}\left(\mathrm{1}+{i}\right)}{\:\sqrt{\mathrm{2}}}\right)\:\:\:….{be}\:{continued}…. \\ $$

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